Chapter 15 Introduction to Graphs (Additional Questions)

In a standard two-dimensional coordinate system (Cartesian plane), the horizontal axis is called the X-axis, and the vertical axis is called the Y-axis.

Therefore, the correct option is (C) X-axis.

In a two-dimensional coordinate system, the point where the horizontal axis (X-axis) and the vertical axis (Y-axis) meet is called the origin.

The coordinates of the origin are always $(0, 0)$. On the X-axis, the y-coordinate is always $0$, and on the Y-axis, the x-coordinate is always $0$. The only point common to both axes is where both coordinates are $0$.

Therefore, the coordinates of the point where the x-axis and y-axis intersect are $(0, 0)$.

The correct option is (B) $(0, 0)$.

In a two-dimensional coordinate system, the plane is divided into four quadrants by the X-axis and the Y-axis. The quadrants are numbered I, II, III, and IV in a counter-clockwise direction starting from the top-right.

The sign conventions for the coordinates $(x, y)$ in each quadrant are:

Quadrant I: $(+, +)$ (x is positive, y is positive)

Quadrant II: $(-, +)$ (x is negative, y is positive)

Quadrant III: $(-, -)$ (x is negative, y is negative)

Quadrant IV: $(+, -)$ (x is positive, y is negative)

The given point is $(-5, 2)$. Here, the x-coordinate is $-5$ (negative) and the y-coordinate is $2$ (positive).

Since the x-coordinate is negative and the y-coordinate is positive, the point $(-5, 2)$ lies in Quadrant II.

Therefore, the correct option is (B) Quadrant II.

In a coordinate system, the y-axis is the vertical line. All points on the y-axis have their x-coordinate equal to $0$.

For any point $(x, y)$, if the point lies on the y-axis, its horizontal distance from the origin is zero, which means the x-coordinate must be $0$.

Examples of points on the y-axis are $(0, 1)$, $(0, -5)$, $(0, \pi)$, etc. In all these points, the x-coordinate is $0$.

Therefore, if a point lies on the y-axis, its x-coordinate is always Zero.

The correct option is (C) Zero.

In a two-dimensional coordinate system, the plane is divided into four quadrants based on the signs of the x and y coordinates:

Quadrant I: The region where $x > 0$ (positive) and $y > 0$ (positive). The signs are $(+, +)$.

Quadrant II: The region where $x < 0$ (negative) and $y > 0$ (positive). The signs are $(-, +)$.

Quadrant III: The region where $x < 0$ (negative) and $y < 0$ (negative). The signs are $(-, -)$.

Quadrant IV: The region where $x > 0$ (positive) and $y < 0$ (negative). The signs are $(+, -)$.

The question asks for the signs in Quadrant IV. As explained above, in Quadrant IV, the x-coordinate is positive, and the y-coordinate is negative.

Therefore, the signs of the x and y coordinates in Quadrant IV are $(+, -)$.

The correct option is (D) $(+, -)$.

In a coordinate system, when plotting a point $(x, y)$ starting from the origin $(0, 0)$:

The first coordinate, $x$, indicates the horizontal movement. A positive $x$ means moving to the right, and a negative $x$ means moving to the left.

The second coordinate, $y$, indicates the vertical movement. A positive $y$ means moving up, and a negative $y$ means moving down.

The given point is $(3, -4)$.

The x-coordinate is $3$, which is positive. This means we move $3$ units to the right from the origin.

The y-coordinate is $-4$, which is negative. From the position after moving right, this means we move $4$ units down.

The sentence provided says: "When plotting the point $(3, -4)$, you move 3 units to the right from the origin and then 4 units _________."

Based on the y-coordinate $-4$, the movement is 4 units down.

Therefore, the missing word is Down.

The correct option is (B) Down.

A point $(x, y)$ lies on the x-axis if and only if its y-coordinate is $0$. The x-coordinate can be any real number.

Let's examine each given point:

(A) $(5, 0)$: The y-coordinate is $0$. So, this point lies on the x-axis.

(B) $(0, 3)$: The y-coordinate is $3$ (not $0$). This point lies on the y-axis.

(C) $(-2, 0)$: The y-coordinate is $0$. So, this point lies on the x-axis.

(D) $(0, -1)$: The y-coordinate is $-1$ (not $0$). This point lies on the y-axis.

(E) $(4, 4)$: The y-coordinate is $4$ (not $0$). This point does not lie on either axis (it's in Quadrant I).

The points that have a y-coordinate of $0$ are $(5, 0)$ and $(-2, 0)$.

Therefore, the points that lie on the x-axis are (A) $(5, 0)$ and (C) $(-2, 0)$.

The given equation is $y = x$. This equation represents a linear relationship where the y-coordinate is always equal to the x-coordinate for any point on the graph.

Let's consider the points on the axes:

A point on the x-axis has coordinates of the form $(x, 0)$. For such a point to lie on the line $y=x$, we must have $0 = x$. This means the only point on the x-axis that lies on the line $y=x$ is $(0, 0)$.

A point on the y-axis has coordinates of the form $(0, y)$. For such a point to lie on the line $y=x$, we must have $y = 0$. This means the only point on the y-axis that lies on the line $y=x$ is $(0, 0)$.

The point $(0, 0)$ is the origin, which is the intersection point of the x-axis and the y-axis.

Since the point $(0, 0)$ satisfies the equation $y=x$ (as $0=0$), the graph of $y=x$ passes through the origin.

The line $y=x$ does not pass through the x-axis *only* (as it also passes through the y-axis at the same point) or the y-axis *only*. It specifically passes through the origin.

Therefore, the graph of the equation $y = x$ is a straight line that passes through the Origin.

The correct option is (C) Origin.

Assertion (A): The point is $(3, -5)$. The x-coordinate is $3$, which is positive, and the y-coordinate is $-5$, which is negative.

In a coordinate system, Quadrant I has $(+,+)$, Quadrant II has $(-,+)$, Quadrant III has $(-,-)$, and Quadrant IV has $(+,-)$ signs for the coordinates $(x, y)$.

Since the x-coordinate is positive and the y-coordinate is negative, the point $(3, -5)$ lies in Quadrant IV. Therefore, Assertion (A) is true.

Reason (R): The statement is that in Quadrant IV, the x-coordinate is positive and the y-coordinate is negative. This is the correct definition of the signs of coordinates for any point located in Quadrant IV.

Therefore, Reason (R) is true.

Now, let's check if Reason (R) is the correct explanation for Assertion (A). Assertion (A) says the point $(3, -5)$ is in Quadrant IV. The reason given (R) explains the condition for a point to be in Quadrant IV based on the signs of its coordinates. The point $(3, -5)$ fits this condition ($x=3 > 0$ and $y=-5 < 0$). Therefore, Reason (R) correctly explains why the point $(3, -5)$ is in Quadrant IV.

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

The correct option is (A) Both A and R are true, and R is the correct explanation of A.

We are given data points relating Time (on the x-axis) and Distance (on the y-axis). The points are $(1, 40)$, $(2, 80)$, $(3, 120)$, and $(4, 160)$.

Let $x$ represent Time and $y$ represent Distance. The relationship between distance, speed, and time is $\text{Distance} = \text{Speed} \times \text{Time}$. If the speed is constant, say $v$, then the equation is $y = vx$. This is a linear equation.

Let's calculate the speed from the given data points:

For $(1, 40)$: Speed = $\frac{40 \text{ km}}{1 \text{ hour}} = 40 \text{ km/h}$

For $(2, 80)$: Speed = $\frac{80 \text{ km}}{2 \text{ hours}} = 40 \text{ km/h}$

For $(3, 120)$: Speed = $\frac{120 \text{ km}}{3 \text{ hours}} = 40 \text{ km/h}$

For $(4, 160)$: Speed = $\frac{160 \text{ km}}{4 \text{ hours}} = 40 \text{ km/h}$

The speed is constant at $40 \text{ km/h}$.

The relationship is therefore $y = 40x$. This is a linear equation of the form $y = mx + c$, where the slope $m = 40$ and the y-intercept $c = 0$.

A linear equation of the form $y = mx$ represents a straight line that passes through the origin $(0, 0)$. When time $x=0$, the distance covered $y = 40 \times 0 = 0$, which corresponds to the van starting from the warehouse (origin).

Assuming constant speed implies a continuous linear relationship between time and distance starting from the origin. Therefore, the graph is expected to be a straight line passing through the origin.

The correct option is (B) A straight line passing through the origin.

From the case study in Question 10, we determined that the relationship between Distance ($y$) and Time ($x$) is linear, with a constant speed of $40 \text{ km/h}$. The equation representing this relationship is $y = 40x$.

We are asked to find the estimated distance travelled after $2.5$ hours. This means we need to find the value of $y$ when $x = 2.5$.

Substitute $x = 2.5$ into the equation $y = 40x$:

$y = 40 \times 2.5$

$y = 40 \times \frac{5}{2}$

$y = \cancel{40}^{20} \times \frac{5}{\cancel{2}_{1}}$

$y = 20 \times 5$

$y = 100$

So, the estimated distance travelled after $2.5$ hours is $100 \text{ km}$.

Therefore, the correct option is (A) $100 \text{ km}$.

From the Case Study in Question 10, the data shows the distance travelled by the van at different times.

Speed is defined as the distance travelled per unit of time.

$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$

Using the data from the table:

At Time = 1 hour, Distance = 40 km. Speed = $\frac{40 \text{ km}}{1 \text{ hour}} = 40 \text{ km/hr}$.

At Time = 2 hours, Distance = 80 km. Speed = $\frac{80 \text{ km}}{2 \text{ hours}} = 40 \text{ km/hr}$.

At Time = 3 hours, Distance = 120 km. Speed = $\frac{120 \text{ km}}{3 \text{ hours}} = 40 \text{ km/hr}$.

At Time = 4 hours, Distance = 160 km. Speed = $\frac{160 \text{ km}}{4 \text{ hours}} = 40 \text{ km/hr}$.

The speed of the delivery van is constant at $40 \text{ km/hr}$.

Therefore, the correct option is (A) $40 \text{ km/hr}$.

A straight line graph represents a linear relationship between two quantities, say $x$ and $y$. The general equation of a straight line is $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

If the straight line passes through the origin, it means that the point $(0, 0)$ lies on the line. Substituting these coordinates into the equation $y = mx + c$:

$0 = m(0) + c$

$0 = 0 + c$

$c = 0$

So, a straight line passing through the origin has the equation $y = mx$, where $m$ is the slope (a constant value, provided the line is not vertical). We can write this as $y = kx$, where $k$ is the constant slope.

The definition of direct proportion states that two quantities are in direct proportion if their ratio is constant. This can be written as $\frac{y}{x} = k$ (for $x \neq 0$) or $y = kx$, where $k$ is the constant of proportionality.

Comparing the equation of a straight line passing through the origin ($y = mx$) with the definition of direct proportion ($y = kx$), we see that they are the same form. The constant slope $m$ is the constant of proportionality $k$.

Therefore, if the graph of a relationship between two quantities is a straight line passing through the origin, the quantities are in direct proportion.

While the relationship is also a linear relationship (as it's a straight line), the specific characteristic of passing through the origin makes it a direct proportion, which is a more precise description in this context.

Inverse proportion is represented by a graph of the form $y = k/x$, which is a curve (a hyperbola), not a straight line.

The correct option is (B) Direct proportion.

We need to determine the quadrant for each given point based on the signs of its x and y coordinates.

Recall the sign conventions for the quadrants:

Quadrant I: $(+, +)$

Quadrant II: $(-, +)$

Quadrant III: $(-, -)$

Quadrant IV: $(+, -)$

Let's analyze each point:

(i) $(4, 6)$: The x-coordinate is $4$ (positive) and the y-coordinate is $6$ (positive). Signs are $(+, +)$. This point lies in Quadrant I. So, (i) matches with (a).

(ii) $(-2, 5)$: The x-coordinate is $-2$ (negative) and the y-coordinate is $5$ (positive). Signs are $(-, +)$. This point lies in Quadrant II. So, (ii) matches with (b).

(iii) $(-1, -3)$: The x-coordinate is $-1$ (negative) and the y-coordinate is $-3$ (negative). Signs are $(-, -)$. This point lies in Quadrant III. So, (iii) matches with (c).

(iv) $(7, -2)$: The x-coordinate is $7$ (positive) and the y-coordinate is $-2$ (negative). Signs are $(+, -)$. This point lies in Quadrant IV. So, (iv) matches with (d).

The correct matching is: (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d).

Comparing this with the given options, we find that option (A) matches our result.

The correct option is (A) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d).

The given equation is $y = 3x$. This is a linear equation of the form $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

Comparing $y = 3x$ with $y = mx + c$, we have $m = 3$ and $c = 0$.

Let's evaluate each option:

(A) A straight line passing through $(1, 3)$.

To check if the point $(1, 3)$ lies on the line $y = 3x$, substitute $x=1$ and $y=3$ into the equation:

$3 = 3(1)$

$3 = 3$

The equation holds true, so the line $y = 3x$ passes through the point $(1, 3)$. Option (A) is correct.

(B) A straight line passing through the origin.

The origin has coordinates $(0, 0)$. To check if the line $y = 3x$ passes through the origin, substitute $x=0$ and $y=0$ into the equation:

$0 = 3(0)$

$0 = 0$

The equation holds true, so the line $y = 3x$ passes through the origin. This is also evident from the equation being in the form $y=mx$ with $c=0$. Option (B) is correct.

(C) A straight line with a slope of 3.

In the equation $y = 3x$, the coefficient of $x$ is the slope, which is $3$. So, the slope of the line $y = 3x$ is 3. Option (C) is correct.

Since options (A), (B), and (C) are all correct descriptions of the graph of $y = 3x$, the correct answer is that all of the above options could represent the relationship $y = 3x$.

Therefore, the correct option is (D) All of the above.

We are asked to find the distance between the point $(5, 0)$ and the y-axis.

In a two-dimensional coordinate system, the y-axis is the vertical line where the x-coordinate of any point is always $0$.

The distance of a point $(x, y)$ from the y-axis is the absolute value of its x-coordinate, i.e., $|x|$. This is because the distance is measured horizontally along a line perpendicular to the y-axis, and the length of this segment is equal to the absolute value of the x-coordinate.

For the given point $(5, 0)$, the x-coordinate is $5$ and the y-coordinate is $0$.

Using the rule, the distance from the y-axis is the absolute value of the x-coordinate.

Distance $= |x| = |5| = 5$ units.

The point $(5, 0)$ lies on the x-axis, 5 units to the right of the origin. The y-axis is the vertical line passing through the origin ($x=0$). The shortest distance from a point $(x,y)$ to the vertical line $x=0$ is the horizontal distance, which is $|x|$.

Therefore, the distance between the point $(5, 0)$ and the y-axis is $5$ units.

The correct option is (B) 5 units.

In a coordinate system, a point $(x, y)$ lies on the x-axis if its y-coordinate is $0$.

A point $(x, y)$ lies on the y-axis if its x-coordinate is $0$.

The given point is $(-6, 0)$.

The x-coordinate is $-6$.

The y-coordinate is $0$.

Since the y-coordinate is $0$, the point $(-6, 0)$ lies on the x-axis.

Since the x-coordinate is $-6$ (which is not $0$), the point does not lie on the y-axis (unless it is the origin, where both coordinates are $0$).

Therefore, the point $(-6, 0)$ lies only on the x-axis.

The correct option is (A) X-axis.

In a coordinate system, the x-axis represents the values of the independent variable, and the y-axis represents the values of the dependent variable.

In this question, we are plotting temperature against time, with time on the x-axis and temperature on the y-axis. This means that time is the independent variable and temperature is the dependent variable (temperature changes over time).

Therefore, the x-axis will represent the different values of time at which the temperature is recorded or observed.

The x-axis usually represents the different times of the day in this scenario.

The correct option is (B) The different times of the day.

Coordinate geometry involves the study of geometry using a coordinate system, while graphs are visual representations of relationships between quantities plotted on a coordinate system.

Let's consider each option:

(A) Plotting a person's height over several years: This involves two quantities (height and time). We can represent this data as ordered pairs $(year, height)$ and plot these points on a graph with year on the x-axis and height on the y-axis. This is a typical use of coordinate geometry and graphs to visualize trends over time.

(B) Showing the number of goals scored by a football team in different matches: This involves matching the number of goals to specific matches. This data can be plotted using bar graphs or line graphs on a coordinate system, with matches (or match number) on the x-axis and the number of goals on the y-axis. This is a common application of graphing data.

(C) Representing the growth of bacteria over time: This involves plotting the number of bacteria (or population size) against time. The data can be represented as ordered pairs $(time, population)$ and plotted on a graph, often showing an exponential growth curve. This is a standard application of graphing relationships between variables.

(D) Calculating the area of an irregular field by dividing it into triangles: This is a method for finding the area of a polygon or an irregular shape by decomposing it into simpler shapes (triangles) whose areas can be calculated using standard geometric formulas (e.g., $\frac{1}{2} \times \text{base} \times \text{height}$). While coordinate geometry *can* be used to find the lengths of sides or heights (and thus area) if the vertices' coordinates are known (e.g., using the distance formula or the shoelace formula), the method described itself, "dividing it into triangles," is primarily a technique from basic geometry (triangulation) for area calculation, rather than a typical application that directly involves plotting the entire scenario on a graph to analyze a relationship or trend between two variables in the same way as options (A), (B), and (C).

Options (A), (B), and (C) involve using graphs to visualize relationships or data points between two variables. Option (D) describes a geometric method for area calculation, which doesn't inherently require plotting a graph of the field's area against another variable, although coordinate geometry tools can assist in the calculation within that method.

Therefore, calculating the area of an irregular field by dividing it into triangles is the least typical application of coordinate geometry and graphs among the given options in the context of using graphs to represent and analyze relationships between quantities.

The correct option is (D) Calculating the area of an irregular field by dividing it into triangles.

Given:

The cost of pens is directly proportional to the number of pens bought.

Data points (Number of Pens, Cost): $(2, 20)$, $(4, 40)$, $(6, 60)$, $(8, 80)$.

Solution:

Let $x$ represent the Number of Pens and $y$ represent the Cost in $\textsf{₹}$.

Since the cost is directly proportional to the number of pens, the relationship between $y$ and $x$ can be expressed as:

$y \propto x$

This can be written as an equation:

$y = kx$

where $k$ is the constant of proportionality. In this context, $k$ represents the cost per pen (i.e., the cost of one pen).

We can use any of the given data points $(x, y)$ from the table to find the value of $k$.

Using the first data point $(2, 20)$:

Substitute $x = 2$ and $y = 20$ into the equation $y = kx$:

$20 = k \times 2$

To find $k$, divide both sides by 2:

$k = \frac{20}{2}$

$k = 10$

Let's verify this with another data point, say $(4, 40)$:

$40 = k \times 4$

$k = \frac{40}{4}$

$k = 10$

The value of $k$ is consistent across the data points.

The equation relating cost and the number of pens is $y = 10x$.

The constant of proportionality, $k=10$, represents the cost of one pen.

Alternatively, we can find the cost of one pen by finding the unit cost from any row in the table:

Cost per pen $= \frac{\text{Total Cost}}{\text{Number of Pens}}$

Using the first row: Cost per pen $= \frac{20}{2} = 10 \textsf{₹}$.

Using the second row: Cost per pen $= \frac{40}{4} = 10 \textsf{₹}$.

Using the third row: Cost per pen $= \frac{60}{6} = 10 \textsf{₹}$.

Using the fourth row: Cost per pen $= \frac{80}{8} = 10 \textsf{₹}$.

The cost of one pen is $\textsf{₹}\ 10$.

The correct option is (A) $\textsf{₹}\ 10$.

From the Case Study in Question 20, we established that the cost ($y$) is directly proportional to the number of pens ($x$). The relationship was found to be $y = 10x$, where $10$ is the cost per pen.

The graph of a relationship of the form $y = kx$ (where $k$ is a constant) is always a straight line that passes through the origin $(0, 0)$. This is because when $x=0$, $y = k \times 0 = 0$, so the point $(0, 0)$ is on the graph.

Let's check the given options:

(A) $(2, 20)$: From the table, when $x=2$, $y=20$. Substituting into the equation $y=10x$: $20 = 10 \times 2$, which is true. So, the line passes through $(2, 20)$.

(B) $(8, 80)$: From the table, when $x=8$, $y=80$. Substituting into the equation $y=10x$: $80 = 10 \times 8$, which is true. So, the line passes through $(8, 80)$.

(C) $(0, 0)$: This is the origin. As explained above, for a direct proportion $y=kx$, the line always passes through the origin. Substituting $x=0$, $y=0$ into the equation $y=10x$: $0 = 10 \times 0$, which is true. So, the line passes through $(0, 0)$.

Since the straight line representing the relationship $y = 10x$ passes through the points $(2, 20)$, $(8, 80)$, and $(0, 0)$, all the points listed in options (A), (B), and (C) lie on the graph.

Therefore, the correct option is (D) All of the above.

Given:

From the Case Study in Question 20, the relationship between the number of pens ($x$) and the cost in $\textsf{₹}$ ($y$) is a direct proportion, represented by the equation $y = 10x$, where the cost of one pen is $\textsf{₹}\ 10$.

We are given a total cost of $\textsf{₹}\ 50$, so $y = 50$.

To Find:

The number of pens ($x$) that can be bought for $\textsf{₹}\ 50$.

Solution:

We use the equation $y = 10x$ that relates the cost ($y$) and the number of pens ($x$).

Substitute the given cost, $y = 50$, into the equation:

$50 = 10x$

To find $x$, we need to isolate $x$ by dividing both sides of the equation by $10$:

$\frac{50}{10} = \frac{10x}{10}$

$x = \frac{50}{10}$

$x = 5$

Thus, $5$ pens can be bought for $\textsf{₹}\ 50$.

Alternatively, since the cost of one pen is $\textsf{₹}\ 10$, the number of pens that can be bought for $\textsf{₹}\ 50$ is the total cost divided by the cost per pen:

Number of pens $= \frac{\text{Total Cost}}{\text{Cost per pen}}$

Number of pens $= \frac{\textsf{₹}\ 50}{\textsf{₹}\ 10/\text{pen}}$

Number of pens $= 5$ pens.

The correct option is (B) 5 pens.

In a two-dimensional coordinate system, a point is represented by its coordinates $(x, y)$.

The x-coordinate ($x$) represents the horizontal distance (or displacement) of the point from the vertical axis (the y-axis). This distance is measured parallel to the x-axis.

The y-coordinate ($y$) represents the vertical distance (or displacement) of the point from the horizontal axis (the x-axis). This distance is measured parallel to the y-axis.

The sentence states that the point $(x, y)$ is located $x$ units from the y-axis. This is consistent with the definition of the x-coordinate.

The sentence then states that the point is located $y$ units from the _________. Based on the definition of the y-coordinate, $y$ represents the distance (vertical displacement) from the x-axis.

Therefore, the blank should be filled with "X-axis".

The correct option is (C) X-axis.

We are asked to find the distance between the point $(0, 5)$ and the x-axis.

In a two-dimensional coordinate system, the x-axis is the horizontal line where the y-coordinate of any point is always $0$.

The distance of a point $(x, y)$ from the x-axis is the absolute value of its y-coordinate, i.e., $|y|$. This is because the distance is measured vertically along a line perpendicular to the x-axis, and the length of this segment is equal to the absolute value of the y-coordinate.

For the given point $(0, 5)$, the x-coordinate is $0$ and the y-coordinate is $5$.

Using the rule, the distance from the x-axis is the absolute value of the y-coordinate.

Distance $= |y| = |5| = 5$ units.

The point $(0, 5)$ lies on the y-axis, 5 units above the origin. The x-axis is the horizontal line passing through the origin ($y=0$). The shortest distance from a point $(x,y)$ to the horizontal line $y=0$ is the vertical distance, which is $|y|$.

Therefore, the distance between the point $(0, 5)$ and the x-axis is $5$ units.

The correct option is (B) 5 units.

Given:

The equation of the line is $y = x + 2$.

Solution:

A point $(x, y)$ lies on the line $y = x + 2$ if the coordinates of the point satisfy the equation when substituted into it. We will check each option:

(A) Point $(1, 3)$:

Substitute $x = 1$ and $y = 3$ into the equation $y = x + 2$:

$3 = 1 + 2$

$3 = 3$

The equation is satisfied. So, the point $(1, 3)$ lies on the line.

(B) Point $(0, 2)$:

Substitute $x = 0$ and $y = 2$ into the equation $y = x + 2$:

$2 = 0 + 2$

$2 = 2$

The equation is satisfied. So, the point $(0, 2)$ lies on the line.

(C) Point $(2, 4)$:

Substitute $x = 2$ and $y = 4$ into the equation $y = x + 2$:

$4 = 2 + 2$

$4 = 4$

The equation is satisfied. So, the point $(2, 4)$ lies on the line.

(D) Point $(-1, 1)$:

Substitute $x = -1$ and $y = 1$ into the equation $y = x + 2$:

$1 = -1 + 2$

$1 = 1$

The equation is satisfied. So, the point $(-1, 1)$ lies on the line.

(E) Point $(2, 0)$:

Substitute $x = 2$ and $y = 0$ into the equation $y = x + 2$:

$0 = 2 + 2$

$0 = 4$

The equation is NOT satisfied. So, the point $(2, 0)$ does not lie on the line.

The points that lie on the line $y = x + 2$ are $(1, 3)$, $(0, 2)$, $(2, 4)$, and $(-1, 1)$.

Therefore, the correct options are (A) $(1, 3)$, (B) $(0, 2)$, (C) $(2, 4)$, and (D) $(-1, 1)$.

Assertion (A): The point is $(-3, -3)$. The x-coordinate is $-3$, which is negative, and the y-coordinate is $-3$, which is negative.

The sign convention for Quadrant III is that both the x and y coordinates are negative.

Since the point $(-3, -3)$ has both coordinates negative, it lies in Quadrant III. Therefore, Assertion (A) is true.

Reason (R): The statement is that in Quadrant III, both the x-coordinate and the y-coordinate are negative. This is the correct and defining characteristic of Quadrant III in the Cartesian coordinate system.

Therefore, Reason (R) is true.

Now, we check if Reason (R) is the correct explanation for Assertion (A). Assertion (A) claims that the point $(-3, -3)$ is in Quadrant III. Reason (R) provides the rule that points with negative x and negative y coordinates are in Quadrant III. The point $(-3, -3)$ satisfies this rule, thus Reason (R) explains why Assertion (A) is true.

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

The correct option is (A) Both A and R are true, and R is the correct explanation of A.

Given:

Data table showing Principal (x) and Simple Interest (y) for 1 year at $10\%$ annual interest.

Data points: $(1000, 100)$, $(2000, 200)$, $(3000, 300)$, $(4000, 400)$.

To Find:

The type of relationship between Principal and Simple Interest.

Solution:

Let $P$ be the Principal (on the x-axis) and $SI$ be the Simple Interest (on the y-axis).

The formula for simple interest is:

$SI = \frac{P \times R \times T}{100}$

Given the rate $R = 10\%$ per annum and time $T = 1$ year.

Substituting these values into the formula:

$SI = \frac{P \times 10 \times 1}{100}$

$SI = \frac{10P}{100}$

$SI = 0.1P$

Let $x = P$ (Principal) and $y = SI$ (Simple Interest).

The relationship becomes:

$y = 0.1x$

This equation is of the form $y = kx$, where $k = 0.1$ is a constant.

A relationship of the form $y = kx$, where $k$ is a non-zero constant, indicates that $y$ is directly proportional to $x$. This means that as $x$ increases, $y$ increases at a constant rate, and the ratio $\frac{y}{x}$ is always equal to $k$.

Let's check the ratio $\frac{y}{x}$ using the given data points:

For $(1000, 100)$: $\frac{100}{1000} = 0.1$

For $(2000, 200)$: $\frac{200}{2000} = 0.1$

For $(3000, 300)$: $\frac{300}{3000} = 0.1$

For $(4000, 400)$: $\frac{400}{4000} = 0.1$

The ratio is constant ($0.1$), confirming that the relationship is a direct proportion.

Graphically, a direct proportion $y=kx$ is represented by a straight line that passes through the origin $(0, 0)$. When the Principal is $\textsf{₹}\ 0$, the Simple Interest is $\textsf{₹}\ 0$, which corresponds to the point $(0, 0)$.

Therefore, the type of relationship that exists between Principal and Simple Interest (at a fixed rate and time) is a direct proportion.

The correct option is (B) Direct proportion.

Given:

From the Case Study in Question 27, the relationship between Simple Interest ($y$) and Principal ($x$) for 1 year at $10\%$ annual interest is given by the equation $y = 0.1x$.

We are given a Principal amount of $\textsf{₹}\ 2500$, so $x = 2500$.

To Find:

The estimated simple interest ($y$) on a principal of $\textsf{₹}\ 2500$.

Solution:

We use the equation $y = 0.1x$ that relates the Simple Interest ($y$) and the Principal ($x$).

Substitute the given Principal amount, $x = 2500$, into the equation:

$y = 0.1 \times 2500$

$y = \frac{1}{10} \times 2500$

$y = \frac{2500}{10}$

$y = 250$

Thus, the estimated simple interest on a principal of $\textsf{₹}\ 2500$ is $\textsf{₹}\ 250$.

The correct option is (B) $\textsf{₹}\ 250$.

Given:

From the Case Study in Question 27, the relationship between Simple Interest ($y$) and Principal ($x$) for 1 year at $10\%$ annual interest is given by the equation $y = 0.1x$.

We are given a Simple Interest amount of $\textsf{₹}\ 350$, so $y = 350$.

To Find:

The principal amount ($x$) that would earn a simple interest of $\textsf{₹}\ 350$.

Solution:

We use the equation $y = 0.1x$ that relates the Simple Interest ($y$) and the Principal ($x$).

Substitute the given Simple Interest amount, $y = 350$, into the equation:

$350 = 0.1x$

To find $x$, we need to isolate $x$. We can rewrite $0.1$ as $\frac{1}{10}$:

$350 = \frac{1}{10}x$

Multiply both sides of the equation by $10$ to solve for $x$:

$350 \times 10 = \frac{1}{10}x \times 10$

$3500 = x$

So, $x = 3500$.

Thus, a principal amount of $\textsf{₹}\ 3500$ would earn a simple interest of $\textsf{₹}\ 350$ under the given conditions.

The correct option is (B) $\textsf{₹}\ 3500$.

Given:

The equation of the graph is $y = 5$.

Solution:

The equation $y = 5$ means that for any value of $x$, the corresponding value of $y$ is always equal to $5$. This describes a set of points where the y-coordinate is fixed at $5$, while the x-coordinate can be any real number.

Let's consider a few points that satisfy this equation:

If $x=0$, $y=5$, so the point is $(0, 5)$.

If $x=1$, $y=5$, so the point is $(1, 5)$.

If $x=-3$, $y=5$, so the point is $(-3, 5)$.

When these points are plotted on a coordinate plane, they all lie on a straight line where the vertical distance from the x-axis is always $5$. This forms a horizontal line.

A horizontal line is always parallel to the horizontal axis, which is the X-axis.

The equation of the X-axis itself is $y=0$. A line with the equation $y=c$ (where $c$ is a constant) is a horizontal line parallel to the x-axis, located $|c|$ units away from it.

To check if it passes through the origin $(0, 0)$, we substitute $x=0$ and $y=0$ into the equation $y=5$. We get $0 = 5$, which is false. Therefore, the line $y=5$ does not pass through the origin.

Thus, the graph of $y = 5$ is a straight line parallel to the X-axis.

The correct option is (B) X-axis.

Given:

The process of plotting a point $(x, y)$ on a graph paper.

Solution:

To plot a point $(x, y)$ on a graph paper (Cartesian plane), the standard steps are as follows:

Step 1: Draw two perpendicular lines, one horizontal and one vertical. The horizontal line is the X-axis, and the vertical line is the Y-axis. Their intersection is the origin $(0, 0)$. Mark scales on both axes.

Step 2: Locate the position corresponding to the x-coordinate ($x$) on the X-axis and the position corresponding to the y-coordinate ($y$) on the Y-axis.

Step 3: From the position $x$ on the X-axis, draw a line perpendicular to the X-axis (i.e., a vertical line). From the position $y$ on the Y-axis, draw a line perpendicular to the Y-axis (i.e., a horizontal line). The point where these two perpendicular lines intersect is the location of the point $(x, y)$. This intersection point is the plotted point.

Let's examine the given options based on these steps:

(A) Draw the X and Y axes: This is the necessary first step to set up the coordinate system. This is a typical step.

(B) Locate $x$ on the X-axis and $y$ on the Y-axis: This is a part of the process to find the required coordinates on each axis before finding the intersection. This is a typical step.

(C) Draw perpendicular lines from $x$ and $y$ to intersect: This is the crucial step to find the unique location of the point $(x, y)$ in the plane where the horizontal and vertical displacements meet. This is a typical step.

(D) Join the origin to the point $(x, y)$: Joining the origin to the point $(x, y)$ is done *after* the point has been plotted. This action creates a line segment or vector, but it is not a step in the fundamental process of *plotting* or *locating* the point itself. While this might be done for other purposes (like finding distance from the origin or representing a vector), it's not part of the basic plotting procedure.

The question asks which is NOT a typical step in plotting the point. Based on the standard procedure, joining the origin to the point is not part of the action of plotting the point $(x, y)$ itself; plotting is the act of marking the location of $(x, y)$ on the graph paper.

Therefore, the step that is NOT typically part of plotting a point $(x, y)$ is joining the origin to the point $(x, y)$.

The correct option is (D) Join the origin to the point $(x, y)$.

A point on the y-axis has its x-coordinate equal to $0$. The y-coordinate can be any value.

Therefore, the coordinates of a point on the y-axis are always of the form $(0, y)$.

The correct option is (B) $(0, y)$.

Given: Assertion (A) and Reason (R).

Solution:

Assertion (A): The area of a square with side $s$ is $A = s^2$. The graph of $A$ versus $s$ is the graph of $y = x^2$, which is a parabola, not a straight line. So, Assertion (A) is false.

Reason (R): The area is $s \times s = s^2$. A linear relationship is of the form $y = mx + c$. Since the variable $s$ is squared, $A=s^2$ is a quadratic relationship, not linear. So, Reason (R) is false.

Both Assertion (A) and Reason (R) are false. Examining the options, none state that both A and R are false.

Option (D) states "A is false, but R is true". Assertion (A) is indeed false, but Reason (R) is also false.

Given the provided options, and the clear falsehood of Assertion (A), option (D) is the only choice that correctly identifies A as false, suggesting it is the intended answer despite the error regarding the truth value of R.

The correct option is (D) A is false, but R is true. (Note: Reason (R) is mathematically false, but option (D) is the only choice where A is correctly identified as false).

Given:

The question asks to complete the sentence about the four parts the coordinate axes divide the plane into.

Solution:

In a Cartesian coordinate system, the two perpendicular lines, the x-axis and the y-axis, intersect at the origin and divide the entire plane into four distinct regions.

These four regions are specifically named based on the signs of the x and y coordinates within each region.

The upper-right region is called Quadrant I.

The upper-left region is called Quadrant II.

The lower-left region is called Quadrant III.

The lower-right region is called Quadrant IV.

Therefore, the four parts into which the coordinate axes divide the plane are called Quadrants.

The correct option is (D) Quadrants.

Given:

Data table showing Units Consumed and Bill Amount:

Units Consumed (x): 100, 150, 200, 250, 300

Bill Amount ($\textsf{₹}$) (y): 500, 750, 1000, 1250, 1500

We are plotting Units Consumed on the x-axis and Bill Amount on the y-axis.

To Find:

The cost per unit of electricity.

Solution:

The cost per unit of electricity can be found by dividing the total bill amount by the number of units consumed.

Cost per unit $= \frac{\text{Bill Amount}}{\text{Units Consumed}}$

Using the first data point from the table:

Units Consumed $= 100$, Bill Amount $= \textsf{₹}\ 500$

Cost per unit $= \frac{500}{100} = 5 \textsf{₹}/\text{unit}$

Let's verify this with another data point, say the second one:

Units Consumed $= 150$, Bill Amount $= \textsf{₹}\ 750$

Cost per unit $= \frac{750}{150} = \frac{75}{15} = 5 \textsf{₹}/\text{unit}$

Let's verify with the third data point:

Units Consumed $= 200$, Bill Amount $= \textsf{₹}\ 1000$

Cost per unit $= \frac{1000}{200} = \frac{10}{2} = 5 \textsf{₹}/\text{unit}$

The cost per unit is consistent across all data points in the table.

The cost per unit of electricity is $\textsf{₹}\ 5$ per unit.

This indicates a direct proportional relationship between Bill Amount ($y$) and Units Consumed ($x$), given by the equation $y = 5x$, where $5$ is the constant of proportionality (cost per unit).

The correct option is (B) $\textsf{₹}\ 5$ per unit.

Given:

From the Case Study in Question 35, the cost per unit of electricity is $\textsf{₹}\ 5$.

The relationship between the Bill Amount ($y$) and the Units Consumed ($x$) is given by the equation $y = 5x$ (a direct proportion).

We are given that the family consumed 180 units, so $x = 180$.

To Find:

The estimated bill amount ($y$) when 180 units are consumed.

Solution:

We use the equation $y = 5x$ that relates the Bill Amount ($y$) and the Units Consumed ($x$).

Substitute the given units consumed, $x = 180$, into the equation:

$y = 5 \times 180$

$y = 5 \times (100 + 80)$

$y = 500 + 400$

$y = 900$

Thus, the estimated bill amount for consuming 180 units would be $\textsf{₹}\ 900$.

From the graph (which is a straight line through the origin with slope 5), the point corresponding to $x=180$ would have a y-coordinate of $5 \times 180 = 900$.

The correct option is (B) $\textsf{₹}\ 900$.

Given:

The point is $(0, -4)$.

To Find:

The statement that is FALSE about the point $(0, -4)$.

Solution:

Let's evaluate each statement about the point $(0, -4)$, where the x-coordinate is $0$ and the y-coordinate is $-4$.

(A) It lies on the y-axis.

A point lies on the y-axis if its x-coordinate is $0$. For the point $(0, -4)$, the x-coordinate is $0$. So, the point lies on the y-axis. This statement is true.

(B) Its distance from the origin is 4 units.

The origin is the point $(0, 0)$. The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

The distance between $(0, -4)$ and $(0, 0)$ is $\sqrt{(0 - 0)^2 + (0 - (-4))^2} = \sqrt{0^2 + (4)^2} = \sqrt{0 + 16} = \sqrt{16} = 4$.

The distance from the origin is indeed $4$ units. This statement is true.

Alternatively, for a point on the y-axis $(0, y)$, its distance from the origin $(0, 0)$ is $|y|$. For $(0, -4)$, the distance is $|-4| = 4$.

(C) It is in Quadrant III.

The quadrants are defined by the signs of the coordinates:

Quadrant I: $(+, +)$

Quadrant II: $(-, +)$

Quadrant III: $(-, -)$

Quadrant IV: $(+, -)$

A point on an axis does not lie in any quadrant. The point $(0, -4)$ lies on the y-axis. A point is in Quadrant III if its x-coordinate is negative and its y-coordinate is negative. The x-coordinate of $(0, -4)$ is $0$, which is neither positive nor negative. Therefore, the point $(0, -4)$ is not in Quadrant III. This statement is false.

(D) Its x-coordinate is zero.

For the point $(0, -4)$, the x-coordinate is the first value, which is $0$. This statement is true.

We are looking for the FALSE statement. The statement that is false is (C).

The correct option is (C) It is in Quadrant III.

Given:

A point lies on the x-axis.

To Find:

The general form of the coordinates of such a point.

Solution:

In a two-dimensional coordinate system, the x-axis is the horizontal line.

Any point that lies on the x-axis has a vertical distance from the x-axis that is zero. The vertical distance from the x-axis is represented by the absolute value of the y-coordinate, $|y|$.

For a point to be on the x-axis, its y-coordinate must be $0$.

The x-coordinate can be any real number, as the point can be anywhere along the horizontal line.

So, the coordinates of a point on the x-axis are always of the form $(x, 0)$, where $x$ can be any real number.

Let's look at the options:

(A) $(x, 0)$: This form represents points where the y-coordinate is $0$, which are precisely the points on the x-axis.

(B) $(0, y)$: This form represents points where the x-coordinate is $0$, which are the points on the y-axis.

(C) $(x, x)$: This form represents points where the x-coordinate and y-coordinate are equal, which lie on the line $y=x$.

(D) $(0, 0)$: This represents the origin, which is a single point that lies on both axes, but not the general form for all points on the x-axis.

The general form that describes all points on the x-axis is $(x, 0)$.

The correct option is (A) $(x, 0)$.

Given:

The graph of a relationship between two variables is a straight line.

To Find:

The name of this type of relationship.

Solution:

In mathematics, the graph of a relationship between two variables, say $x$ and $y$, is a visual representation of how the value of one variable changes with the value of the other. Different types of relationships produce different types of graphs.

A straight line graph is characteristic of a relationship where the variables change at a constant rate relative to each other (constant slope). This type of relationship is called a linear relationship.

The general equation for a straight line is $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept. This equation represents a linear function or a linear equation in two variables. Any relationship between two variables that can be expressed in this form is a linear relationship, and its graph is a straight line.

Let's consider the other options:

(A) Quadratic relationships (e.g., $y = ax^2 + bx + c$) produce graphs that are parabolas.

(B) Exponential relationships (e.g., $y = a \cdot b^x$) produce graphs that are curves that increase or decrease rapidly.

(D) Proportional relationship: Direct proportion ($y = kx$) is a specific type of linear relationship where the line passes through the origin ($c=0$). Inverse proportion ($y = k/x$) produces a hyperbolic curve.

The term that universally describes a relationship whose graph is a straight line is linear.

The correct option is (C) Linear.

Given:

Assertion (A): The point $(2, 5)$ is the same as the point $(5, 2)$ in a coordinate plane.

Reason (R): The order of coordinates in an ordered pair matters.

Solution:

Let's evaluate Assertion (A): "The point $(2, 5)$ is the same as the point $(5, 2)$ in a coordinate plane."

In an ordered pair $(x, y)$, the first coordinate $x$ represents the position along the x-axis (horizontal distance from the origin), and the second coordinate $y$ represents the position along the y-axis (vertical distance from the origin).

For the point $(2, 5)$, the x-coordinate is $2$ and the y-coordinate is $5$.

For the point $(5, 2)$, the x-coordinate is $5$ and the y-coordinate is $2$.

Since the coordinates are different (specifically, the x and y values are swapped, and $2 \neq 5$), these two points represent different locations in the coordinate plane.

Therefore, Assertion (A) is false.

Let's evaluate Reason (R): "The order of coordinates in an ordered pair matters."

In mathematics, an ordered pair is a collection of two elements where the order of the elements is important. This is fundamental to the concept of coordinates in a plane, where the first element specifies the position along one axis and the second element specifies the position along the other axis.

For example, the ordered pair $(2, 5)$ is different from the ordered pair $(5, 2)$ because the order of $2$ and $5$ is different. This difference in order corresponds to different locations in the coordinate plane unless the coordinates are identical (e.g., $(3, 3)$). The fact that $(2, 5)$ and $(5, 2)$ are different points confirms that the order matters.

Therefore, Reason (R) is true.

We have determined that Assertion (A) is false and Reason (R) is true.

Now let's check the options:

(A) Both A and R are true... (Incorrect, A is false)

(B) Both A and R are true... (Incorrect, A is false)

(C) A is true, but R is false. (Incorrect, A is false and R is true)

(D) A is false, but R is true. (Correct)

Reason (R) provides the principle that explains why Assertion (A) is false: because the order of coordinates *does* matter, the point $(2, 5)$ is not the same as $(5, 2)$.

The correct option is (D) A is false, but R is true.

A coordinate system is a system used to uniquely determine the position of a point or other geometric element by using one or more numbers, or coordinates. In two dimensions, the most common type is the Cartesian coordinate system, also known as the rectangular coordinate system.

This system consists of two perpendicular number lines, usually called the axes, which intersect at a point called the origin.

The x-axis is the horizontal number line in the Cartesian coordinate system.

It is typically used to represent the first coordinate of a point, often denoted as $x$.

Points on the x-axis have a y-coordinate of zero. Movement along the x-axis is described as horizontal displacement from the origin. Positive values are usually to the right of the origin, and negative values are to the left.

The y-axis is the vertical number line in the Cartesian coordinate system.

It is typically used to represent the second coordinate of a point, often denoted as $y$.

Points on the y-axis have an x-coordinate of zero. Movement along the y-axis is described as vertical displacement from the origin. Positive values are usually upwards from the origin, and negative values are downwards.

Together, the x-axis and y-axis divide the plane into four regions called quadrants. The intersection point of the x-axis and y-axis is the origin, represented by the coordinates $(0, 0)$. Any point in the plane can be uniquely identified by an ordered pair of coordinates $(x, y)$, where $x$ is the distance along the x-axis (abscissa) and $y$ is the distance along the y-axis (ordinate) relative to the origin.

In a Cartesian coordinate system, the origin is the point where the x-axis and the y-axis intersect.

It serves as the reference point from which all other points in the plane are measured.

The coordinates of the origin are (0, 0).

This means that the origin has an x-coordinate of zero and a y-coordinate of zero.

In a Cartesian coordinate system, points are represented by an ordered pair of numbers, typically written as $(x, y)$.

The first number, $x$, represents the horizontal position relative to the origin along the x-axis.

The second number, $y$, represents the vertical position relative to the origin along the y-axis.

In an ordered pair $(x, y)$:

The abscissa is the first coordinate, which is the x-coordinate.

The ordinate is the second coordinate, which is the y-coordinate.

For the given point $(3, -5)$:

The abscissa is the first coordinate, which is $3$.

The ordinate is the second coordinate, which is $-5$.

The quadrant in which a point $(x, y)$ lies depends on the signs of its coordinates $x$ and $y$.

Quadrant I: $x > 0$, $y > 0$

Quadrant II: $x < 0$, $y > 0$

Quadrant III: $x < 0$, $y < 0$

Quadrant IV: $x > 0$, $y < 0$

(a) For the point $(2, 3)$:

$x = 2$, which is positive ($x > 0$).

$y = 3$, which is positive ($y > 0$).

Since both $x$ and $y$ are positive, the point $(2, 3)$ lies in Quadrant I.

(b) For the point $(-4, 1)$:

$x = -4$, which is negative ($x < 0$).

$y = 1$, which is positive ($y > 0$).

Since $x$ is negative and $y$ is positive, the point $(-4, 1)$ lies in Quadrant II.

(c) For the point $(-2, -5)$:

$x = -2$, which is negative ($x < 0$).

$y = -5$, which is negative ($y < 0$).

Since both $x$ and $y$ are negative, the point $(-2, -5)$ lies in Quadrant III.

(d) For the point $(3, -6)$:

$x = 3$, which is positive ($x > 0$).

$y = -6$, which is negative ($y < 0$).

Since $x$ is positive and $y$ is negative, the point $(3, -6)$ lies in Quadrant IV.

Points with coordinates of the form $(x, 0)$ have their y-coordinate equal to zero.

Any point with a y-coordinate of $0$ lies on the x-axis (the horizontal axis).

The value of $x$ determines the specific position on the x-axis.

Points with coordinates of the form $(0, y)$ have their x-coordinate equal to zero.

Any point with an x-coordinate of $0$ lies on the y-axis (the vertical axis).

The value of $y$ determines the specific position on the y-axis.

The only point that lies on both axes is the origin, which has coordinates $(0, 0)$.

Solution:

To plot the points A$(2, 4)$ and B$(2, -3)$ on a graph sheet:

First, draw the x-axis (horizontal) and the y-axis (vertical) intersecting at the origin $(0, 0)$.

For point A$(2, 4)$: Start from the origin, move 2 units to the right along the x-axis (since the x-coordinate is positive 2), and then move 4 units upwards parallel to the y-axis (since the y-coordinate is positive 4). Mark this point as A$(2, 4)$.

For point B$(2, -3)$: Start from the origin, move 2 units to the right along the x-axis (since the x-coordinate is positive 2), and then move 3 units downwards parallel to the y-axis (since the y-coordinate is negative 3). Mark this point as B$(2, -3)$.

Connect the plotted points A and B with a straight line segment.

Let's examine the coordinates of points A and B.

Point A has coordinates $(2, 4)$.

Point B has coordinates $(2, -3)$.

We observe that the x-coordinate is the same for both points (it is 2).

When two points have the same x-coordinate, the line segment connecting them is a vertical line.

A vertical line in the Cartesian coordinate system is parallel to the y-axis.

So, the line segment AB is a vertical line segment parallel to the y-axis.

To find the length of the line segment AB, since it is a vertical line, we can find the absolute difference of the y-coordinates.

Let the coordinates of A be $(x_1, y_1)$ and B be $(x_2, y_2)$.

Here, $x_1 = 2$, $y_1 = 4$, $x_2 = 2$, $y_2 = -3$.

Length of AB = $|y_2 - y_1|$

Length of AB = $|-3 - 4|$

Length of AB = $|-7|$

The absolute value of $-7$ is $7$.

Therefore, the length of AB is $7$ units.

Solution:

To plot the points C$(-3, 1)$ and D$(5, 1)$ on a graph sheet:

First, draw the x-axis (horizontal) and the y-axis (vertical) intersecting at the origin $(0, 0)$.

For point C$(-3, 1)$: Start from the origin, move 3 units to the left along the x-axis (since the x-coordinate is negative 3), and then move 1 unit upwards parallel to the y-axis (since the y-coordinate is positive 1). Mark this point as C$(-3, 1)$.

For point D$(5, 1)$: Start from the origin, move 5 units to the right along the x-axis (since the x-coordinate is positive 5), and then move 1 unit upwards parallel to the y-axis (since the y-coordinate is positive 1). Mark this point as D$(5, 1)$.

Connect the plotted points C and D with a straight line segment.

Let's examine the coordinates of points C and D.

Point C has coordinates $(-3, 1)$.

Point D has coordinates $(5, 1)$.

We observe that the y-coordinate is the same for both points (it is 1).

When two points have the same y-coordinate, the line segment connecting them is a horizontal line.

A horizontal line in the Cartesian coordinate system is parallel to the x-axis.

So, the line segment CD is a horizontal line segment parallel to the x-axis.

To find the length of the line segment CD, since it is a horizontal line, we can find the absolute difference of the x-coordinates.

Let the coordinates of C be $(x_1, y_1)$ and D be $(x_2, y_2)$.

Here, $x_1 = -3$, $y_1 = 1$, $x_2 = 5$, $y_2 = 1$.

Length of CD = $|x_2 - x_1|$

Length of CD = $|5 - (-3)|$

Length of CD = $|5 + 3|$

Length of CD = $|8|$

The absolute value of $8$ is $8$.

Therefore, the length of CD is $8$ units.

Unfortunately, the graph containing the points P and Q is not visible or provided. Therefore, I cannot read the exact coordinates of points P and Q.

However, I can explain the general method for reading the coordinates of a point from a Cartesian coordinate system graph:

1. Start at the origin $(0, 0)$.

2. To find the x-coordinate (abscissa), move horizontally from the origin to the point along the x-axis or parallel to the x-axis. Note the value on the x-axis where this horizontal line meets the axis. This value is the x-coordinate ($x$). Move right for positive values, left for negative values.

3. To find the y-coordinate (ordinate), move vertically from the origin to the point along the y-axis or parallel to the y-axis. Note the value on the y-axis where this vertical line meets the axis. This value is the y-coordinate ($y$). Move up for positive values, down for negative values.

4. The coordinates of the point are written as an ordered pair $(x, y)$.

Based on the assumption provided in the question about the location of points P and Q in different quadrants, here is how the signs of the coordinates relate to the quadrants:

If P is in Quadrant I (Q1), its coordinates $(x_P, y_P)$ would have $x_P > 0$ and $y_P > 0$.

If Q is in Quadrant II (Q2), its coordinates $(x_Q, y_Q)$ would have $x_Q < 0$ and $y_Q > 0$.

If Q is in Quadrant III (Q3), its coordinates $(x_Q, y_Q)$ would have $x_Q < 0$ and $y_Q < 0$.

If Q is in Quadrant IV (Q4), its coordinates $(x_Q, y_Q)$ would have $x_Q > 0$ and $y_Q < 0$.

To provide the specific coordinates for the points P and Q, the corresponding graph is needed.

Solution:

To plot the points A$(1, 2)$, B$(1, 5)$, C$(5, 5)$, and D$(5, 2)$ on a graph sheet:

Draw the x-axis and y-axis.

Plot A$(1, 2)$: Move 1 unit right from origin, then 2 units up.

Plot B$(1, 5)$: Move 1 unit right from origin, then 5 units up.

Plot C$(5, 5)$: Move 5 units right from origin, then 5 units up.

Plot D$(5, 2)$: Move 5 units right from origin, then 2 units up.

Connect the points in order: A to B, B to C, C to D, and D to A.

Consider the segments formed:

Segment AB: Points are A$(1, 2)$ and B$(1, 5)$. The x-coordinates are the same ($1$). This is a vertical segment. Length AB = $|5 - 2| = |3| = 3$ units.

Segment BC: Points are B$(1, 5)$ and C$(5, 5)$. The y-coordinates are the same ($5$). This is a horizontal segment. Length BC = $|5 - 1| = |4| = 4$ units.

Segment CD: Points are C$(5, 5)$ and D$(5, 2)$. The x-coordinates are the same ($5$). This is a vertical segment. Length CD = $|2 - 5| = |-3| = 3$ units.

Segment DA: Points are D$(5, 2)$ and A$(1, 2)$. The y-coordinates are the same ($2$). This is a horizontal segment. Length DA = $|1 - 5| = |-4| = 4$ units.

The quadrilateral ABCD has opposite sides AB and CD of equal length ($3$ units), and opposite sides BC and DA of equal length ($4$ units).

Also, since AB and CD are vertical and BC and DA are horizontal, they are perpendicular to each other at the vertices (e.g., AB is perpendicular to BC at B). This means all interior angles are right angles ($90^\circ$).

A quadrilateral with opposite sides equal and parallel, and all angles $90^\circ$, is a rectangle.

So, the shape formed is a rectangle.

The perimeter of a rectangle is given by the formula:

Perimeter = $2 \times (\text{Length} + \text{Width})$

Using the lengths we found, Length $= 4$ (BC or DA) and Width $= 3$ (AB or CD).

Perimeter = $2 \times (4 + 3)$ units

Perimeter = $2 \times (7)$ units

Perimeter = $14$ units.

The perimeter of the rectangle ABCD is $14$ units.

Let the quantity of apples in kg be represented by $x$ and the cost in $\textsf{₹}$ be represented by $y$.

Given that the cost of 1 kg of apples is $\textsf{₹}100$.

The relationship between the quantity and the cost can be expressed as $y = 100x$.

Table for the cost of 1, 2, 3, and 4 kg of apples:

A relationship between two variables $x$ and $y$ is considered linear if the graph of the relationship is a straight line.

Mathematically, a linear relationship can be expressed in the form $y = mx + c$, where $m$ and $c$ are constants. $m$ is the slope (rate of change), and $c$ is the y-intercept (value of $y$ when $x=0$).

In this case, the cost $y$ is directly proportional to the quantity $x$, and the relationship is given by $y = 100x$.

Comparing this to the linear form $y = mx + c$, we have $m = 100$ and $c = 0$.

The rate of change of cost with respect to quantity is constant ($100 \textsf{₹}$ per kg).

Therefore, this is a linear relationship.

This is because the cost increases at a constant rate for every additional kilogram of apples purchased. The graph of this relationship would be a straight line passing through the origin $(0, 0)$ with a slope of $100$.

Solution:

The given relationship between the distance covered ($d$) and the time taken ($t$) is:

$d = 50t$

A linear relationship between two variables, let's say $x$ and $y$, is generally represented by an equation that can be written in the form:

$y = mx + c$

where $m$ and $c$ are constants.

Comparing the given equation $d = 50t$ with the standard linear form $y = mx + c$, we can see that it fits this form:

- The dependent variable $d$ corresponds to $y$.

- The independent variable $t$ corresponds to $x$.

- The constant speed 50 corresponds to the slope $m$.

- The constant term $c$ is 0 in this case ($d = 50t + 0$).

Since the equation $d = 50t$ can be written in the form $y = mx + c$ with $m=50$ and $c=0$, it represents a linear relationship between $d$ and $t$. Graphically, this relationship would be a straight line passing through the origin $(0,0)$ with a slope of 50.

Therefore, yes, the relationship between distance and time given by $d = 50t$ is a linear relationship because the equation is in the form of a linear equation $y = mx + c$.

Solution:

The graph shows the temperature ($T$) on the y-axis and the time ($t$) on the x-axis.

When the graph is a straight line, it indicates a linear relationship between the two variables.

The equation of a straight line is generally given by $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept.

In this case, the graph is a straight line passing through the origin. A straight line passing through the origin has a y-intercept of 0, i.e., $c = 0$.

So, the equation of the line is of the form:

$y = mx$

Substituting the variables from the problem, where $y$ represents temperature ($T$) and $x$ represents time ($t$), the relationship is:

$T = mt$

Here, $m$ is the constant slope of the line.

This equation tells us two main things about the temperature:

1. Since the line passes through the origin $(0,0)$, it means when the time $t=0$, the temperature $T = m \times 0 = 0$. Thus, the initial temperature (at time $t=0$) is $0^\circ$ (or 0 units, depending on the temperature scale used).

2. The temperature changes with respect to time at a constant rate. This rate is represented by the slope $m$. If $m$ is positive, the temperature increases constantly with time. If $m$ is negative, the temperature decreases constantly with time. If $m=0$, the temperature remains constant at 0.

In summary, a straight-line graph passing through the origin indicates that the temperature has an initial value of zero and changes at a constant rate over time.

Solution:

In a distance-time graph, time ($t$) is plotted on the x-axis and distance ($d$) is plotted on the y-axis.

A vertical line on any graph represents a situation where the value of the x-axis variable remains constant, while the value of the y-axis variable changes.

If a distance-time graph were a vertical line, it would mean that for a single, specific instant in time (a constant value on the x-axis), the distance covered (on the y-axis) is changing or can take on multiple values.

In reality, an object can only be at one specific location (distance from a reference point) at any given moment in time.

Therefore, it is physically impossible for an object to be at different distances simultaneously at the same instant.

Hence, a distance-time graph cannot be a vertical line.

Solution:

In a graph representing the cost of items, the x-axis usually represents the number of items, and the y-axis usually represents the total cost.

Each point on the graph is in the form $(x, y)$, where $x$ is the value on the x-axis and $y$ is the value on the y-axis.

The given points are $(1, 5), (2, 10), (3, 15), (4, 20)$.

Considering the point $(2, 10)$, the x-coordinate is 2 and the y-coordinate is 10.

Therefore, if the x-axis represents the number of items and the y-axis represents the cost, the point $(2, 10)$ signifies that the cost of 2 items is 10 units (e.g., $\textsf{₹}10$, $\$10$, etc., depending on the currency being used).

Solution:

To plot the points A$(0, 5)$, B$(5, 0)$, C$(0, -5)$, and D$(-5, 0)$ on a graph sheet, we follow these steps:

1. Draw the x-axis and the y-axis, intersecting at the origin $(0, 0)$.

2. Mark a suitable scale on both axes. Let 1 unit on the graph represent 1 unit on the coordinate plane.

3. Plot point A$(0, 5)$: The x-coordinate is 0, and the y-coordinate is 5. This point lies on the y-axis, 5 units above the origin.

4. Plot point B$(5, 0)$: The x-coordinate is 5, and the y-coordinate is 0. This point lies on the x-axis, 5 units to the right of the origin.

5. Plot point C$(0, -5)$: The x-coordinate is 0, and the y-coordinate is -5. This point lies on the y-axis, 5 units below the origin.

6. Plot point D$(-5, 0)$: The x-coordinate is -5, and the y-coordinate is 0. This point lies on the x-axis, 5 units to the left of the origin.

Now, join the points in order: A to B, B to C, C to D, and D back to A.

Let's examine the lengths of the line segments using the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

Length of AB = $\sqrt{(5-0)^2 + (0-5)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50}$

Length of BC = $\sqrt{(0-5)^2 + (-5-0)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50}$

Length of CD = $\sqrt{(-5-0)^2 + (0-(-5))^2} = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50}$

Length of DA = $\sqrt{(0-(-5))^2 + (5-0)^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50}$

All four sides have equal length ($\sqrt{50}$).

Now let's examine the slopes of consecutive line segments. Slope $m = \frac{y_2-y_1}{x_2-x_1}$.

Slope of AB = $\frac{0-5}{5-0} = \frac{-5}{5} = -1$

Slope of BC = $\frac{-5-0}{0-5} = \frac{-5}{-5} = 1$

Slope of CD = $\frac{0-(-5)}{-5-0} = \frac{5}{-5} = -1$

Slope of DA = $\frac{5-0}{0-(-5)} = \frac{5}{5} = 1$

The product of the slopes of adjacent sides (AB and BC) is $(-1) \times 1 = -1$. This indicates that AB is perpendicular to BC. Similarly, BC is perpendicular to CD, CD is perpendicular to DA, and DA is perpendicular to AB.

The quadrilateral formed by joining these points has all four sides of equal length and all four internal angles are $90^\circ$ (because adjacent sides are perpendicular).

Therefore, the shape formed by joining the points A$(0, 5)$, B$(5, 0)$, C$(0, -5)$, and D$(-5, 0)$ in order is a square.

Solution:

The difference between plotting discrete data and continuous data on a graph lies in how the points are represented and whether they are connected by lines.

Discrete Data:

Discrete data can only take on specific, distinct values, usually whole numbers or categories. There are gaps between possible data points.

When plotting discrete data, individual points are plotted on the graph. These points typically represent measured values at specific, separate instances or categories.

Because the data cannot take on values between the plotted points (or those values are not meaningful in the context), the points are generally not joined by a continuous line.

Examples of graphs for discrete data include bar graphs, scatter plots where the points are not connected, or points representing specific items/counts.

Continuous Data:

Continuous data can take on any value within a given range. There are no gaps between potential data points; theoretically, it can be measured to any level of precision.

When plotting continuous data, individual points representing measurements at different times or conditions are plotted. Because the variable can take on values between the plotted points, and the change is generally smooth, the points are typically joined by a continuous line.

The line represents the values that the data is assumed to take between the observed points.

Examples of graphs for continuous data include line graphs (like temperature over time, distance covered over time), histograms (where bars touch), and smooth curves.

In summary, for discrete data, points are plotted individually and usually not connected, reflecting distinct values. For continuous data, points are plotted and typically connected by a line, reflecting a range of possible values.

Solution:

The given relationship between the area ($A$) of a square and its side length ($s$) is:

$A = s^2$

A linear relationship between two variables, let's say $x$ and $y$, is an equation that can be written in the form:

$y = mx + c$

where $m$ and $c$ are constants. In this form, the independent variable ($x$) is raised to the power of 1.

In the given equation $A = s^2$, the dependent variable is $A$ (area) and the independent variable is $s$ (side length).

The side length $s$ is raised to the power of 2 ($s^2$).

Since the variable $s$ is squared, the equation $A = s^2$ is a **quadratic** relationship, not a linear relationship.

The graph of $A = s^2$ (considering $s \ge 0$ since side length cannot be negative) is a curve (specifically, a portion of a parabola), not a straight line.

For example, if we double the side length from $s=2$ to $s=4$, the area changes from $A=2^2=4$ to $A=4^2=16$. The area quadrupled, it did not just double, which would happen in a linear relationship ($A = ms$, where $A$ doubles if $s$ doubles).

Therefore, the relationship between the area of a square and its side length given by $A = s^2$ is not linear because the side length is squared in the equation.

Solution:

The given relationship between the perimeter ($P$) of a square and its side length ($s$) is:

$P = 4s$

A linear relationship between two variables, let's say $x$ and $y$, can be represented by an equation in the form:

$y = mx + c$

where $m$ and $c$ are constants. In this form, the independent variable ($x$) is raised to the power of 1.

Comparing the given equation $P = 4s$ with the standard linear form $y = mx + c$, we can see that it fits this form:

- The dependent variable $P$ corresponds to $y$.

- The independent variable $s$ corresponds to $x$.

- The constant 4 corresponds to the slope $m$.

- The constant term $c$ is 0 in this case ($P = 4s + 0$).

Since the equation $P = 4s$ can be written in the form $y = mx + c$ with $m=4$ and $c=0$, and the independent variable $s$ is raised to the power of 1, it represents a linear relationship between $P$ and $s$. The graph of this relationship would be a straight line passing through the origin $(0,0)$ with a slope of 4.

Therefore, yes, the relationship between the perimeter of a square and its side length given by $P = 4s$ is a linear relationship because the equation is in the form of a linear equation $y = mx + c$, where the independent variable is raised to the power of 1.

Solution:

In a graph, the x-axis represents one quantity (often the independent variable), and the y-axis represents another quantity (often the dependent variable).

A horizontal line on a graph is a line where the y-coordinate remains the same for every point on the line, regardless of the x-coordinate.

This means that as the quantity on the x-axis changes or varies, the value of the quantity on the y-axis does not change; it stays constant.

Therefore, if a graph is a horizontal line, it indicates that the quantity on the y-axis is constant with respect to the quantity on the x-axis.

For example, if a graph shows the speed of a car (y-axis) versus time (x-axis) and it's a horizontal line, it means the car is traveling at a constant speed over that period of time.

Solution:

The coordinate plane is divided into four regions by the x-axis and the y-axis. These regions are called quadrants.

Starting from the upper right and moving counter-clockwise, the four quadrants are named as follows:

1. First Quadrant (I)

2. Second Quadrant (II)

3. Third Quadrant (III)

4. Fourth Quadrant (IV)

The signs of the coordinates $(x, y)$ in each quadrant are described below:

1. First Quadrant (I): In this quadrant, the x-coordinate is positive, and the y-coordinate is positive. So, the signs are $(+, +)$.

2. Second Quadrant (II): In this quadrant, the x-coordinate is negative, and the y-coordinate is positive. So, the signs are $(-, +)$.

3. Third Quadrant (III): In this quadrant, the x-coordinate is negative, and the y-coordinate is negative. So, the signs are $(-, -)$.

4. Fourth Quadrant (IV): In this quadrant, the x-coordinate is positive, and the y-coordinate is negative. So, the signs are $(+, -)$.

Points lying on the axes themselves are not considered to be in any quadrant.

Solution:

If a point lies on the y-axis, its position is solely determined by its distance from the origin along the y-axis. The x-coordinate represents the horizontal distance from the y-axis.

For any point lying on the y-axis, the horizontal distance from the y-axis is always zero.

Therefore, if a point lies on the y-axis, its x-coordinate is 0.

The x-axis is the horizontal line where the y-coordinate is 0.

Points located above the x-axis have a positive y-coordinate.

Points located on the x-axis have a y-coordinate of 0.

Points located below the x-axis have a negative y-coordinate.

Therefore, if a point lies below the x-axis, the sign of its y-coordinate is negative.

Solution:

Given the points of the vertices of a square: $(0, 0)$, $(a, 0)$, $(a, a)$, and $(0, a)$.

Let's denote these points as O$(0, 0)$, A$(a, 0)$, B$(a, a)$, and C$(0, a)$.

To find the side length of the square, we can calculate the distance between any two adjacent vertices.

Let's find the distance between point O$(0, 0)$ and point A$(a, 0)$. The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

Side length OA = $\sqrt{(a-0)^2 + (0-0)^2} = \sqrt{a^2 + 0^2} = \sqrt{a^2}$.

Assuming $a > 0$ for a non-degenerate square, $\sqrt{a^2} = a$. If $a < 0$, the side length would be $|a|$. We assume $a$ represents a positive length here.

Let's verify with another side, say AB:

Side length AB = $\sqrt{(a-a)^2 + (a-0)^2} = \sqrt{0^2 + a^2} = \sqrt{a^2} = a$ (assuming $a>0$).

So, the side length of the square is $a$ units.

The area of a square is calculated by squaring the side length.

Area = $(\text{side length})^2$

Area = $(a)^2$

Area = $a^2$ square units.

Therefore, the side length of the square is $a$ and its area is $a^2$ square units.

Given:

The coordinates of the vertices of a quadrilateral PQRS are P$(1, 1)$, Q$(1, 4)$, R$(5, 4)$, and S$(5, 1)$.

To Find:

1. The type of quadrilateral PQRS.

2. The perimeter of PQRS.

3. The area of PQRS.

4. The coordinates of the point of intersection of the diagonals PR and QS.

Solution:

First, we plot the given points on a graph sheet and join them in order P to Q, Q to R, R to S, and S to P.

P$(1, 1)$ is located 1 unit right and 1 unit up from the origin.

Q$(1, 4)$ is located 1 unit right and 4 units up from the origin.

R$(5, 4)$ is located 5 units right and 4 units up from the origin.

S$(5, 1)$ is located 5 units right and 1 unit up from the origin.

To determine the type of quadrilateral, we find the lengths of the sides using the distance formula, $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

Length of PQ = $\sqrt{(1-1)^2 + (4-1)^2} = \sqrt{0^2 + 3^2} = \sqrt{0 + 9} = \sqrt{9} = 3$ units.

Length of QR = $\sqrt{(5-1)^2 + (4-4)^2} = \sqrt{4^2 + 0^2} = \sqrt{16 + 0} = \sqrt{16} = 4$ units.

Length of RS = $\sqrt{(5-5)^2 + (1-4)^2} = \sqrt{0^2 + (-3)^2} = \sqrt{0 + 9} = \sqrt{9} = 3$ units.

Length of SP = $\sqrt{(1-5)^2 + (1-1)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16 + 0} = \sqrt{16} = 4$ units.

We see that PQ = RS = 3 and QR = SP = 4. Since opposite sides are equal, PQRS is a parallelogram.

Now, let's check if it has a right angle by finding the slopes of adjacent sides using the slope formula, $m = \frac{y_2-y_1}{x_2-x_1}$.

Slope of PQ = $\frac{4-1}{1-1} = \frac{3}{0}$ (undefined). This indicates a vertical line.

Slope of QR = $\frac{4-4}{5-1} = \frac{0}{4} = 0$. This indicates a horizontal line.

Since PQ is a vertical line and QR is a horizontal line, they are perpendicular to each other. Thus, $\angle$PQR = $90^\circ$.

A parallelogram with one right angle is a rectangle.

Thus, the quadrilateral PQRS is a rectangle.

Perimeter of PQRS:

The perimeter of a rectangle is given by $2 \times (\text{Length} + \text{Width})$.

Length = QR = SP = 4 units.

Width = PQ = RS = 3 units.

Perimeter = $2 \times (4 + 3) = 2 \times 7 = 14$ units.

The perimeter of PQRS is 14 units.

Area of PQRS:

The area of a rectangle is given by $\text{Length} \times \text{Width}$.

Area = $4 \times 3 = 12$ square units.

The area of PQRS is 12 square units.

Coordinates of the point of intersection of the diagonals PR and QS:

In a rectangle, the diagonals bisect each other. The point of intersection of the diagonals is the midpoint of both diagonals.

Let's find the midpoint of diagonal PR using the midpoint formula, $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

For diagonal PR, P$(1, 1)$ and R$(5, 4)$.

Midpoint of PR = $\left(\frac{1+5}{2}, \frac{1+4}{2}\right) = \left(\frac{6}{2}, \frac{5}{2}\right) = (3, 2.5)$.

Let's verify by finding the midpoint of diagonal QS.

For diagonal QS, Q$(1, 4)$ and S$(5, 1)$.

Midpoint of QS = $\left(\frac{1+5}{2}, \frac{4+1}{2}\right) = \left(\frac{6}{2}, \frac{5}{2}\right) = (3, 2.5)$.

Both midpoints are the same, which confirms the diagonals bisect each other at this point.

The coordinates of the point of intersection of the diagonals are $(3, 2.5)$ or $(3, \frac{5}{2})$.

Given:

The coordinates of the vertices of a quadrilateral are A$(0, 0)$, B$(5, 0)$, C$(2, 3)$, and D$(7, 3)$.

To Find:

1. The type of quadrilateral formed by joining AB, BC, CD, and DA.

2. The lengths of its sides.

3. The length of the diagonal AC.

Solution:

First, we plot the given points on a graph sheet and join them in order A to B, B to C, C to D, and D to A.

A$(0, 0)$ is the origin.

B$(5, 0)$ is located 5 units right from the origin on the x-axis.

C$(2, 3)$ is located 2 units right and 3 units up from the origin.

D$(7, 3)$ is located 7 units right and 3 units up from the origin.

To find the lengths of the sides, we can use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

Length of side AB (connecting A$(0, 0)$ and B$(5, 0)$):

This is a horizontal line segment. The length can be found by the difference in x-coordinates:

AB = $|5 - 0| = 5$ units.

Alternatively, using the distance formula:

AB = $\sqrt{(5-0)^2 + (0-0)^2} = \sqrt{5^2 + 0^2} = \sqrt{25} = 5$ units.

Length of side BC (connecting B$(5, 0)$ and C$(2, 3)$):

BC = $\sqrt{(2-5)^2 + (3-0)^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18}$ units.

We can simplify $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ units.

Length of side CD (connecting C$(2, 3)$ and D$(7, 3)$):

This is a horizontal line segment. The length can be found by the difference in x-coordinates:

CD = $|7 - 2| = 5$ units.

Alternatively, using the distance formula:

CD = $\sqrt{(7-2)^2 + (3-3)^2} = \sqrt{5^2 + 0^2} = \sqrt{25} = 5$ units.

Length of side DA (connecting D$(7, 3)$ and A$(0, 0)$):

DA = $\sqrt{(0-7)^2 + (0-3)^2} = \sqrt{(-7)^2 + (-3)^2} = \sqrt{49 + 9} = \sqrt{58}$ units.

Lengths of the sides are: AB = 5, BC = $\sqrt{18}$ (or $3\sqrt{2}$), CD = 5, DA = $\sqrt{58}$.

Now, let's determine the type of quadrilateral ABCD.

We observe that AB = CD = 5. These are opposite sides.

Let's check if AB is parallel to CD. The y-coordinates of A and B are both 0, so AB lies on the x-axis (or is parallel to it, if we consider the x-axis itself). The y-coordinates of C and D are both 3. This means the line segment CD is horizontal, 3 units above the x-axis. Since both AB and CD are horizontal, they are parallel to each other.

Since opposite sides AB and CD are parallel and equal in length, the quadrilateral ABCD is a parallelogram.

It is not a rectangle because adjacent sides are not perpendicular (e.g., AB is horizontal, but BC has a slope of $\frac{3-0}{2-5} = -1$, which is not vertical). It is not a rhombus because all sides are not equal (e.g., AB=5, BC=$\sqrt{18} \approx 4.24$).

Length of the diagonal AC (connecting A$(0, 0)$ and C$(2, 3)$):

AC = $\sqrt{(2-0)^2 + (3-0)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$ units.

Summary of findings:

The quadrilateral ABCD is a parallelogram.

The lengths of its sides are AB = 5, BC = $\sqrt{18}$ (or $3\sqrt{2}$), CD = 5, and DA = $\sqrt{58}$ units.

The length of the diagonal AC is $\sqrt{13}$ units.

Solution:

The car is traveling at a constant speed of 40 km/hr.

The relationship between distance ($d$), speed ($v$), and time ($t$) is given by:

$d = v \times t$

Given the speed $v = 40$ km/hr, the relationship becomes:

$d = 40t$

We need to calculate the distance covered for different values of time $t = 0, 1, 2, 3, 4$ hours.

For $t = 0$ hours, $d = 40 \times 0 = 0$ km. Point: $(0, 0)$.

For $t = 1$ hour, $d = 40 \times 1 = 40$ km. Point: $(1, 40)$.

For $t = 2$ hours, $d = 40 \times 2 = 80$ km. Point: $(2, 80)$.

For $t = 3$ hours, $d = 40 \times 3 = 120$ km. Point: $(3, 120)$.

For $t = 4$ hours, $d = 40 \times 4 = 160$ km. Point: $(4, 160)$.

We can put these values in a table:

To draw the graph, we plot these points on a graph sheet with time ($t$) on the x-axis and distance ($d$) on the y-axis.

When these points are plotted and joined, they form a straight line.

This graph is a straight line because the relationship between distance ($d$) and time ($t$) is linear ($d = 40t$). A linear relationship between two variables results in a straight line when plotted on a graph. The equation $d = 40t$ is in the form $y = mx + c$, where $d$ is $y$, $t$ is $x$, $m = 40$ (the speed, which is constant and represents the slope), and $c = 0$ (the y-intercept). The straight line passes through the origin $(0,0)$, indicating that at time $t=0$, the distance covered is 0.

This indicates that for a car traveling at a constant speed, the distance covered is directly proportional to the time taken.

Given:

The cost of sugar for different weights is given in the table.

To Draw:

A graph representing the data with weight on the x-axis and cost on the y-axis.

To Find:

1. The cost of 3.5 kg of sugar using the graph.

2. The weight of sugar that can be bought for $\textsf{₹}175$ using the graph.

Solution:

From the given table, we can list the points $(Weight, Cost)$ to be plotted:

Steps to draw the graph:

1. Draw the x-axis and label it as "Weight of Sugar (kg)".

2. Draw the y-axis and label it as "Cost ($\textsf{₹}$)".

3. Choose appropriate scales for both axes. For the x-axis, we can take 1 unit = 1 kg. For the y-axis, we can take 1 unit = $\textsf{₹}50$ (or $\textsf{₹}25$ for better precision in reading intermediate values). Let's assume a scale where each major grid line represents 1 unit on the x-axis and $\textsf{₹}50$ on the y-axis.

4. Plot the points (1, 50), (2, 100), (3, 150), (4, 200), and (5, 250) on the graph sheet.

5. Join the plotted points. As observed from the data, the cost is directly proportional to the weight (50 $\textsf{₹}$ per kg). So, the points will lie on a straight line passing through the origin $(0, 0)$ since 0 kg of sugar costs $\textsf{₹}0$.

Using the graph to find the cost of 3.5 kg of sugar:

1. Locate 3.5 on the x-axis (Weight). This will be halfway between 3 and 4.

2. Draw a vertical line from the point 3.5 on the x-axis upwards until it meets the graph line.

3. From the point where the vertical line meets the graph, draw a horizontal line to the left until it meets the y-axis (Cost).

4. Read the value on the y-axis where the horizontal line meets it. This value will be $\textsf{₹}175$.

So, the cost of 3.5 kg of sugar is $\textsf{₹}175$.

Using the graph to find the weight of sugar for $\textsf{₹}175$:

1. Locate 175 on the y-axis (Cost). This will be halfway between 150 and 200.

2. Draw a horizontal line from the point 175 on the y-axis to the right until it meets the graph line.

3. From the point where the horizontal line meets the graph, draw a vertical line downwards until it meets the x-axis (Weight).

4. Read the value on the x-axis where the vertical line meets it. This value will be 3.5 kg.

So, the weight of sugar that can be bought for $\textsf{₹}175$ is 3.5 kg.

Verification (optional, based on the linear relationship):

The price per kg = $\frac{50 \textsf{₹}}{1 \text{ kg}} = 50$ $\textsf{₹}$/kg.

Cost of 3.5 kg = $3.5 \text{ kg} \times 50 \textsf{₹}/\text{kg} = 175 \textsf{₹}$.

Weight for $\textsf{₹}175 = \frac{175 \textsf{₹}}{50 \textsf{₹}/\text{kg}} = 3.5 \text{ kg}$.

Given:

The temperature of a city at different times of the day is given in the table.

To Draw:

A line graph representing the data with time on the x-axis and temperature on the y-axis.

To Find:

The estimated temperature at 12 noon using the graph.

Solution:

We represent the given data as points (Time, Temperature) to be plotted on the graph. It's helpful to represent time numerically, for instance, using hours from midnight (24-hour format). 6 AM is hour 6, 10 AM is hour 10, 2 PM is hour 14, 6 PM is hour 18, and 10 PM is hour 22.

Steps to draw the line graph:

1. Draw the x-axis (horizontal axis) and label it as "Time (Hours)".

2. Draw the y-axis (vertical axis) and label it as "Temperature ($^\circ\text{C}$)".

3. Choose appropriate scales for both axes. For the x-axis, a scale covering hours from 6 to 22 would be suitable (e.g., 1 unit = 2 hours). For the y-axis, a scale covering temperatures from 20 to 30 would be suitable (e.g., 1 unit = 2$^\circ\text{C}$ or 5$^\circ\text{C}$). Start the y-axis slightly below the lowest temperature (e.g., from 18$^\circ\text{C}$) for better visualization.

4. Plot the points (6, 20), (10, 25), (14, 30), (18, 28), and (22, 22) on the graph sheet according to the chosen scales.

5. Join the plotted points with straight line segments. This forms the line graph.

Using the graph to estimate the temperature at 12 noon:

12 noon corresponds to hour 12 on the x-axis.

1. Locate the point corresponding to hour 12 on the x-axis. This point is exactly halfway between 10 (10 AM) and 14 (2 PM) on the x-axis.

2. From the point representing hour 12 on the x-axis, draw a vertical line upwards until it intersects the line segment connecting the points (10, 25) and (14, 30) on the graph.

3. From the point of intersection on the graph, draw a horizontal line to the left until it meets the y-axis (Temperature).

4. Read the value on the y-axis where the horizontal line meets it. Since 12 is exactly halfway between 10 and 14, the estimated temperature on the straight line segment between (10, 25) and (14, 30) will be halfway between 25$^\circ\text{C}$ and 30$^\circ\text{C}$.

Estimated temperature = $\frac{25 + 30}{2} = \frac{55}{2} = 27.5^\circ\text{C}$.

So, based on the line graph, the estimated temperature at 12 noon is $27.5^\circ\text{C}$.

Given:

The coordinates of the points are E$(-2, 3)$, F$(4, 3)$, G$(4, -1)$, and H$(-2, -1)$.

To Find:

1. The shape formed by joining the points in order.

2. The area of the shape.

Solution:

First, we plot the given points on a graph sheet and join them in order E to F, F to G, G to H, and H to E.

E$(-2, 3)$ is located 2 units left and 3 units up from the origin.

F$(4, 3)$ is located 4 units right and 3 units up from the origin.

G$(4, -1)$ is located 4 units right and 1 unit down from the origin.

H$(-2, -1)$ is located 2 units left and 1 unit down from the origin.

To determine the shape and calculate its area, we find the lengths of the sides.

Length of side EF (connecting E$(-2, 3)$ and F$(4, 3)$):

Since the y-coordinates are the same (both are 3), EF is a horizontal line segment. The length is the absolute difference of the x-coordinates.

EF = $|4 - (-2)| = |4 + 2| = 6$ units.

Length of side FG (connecting F$(4, 3)$ and G$(4, -1)$):

Since the x-coordinates are the same (both are 4), FG is a vertical line segment. The length is the absolute difference of the y-coordinates.

FG = $|-1 - 3| = |-4| = 4$ units.

Length of side GH (connecting G$(4, -1)$ and H$(-2, -1)$):

Since the y-coordinates are the same (both are -1), GH is a horizontal line segment. The length is the absolute difference of the x-coordinates.

GH = $|-2 - 4| = |-6| = 6$ units.

Length of side HE (connecting H$(-2, -1)$ and E$(-2, 3)$):

Since the x-coordinates are the same (both are -2), HE is a vertical line segment. The length is the absolute difference of the y-coordinates.

HE = $|3 - (-1)| = |3 + 1| = 4$ units.

We have EF = 6, FG = 4, GH = 6, HE = 4.

Opposite sides EF and GH are equal (6 units). Opposite sides FG and HE are equal (4 units).

Also, since EF is horizontal and FG is vertical, they are perpendicular, forming a right angle at F. Similarly, all angles are right angles.

A quadrilateral with opposite sides equal and all angles equal to $90^\circ$ is a rectangle.

Thus, the shape formed by joining the points E, F, G, H in order is a rectangle.

Area of the rectangle EFGH:

The area of a rectangle is given by $\text{Length} \times \text{Width}$.

Length = 6 units (e.g., EF or GH)

Width = 4 units (e.g., FG or HE)

Area = $6 \times 4 = 24$ square units.

Summary of findings:

The shape formed is a rectangle.

The area of the rectangle is 24 square units.

Given:

Runs scored by Ram and Shyam in five matches.

To Draw:

A double line graph comparing their performance.

To Find:

Which player had a more consistent performance based on the graph.

Solution:

We need to plot two sets of points on the same graph, one for Ram and one for Shyam, using the Match number on the x-axis and Runs scored on the y-axis.

Points for Ram (Match, Runs): (1, 50), (2, 35), (3, 60), (4, 40), (5, 55)

Points for Shyam (Match, Runs): (1, 40), (2, 50), (3, 30), (4, 65), (5, 45)

Steps to draw the double line graph:

1. Draw the x-axis (horizontal axis) and label it as "Match Number". Use a scale suitable for match numbers 1 through 5 (e.g., 1 unit per match).

2. Draw the y-axis (vertical axis) and label it as "Runs Scored". Choose a scale that covers the range of runs scored by both players (from 30 to 65). A scale starting from 0 or slightly below the minimum score and going up to or slightly above the maximum score would work (e.g., 1 unit = 5 runs or 10 runs).

3. Plot the points for Ram on the graph using one symbol (e.g., dots or crosses) and color (e.g., red).

4. Join the plotted points for Ram with straight line segments. Use a specific line style (e.g., solid red line).

5. Plot the points for Shyam on the same graph using a different symbol and color (e.g., squares or triangles) and color (e.g., blue).

6. Join the plotted points for Shyam with straight line segments using a different line style (e.g., dashed blue line).

7. Provide a legend to distinguish between Ram's line and Shyam's line.

Interpreting consistency from the graph:

Consistency in performance can be judged by how much the scores fluctuate from match to match. On a line graph, this is reflected by the smoothness or jaggedness of the line. A more consistent player will have a line that changes less sharply and stays closer to their average score.

Looking at the plotted points and the lines connecting them:

Ram's scores are 50, 35, 60, 40, 55.

Shyam's scores are 40, 50, 30, 65, 45.

Ram's scores fluctuate between 35 and 60 (a range of 25 runs).

Shyam's scores fluctuate between 30 and 65 (a range of 35 runs).

Visually, the line graph for Ram will likely appear less erratic (smoother) than the line graph for Shyam, which will have steeper rises and drops.

To further quantify, we could calculate the average score for each player and then look at the deviation from the average, but for a simple visual interpretation from a line graph, the range of scores and the visual fluctuation of the line are key.

Conclusion based on the graph:

By observing the line graph, we can see that Ram's score line shows relatively smaller variations compared to Shyam's score line, which exhibits more pronounced peaks and troughs.

Therefore, Ram had a more consistent performance over the five matches compared to Shyam.

Given:

The relationship between simple interest ($I$) and time ($t$) is given by the equation $I = 100t$.

Principal = $\textsf{₹}1000$

Rate of interest = $10\%$ per annum

Note: The formula for simple interest is $I = \frac{P \times R \times T}{100}$. With $P = 1000$ and $R = 10$, we get $I = \frac{1000 \times 10 \times t}{100} = 100t$, which matches the given equation.

To Draw:

A graph of simple interest vs. time.

To Find:

1. The interest after 2.5 years using the graph.

2. The time required to get $\textsf{₹}300$ as interest using the graph.

Solution:

The equation relating simple interest ($I$) and time ($t$) is $I = 100t$. To draw the graph, we need to find some points $(t, I)$ that satisfy this equation. We will take time ($t$) on the x-axis and simple interest ($I$) on the y-axis.

Let's calculate $I$ for a few values of $t$:

Steps to draw the graph:

1. Draw the x-axis (horizontal axis) and label it as "Time (years)".

2. Draw the y-axis (vertical axis) and label it as "Simple Interest ($\textsf{₹}$)".

3. Choose appropriate scales for both axes. For the x-axis, 1 unit = 1 year is suitable. For the y-axis, a scale of 1 unit = $\textsf{₹}100$ (or $\textsf{₹}50$) would be appropriate to accommodate the values.

4. Plot the points (0, 0), (1, 100), (2, 200), (3, 300), (4, 400) on the graph sheet according to the chosen scales.

5. Join the plotted points with a straight line. The relationship $I = 100t$ is linear, so the graph will be a straight line passing through the origin.

Using the graph to find the interest after 2.5 years:

1. Locate 2.5 on the x-axis (Time). This is halfway between 2 and 3.

2. Draw a vertical line from the point 2.5 on the x-axis upwards until it meets the graph line.

3. From the point where the vertical line meets the graph, draw a horizontal line to the left until it meets the y-axis (Simple Interest).

4. Read the value on the y-axis where the horizontal line meets it. This value should correspond to $\textsf{₹}250$.

So, the interest after 2.5 years is $\textsf{₹}250$.

Using the graph to find the time required to get $\textsf{₹}300$ as interest:

1. Locate 300 on the y-axis (Simple Interest).

2. Draw a horizontal line from the point 300 on the y-axis to the right until it meets the graph line.

3. From the point where the horizontal line meets the graph, draw a vertical line downwards until it meets the x-axis (Time).

4. Read the value on the x-axis where the vertical line meets it. This value should correspond to 3 years.

So, the time required to get $\textsf{₹}300$ as interest is 3 years.

Given:

The relationship between the perimeter ($P$) of a rectangle with a fixed breadth of 5 units and its length ($L$) is given by $P = 2L + 10$.

To Determine:

If the relationship between perimeter and length is linear.

To Draw:

A graph of perimeter vs. length for $L = 1, 2, 3, 4, 5$ units.

To Plot:

The points $(L, P)$.

Solution:

The given relationship is $P = 2L + 10$.

A linear relationship between two variables (say, $x$ and $y$) is represented by an equation of the form $y = mx + c$, where $m$ and $c$ are constants.

Comparing $P = 2L + 10$ with $y = mx + c$, we can see that:

- The dependent variable $P$ corresponds to $y$.

- The independent variable $L$ corresponds to $x$.

- The constant $m = 2$.

- The constant $c = 10$.

Since the equation $P = 2L + 10$ is in the form of a linear equation $y = mx + c$, and the variable $L$ is raised to the power of 1, the relationship between the perimeter and the length is indeed a **linear relationship**.

To draw the graph, we calculate the perimeter $P$ for the given values of length $L = 1, 2, 3, 4, 5$ units using the formula $P = 2L + 10$.

Steps to draw the graph:

1. Draw the x-axis (horizontal axis) and label it as "Length $L$ (units)".

2. Draw the y-axis (vertical axis) and label it as "Perimeter $P$ (units)".

3. Choose appropriate scales for both axes. For the x-axis, a scale of 1 unit = 1 unit of length is suitable. For the y-axis, a scale that accommodates values from 12 to 20 (e.g., starting the y-axis from 10 and using a scale of 1 unit = 2 units of perimeter) would be appropriate.

4. Plot the points (1, 12), (2, 14), (3, 16), (4, 18), and (5, 20) on the graph sheet according to the chosen scales.

When plotted, these points will lie on a straight line, which is consistent with the linear relationship we identified.

Solution:

The importance of choosing an appropriate scale while drawing a graph cannot be overstated. The scale determines how the data is visually represented on the graph paper. A poorly chosen scale can lead to a misleading or unreadable graph.

Importance of Choosing an Appropriate Scale:

1. Clarity and Readability: An appropriate scale ensures that the graph is easy to read and understand. The points should be spread out enough to be distinguishable, and the labels on the axes should be clear.

2. Accurate Representation: The scale must accurately reflect the proportions and variations in the data. If the scale is too large, small variations might appear insignificant. If the scale is too small, large variations might seem compressed.

3. Preventing Distortion: A scale that is too large or too small can distort the visual interpretation of the data, making trends or patterns seem more or less prominent than they are in reality.

4. Utilizing Space: The scale should be chosen so that the graph occupies a significant portion of the available graph paper, but without going off the page. This makes the graph informative and easy to analyze.

Factors to Consider When Selecting a Scale for the Axes:

1. Range of Data: Determine the minimum and maximum values for the data that will be plotted on each axis. The scale must span at least this entire range.

2. Size of the Graph Paper: Consider the physical size of the graph sheet available. A larger range of data might require a smaller scale (more units per grid division) to fit on the paper, while a smaller range allows for a larger scale (fewer units per grid division) for better detail.

3. Interval Size: Choose an interval size for the grid divisions that is easy to work with (e.g., 1, 2, 5, 10, 50, 100, etc.). Avoid awkward intervals that make plotting or reading intermediate values difficult.

4. Starting Point (Origin): Decide whether the scale should start from zero. Starting from zero is standard practice and avoids distorting the perception of relative values. However, if the data values are all large and clustered within a narrow range far from zero, starting the axis at a value slightly below the minimum data point (using a 'break' on the axis if necessary) might be appropriate to utilize the graph space effectively and show variations more clearly, but this should be done cautiously to avoid misinterpretation.

5. Ease of Plotting and Reading: The chosen scale should make it straightforward to plot the given data points accurately and to read values from the graph, including estimating values between plotted points (interpolation) or beyond the plotted points (extrapolation).

By carefully considering these factors, one can select a scale that results in a clear, accurate, and easily interpretable graph.

Given:

The coordinates of the points are A$(-3, 0)$, B$(0, 3)$, C$(3, 0)$, and D$(0, -3)$.

To Find:

1. The shape formed by joining the points in order.

2. Whether the shape is a square and justification.

Solution:

First, we plot the given points on a graph sheet and join them in order A to B, B to C, C to D, and D to A.

A$(-3, 0)$ is located 3 units left from the origin on the x-axis.

B$(0, 3)$ is located 3 units up from the origin on the y-axis.

C$(3, 0)$ is located 3 units right from the origin on the x-axis.

D$(0, -3)$ is located 3 units down from the origin on the y-axis.

When these points are joined, a four-sided figure (quadrilateral) ABCD is formed.

To determine the exact type of quadrilateral and whether it is a square, we calculate the lengths of the sides and the diagonals using the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

Length of side AB (connecting A$(-3, 0)$ and B$(0, 3)$):

$AB = \sqrt{(0 - (-3))^2 + (3 - 0)^2} = \sqrt{(0+3)^2 + 3^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Length of side BC (connecting B$(0, 3)$ and C$(3, 0)$):

$BC = \sqrt{(3 - 0)^2 + (0 - 3)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Length of side CD (connecting C$(3, 0)$ and D$(0, -3)$):

$CD = \sqrt{(0 - 3)^2 + (-3 - 0)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Length of side DA (connecting D$(0, -3)$ and A$(-3, 0)$):

$DA = \sqrt{(-3 - 0)^2 + (0 - (-3))^2} = \sqrt{(-3)^2 + (0+3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Since AB = BC = CD = DA = $3\sqrt{2}$ units, all four sides are equal in length. This indicates that the quadrilateral is either a rhombus or a square.

Now, let's find the lengths of the diagonals.

Length of diagonal AC (connecting A$(-3, 0)$ and C$(3, 0)$):

$AC = \sqrt{(3 - (-3))^2 + (0 - 0)^2} = \sqrt{(3+3)^2 + 0^2} = \sqrt{6^2 + 0} = \sqrt{36} = 6$ units.

Length of diagonal BD (connecting B$(0, 3)$ and D$(0, -3)$):

$BD = \sqrt{(0 - 0)^2 + (-3 - 3)^2} = \sqrt{0^2 + (-6)^2} = \sqrt{0 + 36} = \sqrt{36} = 6$ units.

Since AC = BD = 6 units, the lengths of the diagonals are equal.

A quadrilateral with all four sides equal is a rhombus.

A rhombus with equal diagonals is a square.

Alternatively, we can check if adjacent sides are perpendicular by calculating their slopes. The product of the slopes of two perpendicular lines is -1 (provided neither is vertical).

Slope of AB = $\frac{3-0}{0-(-3)} = \frac{3}{3} = 1$

Slope of BC = $\frac{0-3}{3-0} = \frac{-3}{3} = -1$

The product of the slopes of AB and BC is $1 \times (-1) = -1$. This confirms that AB is perpendicular to BC, meaning $\angle$ABC is a right angle ($90^\circ$).

A rhombus with a right angle is a square.

The shape formed by joining the points A, B, C, and D in order is a square.

Yes, it is a square.

Justification: The shape is a square because all four sides are equal in length ($3\sqrt{2}$ units) and the two diagonals are equal in length (6 units). Alternatively, it is a square because all four sides are equal and one (hence all) internal angle is a right angle ($90^\circ$), as shown by the slopes of adjacent sides.

Given:

Cost of rice at different weights for two shops, Shop A and Shop B.

To Draw:

A double line graph representing the cost vs. weight for both shops on the same sheet.

To Find:

Which shop offers a better price for 5 kg of rice using the graph.

Solution:

We will take the Weight of Sugar (kg) on the x-axis and Cost ($\textsf{₹}$) on the y-axis. From the given data, we list the points (Weight, Cost) for each shop:

For Shop A:

For Shop B:

Steps to draw the double line graph:

1. Draw the x-axis (horizontal axis) and label it as "Weight of Rice (kg)".

2. Draw the y-axis (vertical axis) and label it as "Cost ($\textsf{₹}$)".

3. Choose appropriate scales for both axes. For the x-axis, a scale of 1 unit = 1 kg is suitable. For the y-axis, a scale that covers costs up to at least $\textsf{₹}150$ (and potentially up to $\textsf{₹}250$ for 5 kg estimation) would be appropriate (e.g., 1 unit = $\textsf{₹}25$ or $\textsf{₹}50$). Let's assume a scale where major grid lines represent 1 unit on the x-axis and $\textsf{₹}50$ on the y-axis, with subdivisions for $\textsf{₹}25$. Both relationships appear to be linear and pass through the origin $(0,0)$, so we can start the axes at 0.

4. Plot the points for Shop A: (1, 50), (2, 100), (3, 150). Join these points with a straight line. Label this line "Shop A".

5. Plot the points for Shop B: (1, 45), (2, 90), (3, 135). Join these points with a straight line. Label this line "Shop B". Note that 45, 90, and 135 might fall between major grid lines depending on the y-axis scale chosen.

Using the graph to determine the cost for 5 kg of rice:

1. Extend both lines (Shop A and Shop B) on the graph to the right.

2. Locate 5 on the x-axis (Weight).

3. Draw a vertical line from the point 5 on the x-axis upwards until it intersects the line representing Shop A and the line representing Shop B.

4. From each point of intersection on the graph lines, draw a horizontal line to the left until it meets the y-axis (Cost).

5. Read the values on the y-axis where the horizontal lines meet it.

From the graph, the cost for 5 kg of rice is estimated as follows:

For Shop A: The vertical line from $x=5$ intersects the Shop A line at a y-value of $\textsf{₹}250$.

For Shop B: The vertical line from $x=5$ intersects the Shop B line at a y-value of $\textsf{₹}225$.

Comparing the costs:

The cost of 5 kg of rice at Shop A is $\textsf{₹}250$.

The cost of 5 kg of rice at Shop B is $\textsf{₹}225$.

Since $\textsf{₹}225 < \textsf{₹}250$, Shop B offers a lower price for 5 kg of rice.

Therefore, Shop B offers a better price for 5 kg of rice.

Note: Shop A charges $\textsf{₹}50$ per kg ($50/1 = 100/2 = 150/3 = 50$). Shop B charges $\textsf{₹}45$ per kg ($45/1 = 90/2 = 135/3 = 45$). Since $\textsf{₹}45 < \textsf{₹}50$, Shop B has a lower price per kg, which confirms it offers a better price for any weight.