Chapter 3 Coordinate Geometry (Additional Questions)

To determine the quadrant in which a point $(x, y)$ lies, we examine the signs of its coordinates.

The given point is $(2, -3)$.

Here, the x-coordinate is $x = 2$. Since $2 > 0$, the x-coordinate is positive.

The y-coordinate is $y = -3$. Since $-3 < 0$, the y-coordinate is negative.

The four quadrants are defined by the signs of the coordinates as follows:

First Quadrant: $x > 0$ and $y > 0$ (Signs: $+, +$)

Second Quadrant: $x < 0$ and $y > 0$ (Signs: $-, +$)

Third Quadrant: $x < 0$ and $y < 0$ (Signs: $-, -$)

Fourth Quadrant: $x > 0$ and $y < 0$ (Signs: $+, -$)

For the point $(2, -3)$, the signs of the coordinates are $(+, -)$.

Comparing this with the quadrant definitions, we see that a point with positive x-coordinate and negative y-coordinate lies in the Fourth Quadrant.

Therefore, the point $(2, -3)$ lies in the Fourth Quadrant.

The correct option is (D) Fourth Quadrant.

In the Cartesian coordinate system, the origin is the point where the horizontal axis (x-axis) and the vertical axis (y-axis) intersect.

At the origin, the value of both the x-coordinate and the y-coordinate is zero.

Therefore, the coordinates of the origin are $(0, 0)$.

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

In the Cartesian coordinate system, a point is represented by an ordered pair $(x, y)$, where $x$ is the coordinate along the x-axis and $y$ is the coordinate along the y-axis.

The abscissa of a point is its coordinate along the horizontal axis, which is the x-axis.

The ordinate of a point is its coordinate along the vertical axis, which is the y-axis.

For a point $(x, y)$, $x$ is the abscissa and $y$ is the ordinate.

Therefore, the abscissa of a point is its x coordinate.

The correct option is (B) x.

We are given a point with an ordinate of 5 and it lies on the y-axis.

The ordinate of a point is its y-coordinate.

So, the y-coordinate of the point is 5.

A point that lies on the y-axis always has its x-coordinate equal to 0.

Therefore, the x-coordinate of the point is 0.

Combining the x-coordinate and the y-coordinate, the coordinates of the point are $(x, y) = (0, 5)$.

The coordinates of the point are $(0, 5)$.

The correct option is (B) $(0, 5)$.

In a Cartesian coordinate system, the abscissa refers to the x-coordinate of a point, and the ordinate refers to the y-coordinate of a point.

The four quadrants are defined by the signs of the x and y coordinates:

First Quadrant: x-coordinate is positive, y-coordinate is positive ($+, +$)

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

Third Quadrant: x-coordinate is negative, y-coordinate is negative ($-, -$)

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

We are asked about the signs in the second quadrant.

In the second quadrant, the x-coordinate (abscissa) is negative, and the y-coordinate (ordinate) is positive.

Therefore, the signs of the abscissa and ordinate respectively are $(-, +)$.

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

A point lies on the x-axis if and only if its y-coordinate is 0.

Let's examine the y-coordinate of each given point:

(A) For the point $(0, -7)$, the y-coordinate is $-7$. Since $-7 \neq 0$, this point does not lie on the x-axis.

(B) For the point $(3, 0)$, the y-coordinate is $0$. Since $0 = 0$, this point lies on the x-axis.

(C) For the point $(-5, 2)$, the y-coordinate is $2$. Since $2 \neq 0$, this point does not lie on the x-axis.

(D) For the point $(0, 0)$, the y-coordinate is $0$. Since $0 = 0$, this point lies on the x-axis. This point is the origin, which lies on both axes.

We are looking for a point that lies on the x-axis. Both $(3, 0)$ and $(0, 0)$ satisfy this condition.

However, option (B) $(3, 0)$ represents a point on the x-axis that is distinct from the origin, clearly demonstrating the condition $y=0$. Option (D) $(0, 0)$ is the origin, which is also on the x-axis, but typically in such questions, a non-origin point specifically on the axis is expected unless $(0,0)$ is the only choice satisfying the condition.

Considering the standard representation of points on axes, a point on the x-axis has the form $(x, 0)$ where $x$ can be any real number. Option (B) fits this form with $x=3$. Option (D) fits this form with $x=0$. Both are technically on the x-axis.

Given the options, $(3, 0)$ is a clear example of a point on the x-axis (where $y=0$) that is not the origin, making it a direct test of the concept.

The point whose y-coordinate is 0 lies on the x-axis. From the given options, the point $(3, 0)$ has a y-coordinate of 0.

The correct option is (B) $(3, 0)$.

In the Cartesian coordinate system, for a point $(x, y)$, the abscissa is the x-coordinate and the ordinate is the y-coordinate.

The given point is $(6, -1)$.

Here, the x-coordinate is $6$ and the y-coordinate is $-1$.

The ordinate is the y-coordinate of the point.

Therefore, the ordinate of the point $(6, -1)$ is $-1$.

The correct option is (B) -1.

Let the coordinates of the point be $(x, y)$.

The distance of a point $(x, y)$ from the x-axis is given by the absolute value of its y-coordinate, i.e., $|y|$.

The distance of a point $(x, y)$ from the y-axis is given by the absolute value of its x-coordinate, i.e., $|x|$.

We are given that the point is at a distance of 4 units from the x-axis.

So, $|y| = 4$. This implies $y = 4$ or $y = -4$.

We are also given that the point is at a distance of 3 units from the y-axis.

So, $|x| = 3$. This implies $x = 3$ or $x = -3$.

We are told that the point lies in the first quadrant.

In the first quadrant, both the x-coordinate and the y-coordinate are positive ($x > 0$ and $y > 0$).

Considering the possible values of $x$ and $y$ and the condition for the first quadrant:

Since $x > 0$ and $|x| = 3$, we must have $x = 3$.

Since $y > 0$ and $|y| = 4$, we must have $y = 4$.

Therefore, the coordinates of the point are $(3, 4)$.

The coordinates of the point are $(3, 4)$.

The correct option is (B) $(3, 4)$.

The Cartesian plane, also known as the coordinate plane, is formed by two perpendicular number lines: the horizontal x-axis and the vertical y-axis.

These two axes intersect at the origin $(0, 0)$.

The axes divide the plane into four regions.

These four regions are called quadrants.

The quadrants are numbered counterclockwise, starting from the upper right region as the First Quadrant, followed by the Second, Third, and Fourth Quadrants.

Therefore, there are 4 quadrants in the Cartesian plane.

The correct option is (D) 4.

Let's examine each statement:

Statement (A): "The x-axis and y-axis are perpendicular to each other."

In a standard Cartesian coordinate system, the x-axis and y-axis intersect at a right angle (90 degrees). This statement is True.

Statement (B): "The point $(-2, -5)$ lies in the third quadrant."

A point $(x, y)$ lies in the third quadrant if $x < 0$ and $y < 0$. For the point $(-2, -5)$, $x = -2$ (which is negative) and $y = -5$ (which is negative). The signs are $(-, -)$. This matches the condition for the third quadrant. This statement is True.

Statement (C): "The point $(4, 0)$ lies on the y-axis."

A point lies on the y-axis if its x-coordinate is 0. For the point $(4, 0)$, the x-coordinate is $4$. Since $4 \neq 0$, the point $(4, 0)$ does not lie on the y-axis.

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

Statement (D): "The coordinates of the origin are $(0, 0)$."

The origin is the point where the x-axis and y-axis intersect. At this point, both coordinates are zero. This statement is True.

We are looking for the statement that is FALSE.

Statement (C) is false.

The correct option is (C) The point $(4, 0)$ lies on the y-axis.

Let the given point be $(x, y)$.

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

A point lies on the x-axis if its y-coordinate is 0.

In this case, the y-coordinate is $-5$, which is not 0. So, the point does not lie on the x-axis (unless it is the origin, which has $y=0$).

A point lies on the y-axis if its x-coordinate is 0.

In this case, the x-coordinate is $0$. Since the x-coordinate is 0, the point $(0, -5)$ lies on the y-axis.

The point $(0, -5)$ has its x-coordinate equal to zero and its y-coordinate not equal to zero. This is the characteristic of a point lying on the y-axis but not being the origin (unless the y-coordinate was also 0, which it is not here).

Therefore, the point $(0, -5)$ lies on the y-axis.

The correct option is (B) y.

Let the coordinates of the point be $(x, y)$.

The point lies on the x-axis. Any point on the x-axis has its y-coordinate equal to 0.

So, $y = 0$. The point has coordinates of the form $(x, 0)$.

The point lies on the negative direction of the x-axis. This means the x-coordinate is negative, i.e., $x < 0$.

The distance of a point $(x, y)$ from the origin $(0, 0)$ is given by $\sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$.

We are given that the distance from the origin is 5 units.

So, $\sqrt{x^2 + y^2} = 5$.

Since $y = 0$, we have $\sqrt{x^2 + 0^2} = 5$, which simplifies to $\sqrt{x^2} = 5$.

The value of $\sqrt{x^2}$ is $|x|$. So, $|x| = 5$.

This means $x = 5$ or $x = -5$.

However, we know that the point is on the negative direction of the x-axis, so $x < 0$.

Therefore, we must choose $x = -5$.

With $x = -5$ and $y = 0$, the coordinates of the point are $(-5, 0)$.

The coordinates of the point are $(-5, 0)$.

The correct option is (B) $(-5, 0)$.

In the Cartesian coordinate system, the four quadrants are defined by the signs of the x and y coordinates of a point $(x, y)$.

The quadrants are as follows:

First Quadrant: $x > 0$ and $y > 0$ (positive x, positive y)

Second Quadrant: $x < 0$ and $y > 0$ (negative x, positive y)

Third Quadrant: $x < 0$ and $y < 0$ (negative x, negative y)

Fourth Quadrant: $x > 0$ and $y < 0$ (positive x, negative y)

We are given a point $(h, k)$. Here, $h$ is the x-coordinate and $k$ is the y-coordinate.

For the point $(h, k)$ to lie in the fourth quadrant, the x-coordinate must be positive and the y-coordinate must be negative.

So, we must have $h > 0$ and $k < 0$.

Let's check the given options:

(A) $h > 0, k > 0$: This corresponds to the First Quadrant.

(B) $h < 0, k > 0$: This corresponds to the Second Quadrant.

(C) $h < 0, k < 0$: This corresponds to the Third Quadrant.

(D) $h > 0, k < 0$: This corresponds to the Fourth Quadrant.

Thus, the point $(h, k)$ lies in the fourth quadrant if $h > 0$ and $k < 0$.

The correct option is (D) $h > 0, k < 0$.

To determine the quadrant of a point $(x, y)$, we examine the signs of its coordinates.

The given point is $(-10, 2)$.

Here, the x-coordinate is $x = -10$. Since $-10 < 0$, the x-coordinate is negative.

The y-coordinate is $y = 2$. Since $2 > 0$, the y-coordinate is positive.

The quadrants are defined by the signs of $(x, y)$ as follows:

First Quadrant (Q1): $(+, +)$

Second Quadrant (Q2): $(-, +)$

Third Quadrant (Q3): $(-, -)$

Fourth Quadrant (Q4): $(+, -)$

For the point $(-10, 2)$, the signs of the coordinates are $(-, +)$.

This matches the sign convention for the Second Quadrant (Q2).

Therefore, the point $(-10, 2)$ lies in the Second Quadrant.

The correct option is (B) Q2.

Let's analyze the given Assertion (A) and Reason (R).

Assertion (A): The point $(0, 8)$ lies on the y-axis.

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

The given point is $(0, 8)$. The x-coordinate is $0$ and the y-coordinate is $8$.

Since the x-coordinate is $0$, the point $(0, 8)$ indeed lies on the y-axis.

Thus, Assertion (A) is True.

Reason (R): A point on the y-axis has its x-coordinate equal to 0.

This is the definition of a point lying on the y-axis in the Cartesian coordinate system.

Thus, Reason (R) is True.

Now let's check if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that the point $(0, 8)$ is on the y-axis. Reason (R) states the condition for a point to be on the y-axis, which is having an x-coordinate of $0$.

The point $(0, 8)$ satisfies the condition stated in Reason (R) because its x-coordinate is $0$. This condition is precisely why the point $(0, 8)$ lies on the y-axis.

Therefore, Reason (R) is the correct explanation for Assertion (A).

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

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

Let's analyze the given Assertion (A) and Reason (R).

Assertion (A): The point $( -4, -6)$ lies in the second quadrant.

To determine the quadrant of a point $(x, y)$, we look at the signs of $x$ and $y$.

For the point $(-4, -6)$, the x-coordinate is $-4$ and the y-coordinate is $-6$.

Since $-4 < 0$ and $-6 < 0$, both coordinates are negative.

The signs are $(-, -)$.

A point with signs $(-, -)$ lies in the third quadrant.

Therefore, the point $(-4, -6)$ lies in the third quadrant, not the second quadrant.

Thus, Assertion (A) is False.

Reason (R): In the second quadrant, the x-coordinate is negative and the y-coordinate is positive.

The second quadrant is defined by the conditions $x < 0$ and $y > 0$. This means the x-coordinate is negative and the y-coordinate is positive.

Thus, Reason (R) is True.

We have found that Assertion (A) is False and Reason (R) is True.

Looking at 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)

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

To match the points with their locations, we need to determine which quadrant each point lies in, or if it is the origin.

The location of a point $(x, y)$ depends on the signs of its coordinates:

First Quadrant: $x > 0$, $y > 0$ (Signs: $+, +$)

Second Quadrant: $x < 0$, $y > 0$ (Signs: $-, +$)

Third Quadrant: $x < 0$, $y < 0$ (Signs: $-, -$)

Fourth Quadrant: $x > 0$, $y < 0$ (Signs: $+, -$)

Origin: $x = 0$, $y = 0$ (Point: $(0, 0)$)

Let's examine each point:

(i) Point $(5, -1)$:

Here, $x = 5$ which is positive ($5 > 0$).

Here, $y = -1$ which is negative ($-1 < 0$).

The signs are $(+, -)$. This matches the condition for the Fourth Quadrant.

So, (i) matches (b).

(ii) Point $(-3, 4)$:

Here, $x = -3$ which is negative ($-3 < 0$).

Here, $y = 4$ which is positive ($4 > 0$).

The signs are $(-, +)$. This matches the condition for the Second Quadrant.

So, (ii) matches (a).

(iii) Point $(0, 0)$:

Here, $x = 0$ and $y = 0$.

This point is the Origin.

So, (iii) matches (d).

(iv) Point $(-6, -2)$:

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

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

The signs are $(-, -)$. This matches the condition for the Third Quadrant.

So, (iv) matches (c).

The correct matches are:

(i) - (b)

(ii) - (a)

(iii) - (d)

(iv) - (c)

Comparing this with the given options, we find that Option (B) matches our pairings.

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

Given:

Four points (landmarks) in the Cartesian plane: P(3, 2), Q(-4, 1), R(-2, -3), and S(5, -1).

To Find:

Which two landmarks are located in quadrants with the same sign pattern for their coordinates.

Solution:

The location of a point $(x, y)$ in the Cartesian plane is determined by the signs of its coordinates. The four quadrants have the following sign patterns:

First Quadrant (Q1): $x > 0$, $y > 0$ (Sign pattern: $(+, +)$)

Second Quadrant (Q2): $x < 0$, $y > 0$ (Sign pattern: $(-, +)$)

Third Quadrant (Q3): $x < 0$, $y < 0$ (Sign pattern: $(-, -)$)

Fourth Quadrant (Q4): $x > 0$, $y < 0$ (Sign pattern: $(+, -)$)

A point is in a specific quadrant if its coordinates match the corresponding sign pattern.

Let's determine the sign pattern and quadrant for each given point:

For point P(3, 2):

$x = 3$. Since $3 > 0$, the x-coordinate is positive ($+$).

$y = 2$. Since $2 > 0$, the y-coordinate is positive ($+$).

The sign pattern for P is $(+, +)$. This corresponds to the First Quadrant (Q1).

For point Q(-4, 1):

$x = -4$. Since $-4 < 0$, the x-coordinate is negative ($-$).

$y = 1$. Since $1 > 0$, the y-coordinate is positive ($+$).

The sign pattern for Q is $(-, +)$. This corresponds to the Second Quadrant (Q2).

For point R(-2, -3):

$x = -2$. Since $-2 < 0$, the x-coordinate is negative ($-$).

$y = -3$. Since $-3 < 0$, the y-coordinate is negative ($-$).

The sign pattern for R is $(-, -)$. This corresponds to the Third Quadrant (Q3).

For point S(5, -1):

$x = 5$. Since $5 > 0$, the x-coordinate is positive ($+$).

$y = -1$. Since $-1 < 0$, the y-coordinate is negative ($-$).

The sign pattern for S is $(+, -)$. This corresponds to the Fourth Quadrant (Q4).

Summary of points and their sign patterns/quadrants:

P(3, 2): $(+, +)$ in Q1

Q(-4, 1): $(-, +)$ in Q2

R(-2, -3): $(-, -)$ in Q3

S(5, -1): $(+, -)$ in Q4

We are looking for two landmarks that have the same sign pattern for their coordinates, which means they are in the same quadrant.

Let's check the pairs given in the options:

(A) P and Q: Sign patterns are $(+, +)$ and $(-, +)$. These are different.

(B) Q and R: Sign patterns are $(-, +)$ and $(-, -)$. These are different.

(C) R and S: Sign patterns are $(-, -)$ and $(+, -)$. These are different.

(D) P and S: Sign patterns are $(+, +)$ and $(+, -)$. These are different.

Based on the given points and the standard definition of quadrant sign patterns, none of the pairs of points listed in the options are located in the same quadrant or have the same sign pattern.

Given the structure of a multiple-choice question, there might be an error in the question's coordinates or the options provided, as typically one option should be correct. However, based strictly on the input, no two landmarks from the given options have the same sign pattern for their coordinates.

In the Cartesian coordinate system, two perpendicular lines are used to locate points in a plane.

The horizontal line is called the x-axis.

The vertical line is called the y-axis.

The point where the x-axis and y-axis intersect is called the origin.

The regions into which the plane is divided by the axes are called quadrants.

Therefore, the name of the horizontal line in the Cartesian system is the x-axis.

The correct option is (B) X-axis.

Given:

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

To Find:

The distance of the point $(-3, 4)$ from the y-axis.

Solution:

Let a point in the Cartesian plane be $(x, y)$.

The distance of the point $(x, y)$ from the y-axis is given by the absolute value of its x-coordinate, which is $|x|$.

The distance of the point $(x, y)$ from the x-axis is given by the absolute value of its y-coordinate, which is $|y|$.

For the given point $(-3, 4)$:

The x-coordinate is $x = -3$.

The y-coordinate is $y = 4$.

We need to find the distance from the y-axis.

Distance from y-axis $= |x|$

Distance from y-axis $= |-3|$

The absolute value of $-3$ is $3$.

Distance from y-axis $= 3$ units.

The distance of the point $(-3, 4)$ from the y-axis is 3 units.

The correct option is (C) $3$ units.

To determine which point lies on the y-axis, we need to recall the property of points located on the y-axis in the Cartesian coordinate system.

A point lies on the y-axis if and only if its x-coordinate is 0.

Let's examine the x-coordinate of each point given in the options:

(A) Point $(5, 0)$: The x-coordinate is $5$. Since $5 \neq 0$, this point does not lie on the y-axis. (This point lies on the x-axis).

(B) Point $(-8, 0)$: The x-coordinate is $-8$. Since $-8 \neq 0$, this point does not lie on the y-axis. (This point lies on the x-axis).

(C) Point $(0, 6)$: The x-coordinate is $0$. Since the x-coordinate is $0$, this point lies on the y-axis.

(D) Point $(2, 3)$: The x-coordinate is $2$. Since $2 \neq 0$, this point does not lie on the y-axis. (This point lies in the First Quadrant).

Based on the condition that the x-coordinate must be 0 for a point to lie on the y-axis, the point $(0, 6)$ is the only one that satisfies this condition among the given options.

The correct option is (C) $(0, 6)$.

To determine the quadrant of a point $(x, y)$, we examine the signs of its coordinates.

The given point is $(-5, -9)$.

Here, the x-coordinate is $x = -5$. Since $-5 < 0$, the x-coordinate is negative.

The y-coordinate is $y = -9$. Since $-9 < 0$, the y-coordinate is negative.

The quadrants are defined by the signs of $(x, y)$ as follows:

First Quadrant (Q1): $(+, +)$

Second Quadrant (Q2): $(-, +)$

Third Quadrant (Q3): $(-, -)$

Fourth Quadrant (Q4): $(+, -)$

For the point $(-5, -9)$, the signs of the coordinates are $(-, -)$.

This matches the sign convention for the Third Quadrant (Q3).

Therefore, the point $(-5, -9)$ lies in the Third Quadrant.

The correct option is (C) Q3.

In the Cartesian coordinate system, for a point $(x, y)$, the abscissa is the x-coordinate and the ordinate is the y-coordinate.

The x-axis is the horizontal line in the coordinate plane.

Any point that lies on the x-axis has its vertical distance from the x-axis equal to zero.

The y-coordinate of a point represents its vertical distance from the x-axis (with sign indicating direction).

Therefore, for any point on the x-axis, the y-coordinate is always 0.

The ordinate of a point is its y-coordinate.

Thus, if a point lies on the x-axis, its ordinate is always 0.

Let's consider the options:

(A) 1: This is not always true. For example, the point $(2, 0)$ is on the x-axis, and its ordinate is 0, not 1.

(B) 0: This is always true for any point on the x-axis.

(C) Equal to its abscissa: This is only true for points on the line $y=x$, and the only such point on the x-axis is the origin $(0, 0)$. For other points on the x-axis, like $(5, 0)$, the abscissa is 5 and the ordinate is 0, which are not equal.

(D) Positive: This is not always true. For example, the point $(-3, 0)$ is on the x-axis, and its ordinate is 0, which is neither positive nor negative. Even if the ordinate was allowed to be non-zero (which it is not on the x-axis), points could be on the negative y-axis direction (which would be negative, but those are not on the x-axis anyway). The y-coordinate on the x-axis is strictly 0.

The correct statement is that if a point lies on the x-axis, its ordinate is always 0.

The correct option is (B) 0.

Given:

A point lies on the positive y-axis.

The distance of the point from the origin is 3 units.

To Find:

The coordinates of the point.

Solution:

A point lying on the y-axis has its x-coordinate equal to 0. Therefore, the coordinates of the point are of the form $(0, y)$.

The point is on the positive y-axis. This means that the y-coordinate must be positive, i.e., $y > 0$.

The distance of a point $(x, y)$ from the origin $(0, 0)$ is given by the formula $\sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$.

For a point $(0, y)$ on the y-axis, the distance from the origin is $\sqrt{0^2 + y^2} = \sqrt{y^2} = |y|$.

We are given that the distance from the origin is 3 units.

So, $|y| = 3$. This means $y = 3$ or $y = -3$.

Since the point is on the positive y-axis, the y-coordinate must be positive.

Thus, we must have $y = 3$.

The x-coordinate is $0$ and the y-coordinate is $3$.

The coordinates of the point are $(0, 3)$.

The coordinates of the point are $(0, 3)$.

The correct option is (B) $(0, 3)$.

In the Cartesian coordinate system, the x-axis and the y-axis are two perpendicular lines.

The point where these two axes cross each other is the reference point of the coordinate system.

This point has coordinates $(0, 0)$, meaning the x-coordinate is 0 and the y-coordinate is 0 at this location.

Let's consider the terms in the options:

Abscissa: This is the x-coordinate of a point.

Ordinate: This is the y-coordinate of a point.

Quadrant: This is one of the four regions that the plane is divided into by the x and y axes.

Origin: This is the point where the x-axis and y-axis intersect.

Based on the definitions, the point where the x-axis and y-axis intersect is called the Origin.

The correct option is (D) Origin.

Given:

A list of points: $(2, 5)$, $(-1, 3)$, $(4, -2)$, $(\sqrt{2}, \sqrt{3})$, and $(0, 6)$.

To Find:

Which of the given points lie in the first quadrant.

Solution:

A point $(x, y)$ lies in the first quadrant if and only if its x-coordinate is positive and its y-coordinate is positive. Mathematically, this means $x > 0$ and $y > 0$.

Let's examine each point:

(A) Point $(2, 5)$:

The x-coordinate is $x = 2$. Since $2 > 0$, the x-coordinate is positive.

The y-coordinate is $y = 5$. Since $5 > 0$, the y-coordinate is positive.

Both coordinates are positive. Thus, the point $(2, 5)$ lies in the first quadrant.

(B) Point $(-1, 3)$:

The x-coordinate is $x = -1$. Since $-1 < 0$, the x-coordinate is negative.

The y-coordinate is $y = 3$. Since $3 > 0$, the y-coordinate is positive.

The x-coordinate is negative and the y-coordinate is positive. This point lies in the second quadrant.

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

The x-coordinate is $x = 4$. Since $4 > 0$, the x-coordinate is positive.

The y-coordinate is $y = -2$. Since $-2 < 0$, the y-coordinate is negative.

The x-coordinate is positive and the y-coordinate is negative. This point lies in the fourth quadrant.

(D) Point $(\sqrt{2}, \sqrt{3})$:

The x-coordinate is $x = \sqrt{2}$. Since $\sqrt{2} \approx 1.414 > 0$, the x-coordinate is positive.

The y-coordinate is $y = \sqrt{3}$. Since $\sqrt{3} \approx 1.732 > 0$, the y-coordinate is positive.

Both coordinates are positive. Thus, the point $(\sqrt{2}, \sqrt{3})$ lies in the first quadrant.

(E) Point $(0, 6)$:

The x-coordinate is $x = 0$.

The y-coordinate is $y = 6$. Since $6 \neq 0$, the y-coordinate is non-zero.

A point with an x-coordinate of 0 lies on the y-axis. Since the y-coordinate is positive ($6 > 0$), this point lies on the positive y-axis. Points on the axes are not considered to be in any quadrant.

From the analysis, the points that lie in the first quadrant are those where both coordinates are positive.

These are point (A) $(2, 5)$ and point (D) $(\sqrt{2}, \sqrt{3})$.

The correct options are (A) and (D).

We are given a pair of linear equations in two variables: $x = a$ and $y = b$, where $a$ and $b$ are constants.

Consider the equation $x = a$.

This equation represents all points in the Cartesian plane whose x-coordinate is equal to $a$, regardless of the value of the y-coordinate.

Graphically, this is a vertical line that is parallel to the y-axis and passes through the point $(a, 0)$ on the x-axis.

For example, if $a=3$, the equation $x=3$ represents a vertical line passing through $(3, 0)$.

Consider the equation $y = b$.

This equation represents all points in the Cartesian plane whose y-coordinate is equal to $b$, regardless of the value of the x-coordinate.

Graphically, this is a horizontal line that is parallel to the x-axis and passes through the point $(0, b)$ on the y-axis.

For example, if $b=2$, the equation $y=2$ represents a horizontal line passing through $(0, 2)$.

Now consider the pair of equations $x = a$ and $y = b$ together.

Graphically, we have a vertical line ($x=a$) and a horizontal line ($y=b$).

Unless one of the lines coincides with an axis (which happens if $a=0$ or $b=0$), a vertical line and a horizontal line in the same plane will always intersect at exactly one point.

The point of intersection must satisfy both equations simultaneously.

From the first equation, the x-coordinate of the intersection point must be $x = a$.

From the second equation, the y-coordinate of the intersection point must be $y = b$.

Therefore, the coordinates of the point of intersection are $(a, b)$.

The two lines represented by $x=a$ and $y=b$ are perpendicular (a special case of intersecting lines) and intersect at the point $(a, b)$.

Let's examine the options:

(A) Parallel: The lines $x=a$ and $y=b$ are perpendicular unless $a=0$ and $b=0$, in which case they are the axes themselves, which are perpendicular. They are not parallel to each other (unless one is undefined, which is not the case here).

(B) Intersecting at $(a, b)$: As derived, the lines intersect at the point whose x-coordinate is $a$ and y-coordinate is $b$, which is the point $(a, b)$. This is correct.

(C) Coincident: Coincident lines are the same line. $x=a$ and $y=b$ represent a vertical line and a horizontal line, respectively. They are only the same line if both equations represent the entire plane, which they do not, or if they are the same axis, which only happens if $a=0$ and $b=0$, making them the x and y axes, which are distinct.

(D) Intersecting at $(b, a)$: The x-coordinate of the intersection is determined by $x=a$, not $b$. The y-coordinate is determined by $y=b$, not $a$. The intersection point is $(a, b)$, not $(b, a)$ (unless $a=b$).

The pair of equations $x=a$ and $y=b$ graphically represents two lines which are intersecting at $(a, b)$.

The correct option is (B) Intersecting at $(a, b)$.

Given:

The point is $(p, q)$.

The point lies in the third quadrant.

To Find:

The conditions on the coordinates $p$ and $q$ for the point to be in the third quadrant.

Solution:

In the Cartesian coordinate system, the four quadrants are defined based on the signs of the x-coordinate and the y-coordinate.

For a point $(x, y)$, the location is determined as follows:

First Quadrant: $x > 0$ and $y > 0$ (Sign pattern: $+, +$)

Second Quadrant: $x < 0$ and $y > 0$ (Sign pattern: $-, +$)

Third Quadrant: $x < 0$ and $y < 0$ (Sign pattern: $-, -$)

Fourth Quadrant: $x > 0$ and $y < 0$ (Sign pattern: $+, -$)

The given point is $(p, q)$. Here, $p$ is the x-coordinate and $q$ is the y-coordinate.

We are told that the point $(p, q)$ lies in the third quadrant.

According to the definition of the third quadrant, both the x-coordinate and the y-coordinate must be negative.

Therefore, for the point $(p, q)$ to be in the third quadrant, the following conditions must be met:

The x-coordinate $p$ must be negative, so $p < 0$.

The y-coordinate $q$ must be negative, so $q < 0$.

The conditions for the point $(p, q)$ to lie in the third quadrant are $p < 0$ and $q < 0$.

Let's compare this with the given options:

(A) $p > 0, q > 0$: First Quadrant

(B) $p < 0, q > 0$: Second Quadrant

(C) $p < 0, q < 0$: Third Quadrant

(D) $p > 0, q < 0$: Fourth Quadrant

The correct option that represents the conditions for the third quadrant is (C).

The correct option is (C) $p < 0, q < 0$.

Given:

The vertices of a rectangular room are A(1, 1), B(5, 1), C(5, 4), and D(1, 4).

To Find:

Which quadrant the entire room is located in.

Solution:

To determine the quadrant a point lies in, we examine the signs of its x and y coordinates.

The quadrants are defined by the signs as follows:

First Quadrant (Q1): $x > 0$, $y > 0$

Second Quadrant (Q2): $x < 0$, $y > 0$

Third Quadrant (Q3): $x < 0$, $y < 0$

Fourth Quadrant (Q4): $x > 0$, $y < 0$

Let's analyze the coordinates of each vertex:

For vertex A(1, 1): $x = 1$, $y = 1$. Since $1 > 0$ and $1 > 0$, the signs are $(+, +)$. Point A is in the First Quadrant.

For vertex B(5, 1): $x = 5$, $y = 1$. Since $5 > 0$ and $1 > 0$, the signs are $(+, +)$. Point B is in the First Quadrant.

For vertex C(5, 4): $x = 5$, $y = 4$. Since $5 > 0$ and $4 > 0$, the signs are $(+, +)$. Point C is in the First Quadrant.

For vertex D(1, 4): $x = 1$, $y = 4$. Since $1 > 0$ and $4 > 0$, the signs are $(+, +)$. Point D is in the First Quadrant.

Since all four vertices of the rectangle have positive x-coordinates and positive y-coordinates, all vertices lie in the First Quadrant.

A rectangle is a shape defined by its vertices. If all vertices of the rectangle lie within a specific quadrant, then the entire rectangle (assuming it doesn't cross the axes) is located within that quadrant.

The minimum x-coordinate among the vertices is 1 (from A and D), and the maximum x-coordinate is 5 (from B and C). All x values between 1 and 5 are positive.

The minimum y-coordinate among the vertices is 1 (from A and B), and the maximum y-coordinate is 4 (from C and D). All y values between 1 and 4 are positive.

Since for all points $(x, y)$ within the rectangle, $1 \leq x \leq 5$ and $1 \leq y \leq 4$, we have $x > 0$ and $y > 0$. This confirms that all points within the rectangle lie in the First Quadrant.

Therefore, the entire room is located in the First Quadrant.

The correct option is (A) First Quadrant.

We are asked to identify the name of the system used to locate the position of a point in a plane.

Let's consider the given options:

(A) Number system: A number system is a way of representing numbers. Examples include the decimal system, binary system, etc. While numbers are used in locating points, a number system itself is not the system for locating points in a plane.

(B) Real number line: A real number line is a one-dimensional system used to represent real numbers. It can locate points along a straight line, but not in a two-dimensional plane.

(C) Coordinate geometry: Coordinate geometry, also known as analytic geometry, is a branch of geometry where the position of points in a plane or space is described using coordinates. The Cartesian coordinate system, which uses perpendicular axes (x and y axes) to assign unique ordered pairs $(x, y)$ to each point in a plane, is the primary system used in coordinate geometry to locate points.

(D) Algebraic expression: An algebraic expression is a mathematical phrase that contains variables, numbers, and mathematical operations. It is used to represent quantities or relationships, not as a system for locating points in a plane.

The system specifically designed and used to locate the position of a point in a plane is part of coordinate geometry, which employs a coordinate system (like the Cartesian system) for this purpose.

The correct option is (C) Coordinate geometry.

Given:

A list of points: $(-4, 0)$, $(0, 6)$, $(2, -5)$, and $(0, 0)$.

To Find:

Which of the given points does NOT lie on an axis (either the x-axis or the y-axis).

Solution:

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

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

The origin $(0, 0)$ lies on both the x-axis and the y-axis.

A point does NOT lie on an axis if neither its x-coordinate nor its y-coordinate is 0 (i.e., $x \neq 0$ and $y \neq 0$). Such points lie within one of the four quadrants.

Let's examine each point and determine if it lies on an axis:

(A) Point $(-4, 0)$:

The x-coordinate is $-4$ and the y-coordinate is $0$. Since the y-coordinate is $0$, this point lies on the x-axis.

(B) Point $(0, 6)$:

The x-coordinate is $0$ and the y-coordinate is $6$. Since the x-coordinate is $0$, this point lies on the y-axis.

(C) Point $(2, -5)$:

The x-coordinate is $2$. Since $2 \neq 0$, the x-coordinate is not $0$.

The y-coordinate is $-5$. Since $-5 \neq 0$, the y-coordinate is not $0$.

Since neither the x-coordinate nor the y-coordinate is $0$, this point does NOT lie on an axis. (It lies in the fourth quadrant).

(D) Point $(0, 0)$:

The x-coordinate is $0$ and the y-coordinate is $0$. Since both coordinates are $0$, this point is the origin, which lies on both the x-axis and the y-axis.

The point that does NOT lie on an axis is the one where neither coordinate is zero.

The correct option is (C) $(2, -5)$.

Given:

The point is $(5, -2)$.

To Find:

The distance of the point $(5, -2)$ from the x-axis.

Solution:

Let a point in the Cartesian plane be $(x, y)$.

The distance of the point $(x, y)$ from the x-axis is given by the absolute value of its y-coordinate.

Distance from x-axis $= |y|$.

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

The x-coordinate is $x = 5$.

The y-coordinate is $y = -2$.

We need to find the distance from the x-axis.

Distance from x-axis $= |y| = |-2|$.

The absolute value of $-2$ is $2$.

Distance from x-axis $= 2$ units.

Note that distance is always a non-negative value.

The distance of the point $(5, -2)$ from the x-axis is 2 units.

The correct option is (C) $2$ units.

In the Cartesian coordinate system, the ordinate of a point $(x, y)$ is its y-coordinate.

We are given that the ordinate of a point is 0. This means the y-coordinate of the point is 0.

Let the point be $(x, y)$. We are given $y = 0$. So the point is $(x, 0)$.

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 origin is the point $(0, 0)$. It lies on both axes.

Points in the quadrants have non-zero coordinates (e.g., $x \neq 0$ and $y \neq 0$).

Since the y-coordinate (ordinate) of the point is 0, the point must lie on the x-axis.

Examples of points with ordinate 0 are $(1, 0)$, $(-5, 0)$, $(0, 0)$, etc. All these points lie on the x-axis.

Therefore, if the ordinate of a point is 0, the point lies on the x-axis.

The correct option is (B) x-axis.

To determine which pair of points lies in the same quadrant, we need to identify the quadrant for each point based on the signs of its coordinates $(x, y)$.

Recall the sign patterns for the four quadrants:

First Quadrant (Q1): $x > 0, y > 0$ (Signs: $+, +$)

Second Quadrant (Q2): $x < 0, y > 0$ (Signs: $-, +$)

Third Quadrant (Q3): $x < 0, y < 0$ (Signs: $-, -$)

Fourth Quadrant (Q4): $x > 0, y < 0$ (Signs: $+, -$)

Let's examine each pair of points given in the options:

(A) Points $(-1, 2)$ and $(3, -4)$

For the point $(-1, 2)$:

$x = -1$ (negative), $y = 2$ (positive). The signs are $(-, +)$. This point lies in the Second Quadrant (Q2).

For the point $(3, -4)$:

$x = 3$ (positive), $y = -4$ (negative). The signs are $(+, -)$. This point lies in the Fourth Quadrant (Q4).

Since Q2 and Q4 are different, this pair of points does not lie in the same quadrant.

(B) Points $(-5, -6)$ and $(1, 1)$

For the point $(-5, -6)$:

$x = -5$ (negative), $y = -6$ (negative). The signs are $(-, -)$. This point lies in the Third Quadrant (Q3).

For the point $(1, 1)$:

$x = 1$ (positive), $y = 1$ (positive). The signs are $(+, +)$. This point lies in the First Quadrant (Q1).

Since Q3 and Q1 are different, this pair of points does not lie in the same quadrant.

(C) Points $(2, 5)$ and $(-3, -1)$

For the point $(2, 5)$:

$x = 2$ (positive), $y = 5$ (positive). The signs are $(+, +)$. This point lies in the First Quadrant (Q1).

For the point $(-3, -1)$:

$x = -3$ (negative), $y = -1$ (negative). The signs are $(-, -)$. This point lies in the Third Quadrant (Q3).

Since Q1 and Q3 are different, this pair of points does not lie in the same quadrant.

(D) Points $(4, -2)$ and $(-7, -8)$

For the point $(4, -2)$:

$x = 4$ (positive), $y = -2$ (negative). The signs are $(+, -)$. This point lies in the Fourth Quadrant (Q4).

For the point $(-7, -8)$:

$x = -7$ (negative), $y = -8$ (negative). The signs are $(-, -)$. This point lies in the Third Quadrant (Q3).

Since Q4 and Q3 are different, this pair of points does not lie in the same quadrant.

Based on the standard definition of quadrants and the analysis of the given points, none of the pairs provided in the options lie in the same quadrant.

Therefore, as the question is stated with the given options, no option correctly identifies a pair of points lying in the same quadrant.

If there is an expected correct answer among the options (A, B, C, D), there might be a typo in the points listed in one of the options.

However, strictly following the question and options as written, none of the pairs lie in the same quadrant.

To match the points with their locations, we analyze the coordinates of each point.

(i) Point $(7, 0)$:

The x-coordinate is $7$ ($7 > 0$, positive).

The y-coordinate is $0$.

A point with y-coordinate $0$ and non-zero x-coordinate lies on the x-axis. Since the x-coordinate is positive, the point lies on the Positive x-axis.

This matches option (c) Positive x-axis in Column B.

So, (i) - (c).

(ii) Point $(0, -3)$:

The x-coordinate is $0$.

The y-coordinate is $-3$ ($-3 < 0$, negative).

A point with x-coordinate $0$ and non-zero y-coordinate lies on the y-axis. Since the y-coordinate is negative, the point lies on the Negative y-axis.

This matches option (d) Negative y-axis in Column B.

So, (ii) - (d).

(iii) Point $(-4, 5)$:

The x-coordinate is $-4$ ($-4 < 0$, negative).

The y-coordinate is $5$ ($5 > 0$, positive).

A point with a negative x-coordinate and a positive y-coordinate lies in the Second Quadrant.

This matches option (b) Second Quadrant in Column B.

So, (iii) - (b).

(iv) Point $(1, 1)$:

The x-coordinate is $1$ ($1 > 0$, positive).

The y-coordinate is $1$ ($1 > 0$, positive).

A point with a positive x-coordinate and a positive y-coordinate lies in the First Quadrant.

This matches option (a) First Quadrant in Column B.

So, (iv) - (a).

The correct matches are:

(i) - (c)

(ii) - (d)

(iii) - (b)

(iv) - (a)

Let's check the options provided:

(A) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a) - This matches our findings.

(B) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b) - Incorrect.

(C) (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a) - Incorrect.

(D) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b) - Incorrect.

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

Given:

A point lies on the negative x-axis.

The distance of the point from the origin is 6 units.

To Find:

The coordinates of the point.

Solution:

Let the coordinates of the point be $(x, y)$.

A point that lies on the x-axis always has its y-coordinate equal to 0.

So, for the point on the x-axis, $y = 0$. The coordinates are of the form $(x, 0)$.

The point lies on the negative direction of the x-axis. This means that the x-coordinate must be negative, i.e., $x < 0$.

The distance of a point $(x, y)$ from the origin $(0, 0)$ in the Cartesian plane is given by the distance formula, which simplifies to $\sqrt{x^2 + y^2}$.

We are given that the distance from the origin is 6 units.

So, the distance of the point $(x, 0)$ from the origin is $\sqrt{x^2 + 0^2} = \sqrt{x^2} = |x|$.

We have $|x| = 6$.

This implies that $x = 6$ or $x = -6$.

We know that the point is on the negative x-axis, which means $x < 0$.

Therefore, we must choose $x = -6$.

The y-coordinate is $0$.

So, the coordinates of the point are $(-6, 0)$.

Let's check the options:

(A) $(0, -6)$: This point is on the negative y-axis, 6 units from the origin.

(B) $(-6, 0)$: This point is on the negative x-axis, 6 units from the origin.

(C) $(6, 0)$: This point is on the positive x-axis, 6 units from the origin.

(D) $(0, 6)$: This point is on the positive y-axis, 6 units from the origin.

The coordinates of the point are $(-6, 0)$.

The correct option is (B) $(-6, 0)$.

Given:

The point is $(a, b)$.

The point lies in the third quadrant.

To Find:

The conditions that must be true about $a$ and $b$ for the point to be in the third quadrant.

Solution:

In the Cartesian coordinate system, the location of a point $(x, y)$ in a quadrant is determined by the signs of its x-coordinate and y-coordinate.

The quadrants and their corresponding sign patterns are:

First Quadrant (Q1): x-coordinate is positive, y-coordinate is positive ($+, +$)

Second Quadrant (Q2): x-coordinate is negative, y-coordinate is positive ($-, +$)

Third Quadrant (Q3): x-coordinate is negative, y-coordinate is negative ($-, -$)

Fourth Quadrant (Q4): x-coordinate is positive, y-coordinate is negative ($+, -$)

The given point is $(a, b)$. Here, $a$ represents the x-coordinate and $b$ represents the y-coordinate.

We are told that the point $(a, b)$ lies in the third quadrant.

According to the definition of the third quadrant, the x-coordinate must be negative and the y-coordinate must be negative.

So, for the point $(a, b)$ to be in the third quadrant, we must have:

$a < 0$

$b < 0$

Let's check the provided options:

(A) $a > 0, b > 0$: This corresponds to the First Quadrant.

(B) $a < 0, b > 0$: This corresponds to the Second Quadrant.

(C) $a < 0, b < 0$: This corresponds to the Third Quadrant.

(D) $a > 0, b < 0$: This corresponds to the Fourth Quadrant.

The statement that is true about the point $(a, b)$ if it lies in the third quadrant is $a < 0$ and $b < 0$.

The correct option is (C) $a < 0, b < 0$.

Given:

A point with coordinates $(x, y)$ such that $xy < 0$.

To Find:

The quadrant(s) where the point $(x, y)$ can lie under the given condition.

Solution:

The condition $xy < 0$ means that the product of the x-coordinate and the y-coordinate is negative.

The product of two numbers is negative if and only if one number is positive and the other is negative.

So, the condition $xy < 0$ can occur in two cases:

Case 1: $x$ is positive and $y$ is negative ($x > 0$ and $y < 0$).

The quadrant where the x-coordinate is positive and the y-coordinate is negative is the Fourth Quadrant (Q4).

Case 2: $x$ is negative and $y$ is positive ($x < 0$ and $y > 0$).

The quadrant where the x-coordinate is negative and the y-coordinate is positive is the Second Quadrant (Q2).

Therefore, if $xy < 0$, the point $(x, y)$ can lie in either the Second Quadrant or the Fourth Quadrant.

Let's look at the options:

(A) Q1 or Q2: Q1 has $(+, +)$ signs, $xy > 0$. Q2 has $(-, +)$ signs, $xy < 0$. So Q2 works, but not Q1.

(B) Q2 or Q4: Q2 has $(-, +)$ signs, $xy < 0$. Q4 has $(+, -)$ signs, $xy < 0$. Both Q2 and Q4 satisfy the condition $xy < 0$.

(C) Q1 or Q3: Q1 has $(+, +)$ signs, $xy > 0$. Q3 has $(-, -)$ signs, $xy > 0$. Neither Q1 nor Q3 satisfies the condition $xy < 0$.

(D) Q3 or Q4: Q3 has $(-, -)$ signs, $xy > 0$. Q4 has $(+, -)$ signs, $xy < 0$. So Q4 works, but not Q3.

The quadrants where the point $(x, y)$ can lie if $xy < 0$ are Q2 and Q4.

The correct option is (B) Q2 or Q4.

Given:

The point is $(-2, 0)$.

To Find:

Which part of the Cartesian plane the point $(-2, 0)$ lies on.

Solution:

Let the coordinates of the point be $(x, y)$.

For the given point $(-2, 0)$, the x-coordinate is $x = -2$ and the y-coordinate is $y = 0$.

A point lies on the x-axis if its y-coordinate is 0.

Since the y-coordinate of the point $(-2, 0)$ is $0$, the point lies on the x-axis.

The x-axis is divided into two parts by the origin $(0, 0)$:

The positive x-axis consists of points with coordinates $(x, 0)$ where $x > 0$.

The negative x-axis consists of points with coordinates $(x, 0)$ where $x < 0$.

For the point $(-2, 0)$, the x-coordinate is $-2$.

Since $-2 < 0$, the x-coordinate is negative.

Therefore, the point $(-2, 0)$ lies on the negative x-axis.

The correct option is (B) Negative x-axis.

Given:

The locations of four houses relative to the origin $(0,0)$ are:

House A: (5, -2)

House B: (-3, 4)

House C: (-1, -1)

House D: (2, 5)

To Find:

Which house is located in the quadrant where both coordinates are negative.

Solution:

In the Cartesian coordinate system, the quadrant where both the x-coordinate and the y-coordinate are negative is the Third Quadrant.

The sign pattern for points in the Third Quadrant is $(-, -)$, meaning the x-coordinate ($x$) is less than 0 ($x < 0$) and the y-coordinate ($y$) is less than 0 ($y < 0$).

Let's examine the coordinates of each house and determine the sign pattern:

For House A at (5, -2):

x-coordinate is 5 (positive, $5 > 0$).

y-coordinate is -2 (negative, $-2 < 0$).

The sign pattern is $(+, -)$. This corresponds to the Fourth Quadrant.

For House B at (-3, 4):

x-coordinate is -3 (negative, $-3 < 0$).

y-coordinate is 4 (positive, $4 > 0$).

The sign pattern is $(-, +)$. This corresponds to the Second Quadrant.

For House C at (-1, -1):

x-coordinate is -1 (negative, $-1 < 0$).

y-coordinate is -1 (negative, $-1 < 0$).

The sign pattern is $(-, -)$. This corresponds to the Third Quadrant.

For House D at (2, 5):

x-coordinate is 2 (positive, $2 > 0$).

y-coordinate is 5 (positive, $5 > 0$).

The sign pattern is $(+, +)$. This corresponds to the First Quadrant.

We are looking for the house located in the quadrant where both coordinates are negative (Third Quadrant).

From our analysis, House C has coordinates $(-1, -1)$, where both coordinates are negative.

Therefore, House C is located in the third quadrant.

The correct option is (C) House C.

The Cartesian coordinate system is a system used to locate points in a plane or in space using a set of numerical coordinates. In a two-dimensional system, it consists of two perpendicular number lines that intersect at a point called the origin.

The two-dimensional Cartesian coordinate system is typically represented by two axes: one horizontal and one vertical.

The horizontal axis is called the x-axis.

The vertical axis is called the y-axis.

The origin in the Cartesian plane is the point where the horizontal axis (x-axis) and the vertical axis (y-axis) intersect.

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

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

In a two-dimensional Cartesian coordinate system, a point is represented by an ordered pair of numbers $(x, y)$, where $x$ and $y$ are its coordinates.

The first number, $x$, is called the abscissa. It represents the horizontal distance of the point from the y-axis. It is the x-coordinate of the point.

The second number, $y$, is called the ordinate. It represents the vertical distance of the point from the x-axis. It is the y-coordinate of the point.

For the given point P$(-5, 2)$:

The abscissa is the first coordinate, which is -5.

The ordinate is the second coordinate, which is 2.

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

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 $(5, -4)$, the x-coordinate is positive ($5 > 0$) and the y-coordinate is negative ($-4 < 0$). Therefore, the point $(5, -4)$ lies in Quadrant IV.

(b) For the point $(-2, -7)$, the x-coordinate is negative ($-2 < 0$) and the y-coordinate is negative ($-7 < 0$). Therefore, the point $(-2, -7)$ lies in Quadrant III.

(c) For the point $(-3, 8)$, the x-coordinate is negative ($-3 < 0$) and the y-coordinate is positive ($8 > 0$). Therefore, the point $(-3, 8)$ lies in Quadrant II.

(d) For the point $(6, 1)$, the x-coordinate is positive ($6 > 0$) and the y-coordinate is positive ($1 > 0$). Therefore, the point $(6, 1)$ lies in Quadrant I.

The location of a point $(x, y)$ in the Cartesian plane is determined by the signs of its abscissa ($x$) and ordinate ($y$).

Points with a positive abscissa ($x > 0$) and a negative ordinate ($y < 0$) lie in the region where the x-coordinate is positive and the y-coordinate is negative.

This region is known as Quadrant IV.

Points with a negative abscissa ($x < 0$) and a positive ordinate ($y > 0$) lie in the region where the x-coordinate is negative and the y-coordinate is positive.

This region is known as Quadrant II.

To plot the points A$(3, 0)$ and B$(-2, 0)$ on the Cartesian plane:

The coordinates of point A are $(3, 0)$. The x-coordinate is $3$ and the y-coordinate is $0$. Since the y-coordinate is $0$, the point lies on the x-axis, $3$ units to the right of the origin.

The coordinates of point B are $(-2, 0)$. The x-coordinate is $-2$ and the y-coordinate is $0$. Since the y-coordinate is $0$, the point lies on the x-axis, $2$ units to the left of the origin.

Imagine the x-axis as a number line. Point A is at the position corresponding to the number $3$, and point B is at the position corresponding to the number $-2$.

To find the distance between points A and B, since they lie on the same horizontal line (the x-axis), the distance is the absolute difference of their x-coordinates.

Distance AB = $|x_2 - x_1|$

Let $x_1 = 3$ (x-coordinate of A) and $x_2 = -2$ (x-coordinate of B).

Distance AB = $|-2 - 3|$

Distance AB = $|-5|$

Distance AB = $5$ units.

Alternatively, we can think of the distance from B to the origin as $2$ units (since the x-coordinate is $-2$) and the distance from the origin to A as $3$ units (since the x-coordinate is $3$). The total distance between A and B is the sum of these distances from the origin along the x-axis.

Distance AB = (Distance from O to A) + (Distance from O to B)

Distance AB = $|3| + |-2|$

Distance AB = $3 + 2$

Distance AB = $5$ units.

The distance between A$(3, 0)$ and B$(-2, 0)$ is 5 units.

To plot the points C$(0, 4)$ and D$(0, -5)$ on the Cartesian plane:

The coordinates of point C are $(0, 4)$. The x-coordinate is $0$ and the y-coordinate is $4$. Since the x-coordinate is $0$, the point lies on the y-axis, $4$ units above the origin.

The coordinates of point D are $(0, -5)$. The x-coordinate is $0$ and the y-coordinate is $-5$. Since the x-coordinate is $0$, the point lies on the y-axis, $5$ units below the origin.

Imagine the y-axis as a number line. Point C is at the position corresponding to the number $4$, and point D is at the position corresponding to the number $-5$.

To find the distance between points C and D, since they lie on the same vertical line (the y-axis), the distance is the absolute difference of their y-coordinates.

Distance CD = $|y_2 - y_1|$

Let $y_1 = 4$ (y-coordinate of C) and $y_2 = -5$ (y-coordinate of D).

Distance CD = $|-5 - 4|$

Distance CD = $|-9|$

Distance CD = $9$ units.

Alternatively, we can think of the distance from D to the origin as $5$ units (since the y-coordinate is $-5$) and the distance from the origin to C as $4$ units (since the y-coordinate is $4$). The total distance between C and D is the sum of these distances from the origin along the y-axis.

Distance CD = (Distance from O to C) + (Distance from O to D)

Distance CD = $|4| + |-5|$

Distance CD = $4 + 5$

Distance CD = $9$ units.

The distance between C$(0, 4)$ and D$(0, -5)$ is 9 units.

A point that lies on the x-axis has its y-coordinate equal to $0$. So, the coordinates will be of the form $(x, 0)$.

The origin is the point $(0, 0)$.

A distance of 5 units to the left of the origin means moving 5 units along the negative direction of the x-axis from the origin.

Starting from the origin $(0, 0)$ and moving 5 units to the left along the x-axis, the x-coordinate changes by $-5$.

The new x-coordinate is $0 + (-5) = -5$.

The y-coordinate remains $0$ as the point is on the x-axis.

Therefore, the coordinates of the point are $(-5, 0)$.

The point that lies on the x-axis at a distance of 5 units to the left of the origin is $(-5, 0)$.

A point that lies on the y-axis has its x-coordinate equal to $0$. So, the coordinates will be of the form $(0, y)$.

The origin is the point $(0, 0)$.

A distance of 3 units below the origin means moving 3 units along the negative direction of the y-axis from the origin.

Starting from the origin $(0, 0)$ and moving 3 units down along the y-axis, the y-coordinate changes by $-3$.

The new y-coordinate is $0 + (-3) = -3$.

The x-coordinate remains $0$ as the point is on the y-axis.

Therefore, the coordinates of the point are $(0, -3)$.

The point that lies on the y-axis at a distance of 3 units below the origin is $(0, -3)$.

To plot the points E$(-3, 2)$ and F$(-3, -4)$ on the Cartesian plane:

Point E has coordinates $(-3, 2)$. This means we move 3 units to the left from the origin along the x-axis, and then 2 units up parallel to the y-axis.

Point F has coordinates $(-3, -4)$. This means we move 3 units to the left from the origin along the x-axis, and then 4 units down parallel to the y-axis.

Now consider the line segment EF connecting these two points.

Let's look at the coordinates of E and F:

E is $(-3, 2)$

F is $(-3, -4)$

Observe that the x-coordinate is the same for both points ($x = -3$). When the x-coordinate of two points is the same, the line segment connecting them is a vertical line.

A vertical line is always parallel to the y-axis.

The line segment EF lies on the vertical line $x = -3$.

Therefore, the line segment EF is parallel to the y-axis.

To plot the points G$(1, 5)$ and H$(-4, 5)$ on the Cartesian plane:

Point G has coordinates $(1, 5)$. This means we move 1 unit to the right from the origin along the x-axis, and then 5 units up parallel to the y-axis.

Point H has coordinates $(-4, 5)$. This means we move 4 units to the left from the origin along the x-axis, and then 5 units up parallel to the y-axis.

Now consider the line segment GH connecting these two points.

Let's look at the coordinates of G and H:

G is $(1, 5)$

H is $(-4, 5)$

Observe that the y-coordinate is the same for both points ($y = 5$). When the y-coordinate of two points is the same, the line segment connecting them is a horizontal line.

A horizontal line is always parallel to the x-axis.

The line segment GH lies on the horizontal line $y = 5$.

Therefore, the line segment GH is parallel to the x-axis.

In the Cartesian coordinate system:

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

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

For the given point with coordinates $(0, -7)$:

The x-coordinate is $0$.

The y-coordinate is $-7$.

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

The point $(0, -7)$ lies on the y-axis.

The Cartesian plane is divided into four quadrants based on the intersection of the x-axis and the y-axis at the origin.

The quadrants are defined by the signs of the x and y coordinates of the points within them.

For a point $(x, y)$ lying in the first quadrant:

The x-coordinate (abscissa) is positive ($x > 0$).

The y-coordinate (ordinate) is positive ($y > 0$).

The equation of a line in the Cartesian plane represents the relationship between the x and y coordinates of all points that lie on that line.

For the x-axis:

Any point that lies on the x-axis has its y-coordinate equal to $0$, regardless of its x-coordinate.

For example, points like $(1, 0)$, $(-5, 0)$, $(\sqrt{2}, 0)$ all lie on the x-axis.

Therefore, the equation of the x-axis is $\mathbf{y = 0}$.

For the y-axis:

Any point that lies on the y-axis has its x-coordinate equal to $0$, regardless of its y-coordinate.

For example, points like $(0, 3)$, $(0, -1.5)$, $(0, \pi)$ all lie on the y-axis.

Therefore, the equation of the y-axis is $\mathbf{x = 0}$.

Let the vertices of the rectangle be A$(1, 1)$, B$(1, 4)$, C$(5, 4)$, and D$(5, 1)$.

We can find the lengths of the sides by calculating the distance between adjacent vertices.

Length of side AB:

The points are A$(1, 1)$ and B$(1, 4)$. These points have the same x-coordinate, so the segment is vertical.

Length AB = $|y_2 - y_1| = |4 - 1| = |3| = 3$ units.

Length of side BC:

The points are B$(1, 4)$ and C$(5, 4)$. These points have the same y-coordinate, so the segment is horizontal.

Length BC = $|x_2 - x_1| = |5 - 1| = |4| = 4$ units.

Length of side CD:

The points are C$(5, 4)$ and D$(5, 1)$. These points have the same x-coordinate, so the segment is vertical.

Length CD = $|y_2 - y_1| = |1 - 4| = |-3| = 3$ units.

Length of side DA:

The points are D$(5, 1)$ and A$(1, 1)$. These points have the same y-coordinate, so the segment is horizontal.

Length DA = $|x_2 - x_1| = |1 - 5| = |-4| = 4$ units.

We see that AB = CD = 3 units and BC = DA = 4 units. Since opposite sides are equal in length, this confirms it is a rectangle.

The side lengths of the rectangle are $l = 4$ units and $w = 3$ units.

The perimeter of a rectangle is given by the formula $P = 2(l + w)$.

$P = 2(4 + 3)$

$P = 2(7)$

$P = 14$ units.

The perimeter of the rectangle is 14 units.

Let the vertices of the square be denoted by A, B, C, and D.

We are given that one vertex is at the origin. Let's assume this vertex is A.

So, the coordinates of vertex A are $(0, 0)$.

The sides of the square are along the positive x and y axes, and the side length is 3 units.

This means one side of the square extends from the origin along the positive x-axis for a length of 3 units. The endpoint of this side will be another vertex. Since it's on the positive x-axis, its y-coordinate is $0$. The x-coordinate will be $0 + 3 = 3$.

Let this vertex be B. The coordinates of vertex B are $(3, 0)$.

Similarly, another side of the square extends from the origin along the positive y-axis for a length of 3 units. The endpoint of this side will be another vertex. Since it's on the positive y-axis, its x-coordinate is $0$. The y-coordinate will be $0 + 3 = 3$.

Let this vertex be D. The coordinates of vertex D are $(0, 3)$.

The fourth vertex (C) is formed by the intersection of lines parallel to the axes passing through B and D. It is 3 units to the right of D (or 3 units above B).

Starting from D$(0, 3)$, moving 3 units right (parallel to the x-axis), the x-coordinate becomes $0 + 3 = 3$, and the y-coordinate remains $3$. The coordinates are $(3, 3)$.

Starting from B$(3, 0)$, moving 3 units up (parallel to the y-axis), the y-coordinate becomes $0 + 3 = 3$, and the x-coordinate remains $3$. The coordinates are $(3, 3)$.

So, the coordinates of vertex C are $(3, 3)$.

The coordinates of the vertices of the square are $(0, 0), (3, 0), (3, 3), (0, 3)$.

In the Cartesian coordinate system, a point is represented by an ordered pair $(x, y)$, where $x$ is the x-coordinate and $y$ is the y-coordinate.

The x-coordinate represents the horizontal distance of the point from the y-axis. If the x-coordinate is $0$, it means the point has zero horizontal distance from the y-axis.

Therefore, any point with an x-coordinate of $0$ must lie on the y-axis.

In the Cartesian coordinate system, a point is represented by an ordered pair $(x, y)$, where $x$ is the x-coordinate and $y$ is the y-coordinate.

The y-coordinate represents the vertical distance of the point from the x-axis. If the y-coordinate is $0$, it means the point has zero vertical distance from the x-axis.

Therefore, any point with a y-coordinate of $0$ must lie on the x-axis.

To determine the shape formed by joining the points A$(1, 2)$, B$(3, 2)$, and C$(1, 4)$, we can plot these points on a Cartesian plane and calculate the lengths of the line segments connecting them.

Plotting the points:

Point A$(1, 2)$: Move 1 unit right from the origin on the x-axis, then 2 units up parallel to the y-axis.

Point B$(3, 2)$: Move 3 units right from the origin on the x-axis, then 2 units up parallel to the y-axis.

Point C$(1, 4)$: Move 1 unit right from the origin on the x-axis, then 4 units up parallel to the y-axis.

Calculating side lengths:

We use the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Length of side AB (between A$(1, 2)$ and B$(3, 2)$):

$AB = \sqrt{(3 - 1)^2 + (2 - 2)^2}$

$AB = \sqrt{(2)^2 + (0)^2}$

$AB = \sqrt{4 + 0}$

$AB = \sqrt{4}$

$AB = 2$ units.

Note that points A and B have the same y-coordinate ($y=2$), so the segment AB is horizontal.

Length of side BC (between B$(3, 2)$ and C$(1, 4)$):

$BC = \sqrt{(1 - 3)^2 + (4 - 2)^2}$

$BC = \sqrt{(-2)^2 + (2)^2}$

$BC = \sqrt{4 + 4}$

$BC = \sqrt{8}$

$BC = 2\sqrt{2}$ units.

Length of side CA (between C$(1, 4)$ and A$(1, 2)$):

$CA = \sqrt{(1 - 1)^2 + (2 - 4)^2}$

$CA = \sqrt{(0)^2 + (-2)^2}$

$CA = \sqrt{0 + 4}$

$CA = \sqrt{4}$

$CA = 2$ units.

Note that points C and A have the same x-coordinate ($x=1$), so the segment CA is vertical.

The side lengths are AB = 2, BC = $2\sqrt{2}$, and CA = 2.

Since AB = CA = 2, the triangle ABC has two sides of equal length. This means the triangle is an isosceles triangle.

Let's also check for a right angle. Since AB is a horizontal segment and CA is a vertical segment, they are perpendicular to each other. They meet at vertex A. Therefore, the angle at vertex A ($\angle$BAC) is a right angle ($90^\circ$).

Alternatively, we can check if the Pythagorean theorem holds for the side lengths $2, 2, 2\sqrt{2}$. The longest side is BC with length $2\sqrt{2}$.

$AB^2 + CA^2 = 2^2 + 2^2 = 4 + 4 = 8$

$BC^2 = (2\sqrt{2})^2 = 4 \times 2 = 8$

Since $AB^2 + CA^2 = BC^2$, the triangle satisfies the Pythagorean theorem, meaning it is a right-angled triangle, with the right angle at A (opposite the hypotenuse BC).

Combining the observations:

The triangle has two equal sides (AB = CA).

The triangle has a right angle (at A).

Therefore, the shape formed by joining points A, B, and C is a right-angled isosceles triangle.

If a point lies on a line, its coordinates must satisfy the equation of the line.

The equation of the line is $y = x + 3$.

Consider the first point $(2, y)$.

The x-coordinate of this point is $2$, and the y-coordinate is $y$.

Substitute these coordinates into the equation of the line:

$y = 2 + 3$

$y = 5$

So, for the point $(2, y)$ to lie on the line, the value of $y$ must be $5$.

Consider the second point $(x, 5)$.

The x-coordinate of this point is $x$, and the y-coordinate is $5$.

Substitute these coordinates into the equation of the line:

$5 = x + 3$

Now, solve for $x$:

$x = 5 - 3$

$x = 2$

So, for the point $(x, 5)$ to lie on the line, the value of $x$ must be $2$.

The values of $x$ and $y$ are $x = 2$ and $y = 5$.

Solution:

We are given that one vertex of the rectangle is at the point A$(-1, -2)$.

The length of the rectangle is 6 units, and the breadth is 4 units.

The sides of the rectangle are parallel to the coordinate axes. This means that the sides are either horizontal (parallel to the x-axis) or vertical (parallel to the y-axis).

The distances between adjacent vertices will be either the length (6 units) or the breadth (4 units), corresponding to horizontal or vertical movements from a vertex.

There are two possibilities for the orientation of the rectangle's sides relative to the given vertex and the axes:

Possibility 1: The side parallel to the x-axis (horizontal) has length 6, and the side parallel to the y-axis (vertical) has length 4.

Let the vertices be A, B, C, and D in order.

Let A be the given vertex $(-1, -2)$.

From A$(-1, -2)$, an adjacent vertex can be found by moving horizontally by 6 units. The y-coordinate remains $-2$. The x-coordinate changes by $\pm 6$.

One adjacent vertex is B$(-1 + 6, -2) = (5, -2)$. (Moving right)

Another adjacent vertex can be found by moving vertically by 4 units from A. The x-coordinate remains $-1$. The y-coordinate changes by $\pm 4$.

The other adjacent vertex (forming the corner with A) is D$(-1, -2 + 4) = (-1, 2)$. (Moving up)

The fourth vertex C will have the x-coordinate of B and the y-coordinate of D.

So, C is $(5, 2)$.

The vertices in this configuration (moving right and up from A) are $(-1, -2), (5, -2), (5, 2), (-1, 2)$.

Possibility 2: The side parallel to the x-axis (horizontal) has length 4, and the side parallel to the y-axis (vertical) has length 6.

Let A be the given vertex $(-1, -2)$.

From A$(-1, -2)$, an adjacent vertex can be found by moving horizontally by 4 units. The y-coordinate remains $-2$. The x-coordinate changes by $\pm 4$.

One adjacent vertex is B$(-1 + 4, -2) = (3, -2)$. (Moving right)

Another adjacent vertex can be found by moving vertically by 6 units from A. The x-coordinate remains $-1$. The y-coordinate changes by $\pm 6$.

The other adjacent vertex is D$(-1, -2 + 6) = (-1, 4)$. (Moving up)

The fourth vertex C will have the x-coordinate of B and the y-coordinate of D.

So, C is $(3, 4)$.

The vertices in this configuration (moving right and up from A) are $(-1, -2), (3, -2), (3, 4), (-1, 4)$.

In addition to these two possibilities (where the rectangle extends right and up from the given vertex), there are other possibilities where the rectangle extends left or down. For instance, in Possibility 1, moving left by 6 and up by 4 from A$(-1, -2)$ would give vertices $(-1, -2), (-7, -2), (-7, 2), (-1, 2)$. However, typically one set of vertices for each orientation is sufficient unless otherwise specified.

Therefore, two possible sets of coordinates for the vertices of the rectangle are:

1. Horizontal side length 6, Vertical side length 4: $(-1, -2), (5, -2), (5, 2), (-1, 2)$

2. Horizontal side length 4, Vertical side length 6: $(-1, -2), (3, -2), (3, 4), (-1, 4)$

Solution:

We are given that one vertex of the rectangle is at the point A$(-1, -2)$.

The length of the rectangle is 6 units, and the breadth is 4 units.

The sides of the rectangle are parallel to the coordinate axes. This means that the sides are either horizontal (parallel to the x-axis) or vertical (parallel to the y-axis).

The distances between adjacent vertices will be either the length (6 units) or the breadth (4 units), corresponding to horizontal or vertical movements from a vertex.

There are two possibilities for the orientation of the rectangle's sides relative to the given vertex and the axes:

Possibility 1: The side parallel to the x-axis (horizontal) has length 6, and the side parallel to the y-axis (vertical) has length 4.

Let the vertices be A, B, C, and D in order.

Let A be the given vertex $(-1, -2)$.

From A$(-1, -2)$, an adjacent vertex can be found by moving horizontally by 6 units. The y-coordinate remains $-2$. The x-coordinate changes by $\pm 6$.

One adjacent vertex is B$(-1 + 6, -2) = (5, -2)$. (Moving right)

Another adjacent vertex can be found by moving vertically by 4 units from A. The x-coordinate remains $-1$. The y-coordinate changes by $\pm 4$.

The other adjacent vertex (forming the corner with A) is D$(-1, -2 + 4) = (-1, 2)$. (Moving up)

The fourth vertex C will have the x-coordinate of B and the y-coordinate of D.

So, C is $(5, 2)$.

The vertices in this configuration (moving right and up from A) are $(-1, -2), (5, -2), (5, 2), (-1, 2)$.

Possibility 2: The side parallel to the x-axis (horizontal) has length 4, and the side parallel to the y-axis (vertical) has length 6.

Let A be the given vertex $(-1, -2)$.

From A$(-1, -2)$, an adjacent vertex can be found by moving horizontally by 4 units. The y-coordinate remains $-2$. The x-coordinate changes by $\pm 4$.

One adjacent vertex is B$(-1 + 4, -2) = (3, -2)$. (Moving right)

Another adjacent vertex can be found by moving vertically by 6 units from A. The x-coordinate remains $-1$. The y-coordinate changes by $\pm 6$.

The other adjacent vertex is D$(-1, -2 + 6) = (-1, 4)$. (Moving up)

The fourth vertex C will have the x-coordinate of B and the y-coordinate of D.

So, C is $(3, 4)$.

The vertices in this configuration (moving right and up from A) are $(-1, -2), (3, -2), (3, 4), (-1, 4)$.

In addition to these two possibilities (where the rectangle extends right and up from the given vertex), there are other possibilities where the rectangle extends left or down. For instance, in Possibility 1, moving left by 6 and up by 4 from A$(-1, -2)$ would give vertices $(-1, -2), (-7, -2), (-7, 2), (-1, 2)$. However, typically one set of vertices for each orientation is sufficient unless otherwise specified.

Therefore, two possible sets of coordinates for the vertices of the rectangle are:

1. Horizontal side length 6, Vertical side length 4: $(-1, -2), (5, -2), (5, 2), (-1, 2)$

2. Horizontal side length 4, Vertical side length 6: $(-1, -2), (3, -2), (3, 4), (-1, 4)$

In the Cartesian coordinate system, the plane is divided into four quadrants.

The abscissa is the x-coordinate of a point, and the ordinate is the y-coordinate of a point.

The signs of the coordinates determine the quadrant in which a point lies:

Quadrant I: (positive x, positive y)

Quadrant II: (negative x, positive y)

Quadrant III: (negative x, negative y)

Quadrant IV: (positive x, negative y)

If a point lies in the third quadrant, it is located to the left of the y-axis (where x-values are negative) and below the x-axis (where y-values are negative).

For any point $(x, y)$ in the third quadrant:

The abscissa ($x$) is negative ($x < 0$).

The ordinate ($y$) is negative ($y < 0$).

To plot the point $(4, -2)$ on a graph sheet (Cartesian plane), follow these steps:

1. Start at the origin $(0, 0)$, which is the point where the x-axis and y-axis intersect.

2. The first coordinate is the x-coordinate (abscissa), which is $4$. Move horizontally from the origin along the x-axis. Since the value is positive, move 4 units to the right along the positive x-axis.

3. From the position on the x-axis (at $x=4$), the second coordinate is the y-coordinate (ordinate), which is $-2$. Move vertically from this position. Since the value is negative, move 2 units down parallel to the y-axis.

4. The point where you end up is the location of $(4, -2)$. This point will be in the region where x is positive and y is negative, which is the Quadrant IV.

So, to plot $(4, -2)$: Start at $(0,0)$, move right 4 units, then move down 2 units. Mark this location.

Explanation of Plotting a Point $(x, y)$ on the Cartesian Plane:

The Cartesian plane consists of two perpendicular number lines, the horizontal x-axis and the vertical y-axis, intersecting at the origin $(0, 0)$. A point on this plane is represented by an ordered pair of coordinates $(x, y)$, where $x$ is the abscissa and $y$ is the ordinate.

To plot a point $(x, y)$:

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

2. Move horizontally along the x-axis according to the value of the x-coordinate ($x$). If $x$ is positive, move $|x|$ units to the right. If $x$ is negative, move $|x|$ units to the left. If $x$ is zero, stay on the y-axis.

3. From the position reached in step 2, move vertically parallel to the y-axis according to the value of the y-coordinate ($y$). If $y$ is positive, move $|y|$ units upwards. If $y$ is negative, move $|y|$ units downwards. If $y$ is zero, stay on the x-axis.

4. The final position reached is the location of the point $(x, y)$.

Plotting the given points and identifying their location:

Point A$(3, 5)$:

Here, $x = 3$ (positive) and $y = 5$ (positive).

Start at $(0, 0)$, move 3 units right along the x-axis. From there, move 5 units up parallel to the y-axis.

Since $x > 0$ and $y > 0$, the point A$(3, 5)$ lies in the Quadrant I.

Point B$(-2, 4)$:

Here, $x = -2$ (negative) and $y = 4$ (positive).

Start at $(0, 0)$, move 2 units left along the x-axis. From there, move 4 units up parallel to the y-axis.

Since $x < 0$ and $y > 0$, the point B$(-2, 4)$ lies in the Quadrant II.

Point C$(0, -3)$:

Here, $x = 0$ and $y = -3$ (negative).

Start at $(0, 0)$, move 0 units horizontally (stay on the y-axis). From there, move 3 units down parallel to the y-axis.

Since $x = 0$ and $y \neq 0$, the point C$(0, -3)$ lies on the y-axis.

Point D$(-1, -2)$:

Here, $x = -1$ (negative) and $y = -2$ (negative).

Start at $(0, 0)$, move 1 unit left along the x-axis. From there, move 2 units down parallel to the y-axis.

Since $x < 0$ and $y < 0$, the point D$(-1, -2)$ lies in the Quadrant III.

Point E$(5, 0)$:

Here, $x = 5$ (positive) and $y = 0$.

Start at $(0, 0)$, move 5 units right along the x-axis. From there, move 0 units vertically (stay on the x-axis).

Since $x \neq 0$ and $y = 0$, the point E$(5, 0)$ lies on the x-axis.

Plotting the points:

To plot each point $(x, y)$, start at the origin $(0, 0)$. Move horizontally $|x|$ units (right if $x > 0$, left if $x < 0$). Then, move vertically $|y|$ units (up if $y > 0$, down if $y < 0$).

P$(-4, -2)$: Move 4 units left, then 2 units down.

Q$(-4, 3)$: Move 4 units left, then 3 units up.

R$(2, 3)$: Move 2 units right, then 3 units up.

S$(2, -2)$: Move 2 units right, then 2 units down.

Joining the points:

Join the points in the given order: P to Q, Q to R, R to S, and S to P.

Naming the shape:

Let's examine the coordinates of the vertices:

P$(-4, -2)$, Q$(-4, 3)$, R$(2, 3)$, S$(2, -2)$.

Observe the x and y coordinates:

Points P and Q have the same x-coordinate ($-4$). The segment PQ is vertical.

Points Q and R have the same y-coordinate ($3$). The segment QR is horizontal.

Points R and S have the same x-coordinate ($2$). The segment RS is vertical.

Points S and P have the same y-coordinate ($-2$). The segment SP is horizontal.

Since opposite sides are horizontal and vertical, they are parallel to the axes and perpendicular to adjacent sides. This means the angles at the vertices are $90^\circ$. A quadrilateral with all four angles equal to $90^\circ$ is a rectangle.

Finding the area and perimeter:

We need to find the lengths of the sides.

Length of PQ (vertical): $|y_Q - y_P| = |3 - (-2)| = |3 + 2| = |5| = 5$ units.

Length of QR (horizontal): $|x_R - x_Q| = |2 - (-4)| = |2 + 4| = |6| = 6$ units.

Length of RS (vertical): $|y_S - y_R| = |-2 - 3| = |-5| = 5$ units.

Length of SP (horizontal): $|x_P - x_S| = |-4 - 2| = |-6| = 6$ units.

The lengths of the sides are 5 units and 6 units. Since opposite sides are equal (PQ=RS=5, QR=SP=6), it is indeed a rectangle.

Area of the rectangle = Length $\times$ Breadth

Let Length = 6 units (horizontal sides) and Breadth = 5 units (vertical sides).

Area = $6 \times 5 = 30$ square units.

Perimeter of the rectangle = $2 \times (\text{Length} + \text{Breadth})$

Perimeter = $2 \times (6 + 5) = 2 \times 11 = 22$ units.

The shape formed is a rectangle.

The area of the rectangle is 30 square units.

The perimeter of the rectangle is 22 units.

Solution:

Plotting the given points:

A$(0, 0)$: This is the origin, where the x-axis and y-axis intersect.

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

D$(0, 4)$: The x-coordinate is 0 and the y-coordinate is 4. This point is on the y-axis, 4 units above the origin.

Finding the coordinates of point C:

For ABCD to be a rectangle, opposite sides must be parallel and equal in length, and adjacent sides must be perpendicular. Since A is at the origin and the given points B and D are on the x and y axes respectively, the sides AB and AD lie along the axes.

The side AB is a horizontal segment along the x-axis from $(0, 0)$ to $(5, 0)$. Its length is the difference in x-coordinates: $|5 - 0| = 5$ units.

The side AD is a vertical segment along the y-axis from $(0, 0)$ to $(0, 4)$. Its length is the difference in y-coordinates: $|4 - 0| = 4$ units.

For ABCD to be a rectangle:

The side BC must be parallel to AD and have the same length (4 units). Since AD is vertical, BC must be vertical. Starting from B$(5, 0)$, a vertical segment means the x-coordinate remains 5. Moving upwards by 4 units means the y-coordinate becomes $0 + 4 = 4$. This gives the point $(5, 4)$.

The side CD must be parallel to AB and have the same length (5 units). Since AB is horizontal, CD must be horizontal. Starting from D$(0, 4)$, a horizontal segment means the y-coordinate remains 4. Moving to the right by 5 units means the x-coordinate becomes $0 + 5 = 5$. This gives the point $(5, 4)$.

Both paths lead to the same point. Thus, the coordinates of point C are $(5, 4)$.

Plotting point C$(5, 4)$:

Move 5 units right from the origin along the x-axis, then 4 units up parallel to the y-axis.

Verifying that ABCD is a rectangle:

The vertices are A$(0, 0)$, B$(5, 0)$, C$(5, 4)$, and D$(0, 4)$.

Let's check the lengths of the opposite sides by counting units on the graph sheet:

Length of AB: From A$(0, 0)$ to B$(5, 0)$. This is a horizontal segment. Count the units along the x-axis from 0 to 5. Length AB = 5 units.

Length of CD: From C$(5, 4)$ to D$(0, 4)$. This is a horizontal segment (since both y-coordinates are 4). Count the units along the horizontal line $y=4$ from $x=5$ to $x=0$. Length CD = $|0 - 5| = |-5| = 5$ units.

Length of BC: From B$(5, 0)$ to C$(5, 4)$. This is a vertical segment (since both x-coordinates are 5). Count the units along the vertical line $x=5$ from $y=0$ to $y=4$. Length BC = $|4 - 0| = |4| = 4$ units.

Length of DA: From D$(0, 4)$ to A$(0, 0)$. This is a vertical segment (since both x-coordinates are 0). Count the units along the vertical line $x=0$ (y-axis) from $y=4$ to $y=0$. Length DA = $|0 - 4| = |-4| = 4$ units.

We found that AB = CD = 5 units and BC = DA = 4 units.

Since opposite sides are equal in length and the sides are parallel to the axes (meaning adjacent sides are perpendicular, forming $90^\circ$ angles), the quadrilateral ABCD is indeed a rectangle.

Plotting the points:

M$(-3, 0)$: On the x-axis, 3 units to the left of the origin.

N$(3, 0)$: On the x-axis, 3 units to the right of the origin.

P$(0, 4)$: On the y-axis, 4 units above the origin.

Joining the points:

Joining M, N, and P forms a closed figure with three vertices. This shape is a triangle.

Finding the area of the triangle:

The vertices are M$(-3, 0)$, N$(3, 0)$, and P$(0, 4)$.

Consider the base of the triangle as the segment MN. Since both M and N have a y-coordinate of 0, the segment MN lies on the x-axis.

The length of the base MN is the distance between $(-3, 0)$ and $(3, 0)$.

Base length MN = $|x_N - x_M| = |3 - (-3)| = |3 + 3| = |6| = 6$ units.

The height of the triangle is the perpendicular distance from the third vertex P$(0, 4)$ to the base MN (which lies on the x-axis, the line $y=0$).

The perpendicular distance of a point $(x_0, y_0)$ from the x-axis ($y=0$) is $|y_0|$.

For point P$(0, 4)$, the y-coordinate is 4. The perpendicular distance from P to the x-axis is $|4| = 4$ units.

So, the height of the triangle with respect to base MN is 4 units.

The area of a triangle is given by the formula: Area = $\frac{1}{2} \times \text{base} \times \text{height}$.

Area of triangle MNP = $\frac{1}{2} \times \text{MN} \times \text{height}$

Area = $\frac{1}{2} \times 6 \times 4$

Area = $\frac{1}{\cancel{2}_1} \times \cancel{6}^3 \times 4$

Area = $3 \times 4$

Area = 12 square units.

The shape formed is a triangle.

The area of the triangle MNP is 12 square units.

Plotting the points:

K$(-2, 1)$: Move 2 units left from the origin along the x-axis, then 1 unit up parallel to the y-axis.

L$(3, 1)$: Move 3 units right from the origin along the x-axis, then 1 unit up parallel to the y-axis.

M$(0, 4)$: Move 0 units horizontally (stay on the y-axis), then 4 units up parallel to the y-axis.

Joining the points:

Joining the three points K, L, and M forms a triangle.

Finding the area of the triangle:

The vertices of the triangle are K$(-2, 1)$, L$(3, 1)$, and M$(0, 4)$.

Let's consider the segment KL as the base of the triangle. Points K and L both have a y-coordinate of 1, which means the segment KL is a horizontal line segment on the line $y = 1$.

The length of the base KL is the distance between K$(-2, 1)$ and L$(3, 1)$. Since it's a horizontal segment, the length is the absolute difference of the x-coordinates.

Base length KL $= |x_L - x_K| = |3 - (-2)| = |3 + 2| = |5| = 5$ units.

The height of the triangle with respect to the base KL is the perpendicular distance from the vertex M$(0, 4)$ to the line containing the base (which is the line $y = 1$).

The perpendicular distance from a point $(x_0, y_0)$ to a horizontal line $y = c$ is $|y_0 - c|$.

For point M$(0, 4)$ and the line $y = 1$, the height is $|4 - 1| = |3| = 3$ units.

The area of a triangle is given by the formula: Area = $\frac{1}{2} \times \text{base} \times \text{height}$.

Area of triangle KLM = $\frac{1}{2} \times \text{KL} \times \text{height}$

Area = $\frac{1}{2} \times 5 \times 3$

Area = $\frac{15}{2}$

Area = $7.5$ square units.

The shape formed is a triangle.

The area of the triangle KLM is 7.5 square units.

Plotting the points:

To plot the points on a graph sheet:

A$(1, 2)$: Start at the origin, move 1 unit right, then 2 units up.

B$(4, 2)$: Start at the origin, move 4 units right, then 2 units up.

C$(6, 4)$: Start at the origin, move 6 units right, then 4 units up.

D$(3, 4)$: Start at the origin, move 3 units right, then 4 units up.

Joining the points:

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

Identifying the shape and Justification:

Let's examine the segments formed by joining the points:

Segment AB connects A$(1, 2)$ and B$(4, 2)$. The y-coordinates are the same ($y=2$). This is a horizontal segment. Its length is the difference in x-coordinates: $|4 - 1| = 3$ units.

Segment CD connects C$(6, 4)$ and D$(3, 4)$. The y-coordinates are the same ($y=4$). This is a horizontal segment. Its length is the difference in x-coordinates: $|3 - 6| = |-3| = 3$ units.

Since both AB and CD are horizontal segments, they are parallel to the x-axis and thus parallel to each other. Also, their lengths are equal (both 3 units).

Segment BC connects B$(4, 2)$ and C$(6, 4)$.

Segment DA connects D$(3, 4)$ and A$(1, 2)$.

Let's look at the changes in coordinates:

From B$(4, 2)$ to C$(6, 4)$: x changes by $6-4 = 2$, y changes by $4-2 = 2$. (Move 2 right, 2 up)

From D$(3, 4)$ to A$(1, 2)$: x changes by $1-3 = -2$, y changes by $2-4 = -2$. (Move 2 left, 2 down)

The vector from B to C is $(2, 2)$. The vector from D to A is $(-2, -2)$. These vectors are opposite, indicating that BC is parallel to DA and they have the same length. (Moving 2 right and 2 up is parallel to moving 2 left and 2 down).

Length of BC = $\sqrt{(6-4)^2 + (4-2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4+4} = \sqrt{8}$ units.

Length of DA = $\sqrt{(1-3)^2 + (2-4)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8}$ units.

Since both pairs of opposite sides are parallel (AB || CD and BC || DA) and equal in length (AB = CD and BC = DA), the quadrilateral ABCD is a parallelogram.

Looking at the graph, you can visually confirm that AB and CD are horizontal and equal, and BC and DA appear to be parallel and equal sloped. The angles at the vertices do not look like right angles, so it is not a rectangle or square.

Plotting the points:

To plot the points A$(2, 5)$, B$(6, 5)$, and C$(4, 8)$ on a graph sheet:

A$(2, 5)$: Start at the origin, move 2 units right, then 5 units up.

B$(6, 5)$: Start at the origin, move 6 units right, then 5 units up.

C$(4, 8)$: Start at the origin, move 4 units right, then 8 units up.

Joining the points:

Joining the three points A, B, and C forms a closed figure with three sides. This shape is a triangle.

Finding the lengths of the sides AB and BC:

Length of side AB (between A$(2, 5)$ and B$(6, 5)$):

Points A and B have the same y-coordinate ($y=5$). This means the segment AB is a horizontal line segment. The length of a horizontal segment can be found by counting units along the x-axis or by taking the absolute difference of the x-coordinates.

Length AB = $|x_B - x_A| = |6 - 2| = |4| = 4$ units.

Length of side BC (between B$(6, 5)$ and C$(4, 8)$):

We can use the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Length BC = $\sqrt{(4 - 6)^2 + (8 - 5)^2}$

Length BC = $\sqrt{(-2)^2 + (3)^2}$

Length BC = $\sqrt{4 + 9}$

Length BC = $\sqrt{13}$ units.

Finding the area of the triangle using AB as the base:

We have the base length AB = 4 units.

The base AB lies on the horizontal line $y = 5$ (since both A and B have a y-coordinate of 5).

The height of the triangle with respect to the base AB is the perpendicular distance from the opposite vertex C$(4, 8)$ to the line $y = 5$.

The perpendicular distance from a point $(x_0, y_0)$ to a horizontal line $y = c$ is $|y_0 - c|$.

For point C$(4, 8)$ and the line $y=5$, the height is $|8 - 5| = |3| = 3$ units.

The area of a triangle is given by the formula: Area = $\frac{1}{2} \times \text{base} \times \text{height}$.

Area of triangle ABC = $\frac{1}{2} \times \text{AB} \times \text{height}$

Area = $\frac{1}{2} \times 4 \times 3$

Area = $\frac{1}{\cancel{2}_1} \times \cancel{4}^2 \times 3$

Area = $2 \times 3$

Area = 6 square units.

The shape formed is a triangle.

The length of side AB is 4 units.

The length of side BC is $\sqrt{13}$ units.

The area of the triangle ABC is 6 square units.

Concept of Quadrants:

The Cartesian coordinate plane is formed by two perpendicular number lines, the horizontal x-axis and the vertical y-axis, which intersect at the origin $(0, 0)$. These two axes divide the entire plane into four distinct regions. Each of these regions is called a quadrant.

The quadrants are numbered in a counter-clockwise direction, starting from the upper right region.

1. The upper right region is the First Quadrant.

2. The upper left region is the Second Quadrant.

3. The lower left region is the Third Quadrant.

4. The lower right region is the Fourth Quadrant.

Signs of Coordinates in Each Quadrant:

The sign of the x-coordinate (abscissa) and the y-coordinate (ordinate) of a point determines which quadrant the point lies in.

In the First Quadrant (Quadrant I):

Points in this quadrant are to the right of the y-axis and above the x-axis. Thus, both the x-coordinate and the y-coordinate are positive.

Signs: (+x, +y)

For any point $(x, y)$ in Quadrant I, $x > 0$ and $y > 0$.

In the Second Quadrant (Quadrant II):

Points in this quadrant are to the left of the y-axis and above the x-axis. Thus, the x-coordinate is negative, and the y-coordinate is positive.

Signs: (-x, +y)

For any point $(x, y)$ in Quadrant II, $x < 0$ and $y > 0$.

In the Third Quadrant (Quadrant III):

Points in this quadrant are to the left of the y-axis and below the x-axis. Thus, both the x-coordinate and the y-coordinate are negative.

Signs: (-x, -y)

For any point $(x, y)$ in Quadrant III, $x < 0$ and $y < 0$.

In the Fourth Quadrant (Quadrant IV):

Points in this quadrant are to the right of the y-axis and below the x-axis. Thus, the x-coordinate is positive, and the y-coordinate is negative.

Signs: (+x, -y)

For any point $(x, y)$ in Quadrant IV, $x > 0$ and $y < 0$.

Signs of Coordinates on the Axes:

Points that lie exactly on the axes do not belong to any quadrant.

On the x-axis:

Any point on the x-axis has a y-coordinate of $0$. The x-coordinate can be positive, negative, or zero.

Coordinates: $(x, 0)$ where $x$ is any real number.

Points to the right of the origin have $x > 0$ (positive x-axis). Points to the left of the origin have $x < 0$ (negative x-axis).

On the y-axis:

Any point on the y-axis has an x-coordinate of $0$. The y-coordinate can be positive, negative, or zero.

Coordinates: $(0, y)$ where $y$ is any real number.

Points above the origin have $y > 0$ (positive y-axis). Points below the origin have $y < 0$ (negative y-axis).

The Origin:

The origin is the point where the x-axis and y-axis intersect. Both coordinates are zero.

Coordinates: $(0, 0)$. The origin does not lie in any quadrant.

Plotting the points:

Let the points be P$(-1, -2)$, Q$(-1, 3)$, R$(4, 3)$, and S$(4, -2)$. To plot them:

P$(-1, -2)$: Move 1 unit left from origin, then 2 units down.

Q$(-1, 3)$: Move 1 unit left from origin, then 3 units up.

R$(4, 3)$: Move 4 units right from origin, then 3 units up.

S$(4, -2)$: Move 4 units right from origin, then 2 units down.

Identifying the shape:

Let's examine the sides formed by joining the points in order P-Q-R-S-P.

Segment PQ: Connects P$(-1, -2)$ and Q$(-1, 3)$. Same x-coordinate ($-1$), so it's a vertical segment. Length = $|3 - (-2)| = 5$ units.

Segment QR: Connects Q$(-1, 3)$ and R$(4, 3)$. Same y-coordinate ($3$), so it's a horizontal segment. Length = $|4 - (-1)| = 5$ units.

Segment RS: Connects R$(4, 3)$ and S$(4, -2)$. Same x-coordinate ($4$), so it's a vertical segment. Length = $|-2 - 3| = 5$ units.

Segment SP: Connects S$(4, -2)$ and P$(-1, -2)$. Same y-coordinate ($-2$), so it's a horizontal segment. Length = $|-1 - 4| = 5$ units.

All four sides have the same length (5 units). Since adjacent sides are perpendicular (vertical and horizontal segments are perpendicular), this shape is a square.

Finding the coordinates of the intersection of the diagonals:

The diagonals of the quadrilateral are PR and QS.

Diagonal PR connects P$(-1, -2)$ and R$(4, 3)$.

Diagonal QS connects Q$(-1, 3)$ and S$(4, -2)$.

In a parallelogram (which includes squares, rectangles, and rhombuses), the diagonals bisect each other. This means the intersection point is the midpoint of both diagonals.

We can find the midpoint of either diagonal using the midpoint formula. The midpoint of a segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

Using diagonal PR with P$(-1, -2)$ and R$(4, 3)$:

x-coordinate of midpoint = $\frac{-1 + 4}{2} = \frac{3}{2} = 1.5$

y-coordinate of midpoint = $\frac{-2 + 3}{2} = \frac{1}{2} = 0.5$

The midpoint of PR is $(1.5, 0.5)$.

Using diagonal QS with Q$(-1, 3)$ and S$(4, -2)$:

x-coordinate of midpoint = $\frac{-1 + 4}{2} = \frac{3}{2} = 1.5$

y-coordinate of midpoint = $\frac{3 + (-2)}{2} = \frac{1}{2} = 0.5$

The midpoint of QS is $(1.5, 0.5)$.

The midpoints are the same, as expected.

The shape formed by the points is a square.

The coordinates of the intersection of the diagonals are $(1.5, 0.5)$ or $\left(\frac{3}{2}, \frac{1}{2}\right)$.

Solution:

We are given three vertices of a square: A$(-2, 2)$, B$(-2, -1)$, and C$(1, -1)$. Let the fourth vertex be D$(x, y)$.

First, let's analyze the given points to understand the orientation and side length of the square.

Consider the segment AB with endpoints A$(-2, 2)$ and B$(-2, -1)$. Since the x-coordinates are the same ($-2$), this is a vertical segment.

The length of AB is the absolute difference of the y-coordinates: $|-1 - 2| = |-3| = 3$ units.

Consider the segment BC with endpoints B$(-2, -1)$ and C$(1, -1)$. Since the y-coordinates are the same ($-1$), this is a horizontal segment.

The length of BC is the absolute difference of the x-coordinates: $|1 - (-2)| = |1 + 2| = |3| = 3$ units.

Since AB and BC share the vertex B and are vertical and horizontal respectively, the angle at B is a right angle ($90^\circ$). Also, their lengths are equal (3 units), which is consistent with them being adjacent sides of a square.

For ABCD to be a square, the vertices must be connected in sequence. Given A, B, C, the next vertex is D such that CD is perpendicular to BC and parallel to AB, and AD is perpendicular to AB and parallel to BC. Also, the side lengths CD and AD must be 3 units.

Since BC is a horizontal segment from $x=-2$ to $x=1$ (a change of $+3$ in x), the segment AD must be a horizontal segment of length 3 units, starting from A$(-2, 2)$. Moving horizontally by $+3$ from A means adding 3 to the x-coordinate: $-2 + 3 = 1$. The y-coordinate remains the same, $2$. This gives the point $(1, 2)$.

Let's verify with the other side. Since AB is a vertical segment from $y=2$ to $y=-1$ (a change of $-3$ in y), the segment CD must be a vertical segment of length 3 units, starting from C$(1, -1)$. Moving vertically by $+3$ (to go up, parallel to the direction from B to A) from C means adding 3 to the y-coordinate: $-1 + 3 = 2$. The x-coordinate remains the same, $1$. This gives the point $(1, 2)$.

Both approaches give the same coordinates for D.

The coordinates of the fourth vertex D are $(1, 2)$.

Plotting the points:

A$(-2, 2)$: Move 2 units left, 2 units up.

B$(-2, -1)$: Move 2 units left, 1 unit down.

C$(1, -1)$: Move 1 unit right, 1 unit down.

D$(1, 2)$: Move 1 unit right, 2 units up.

Verifying that ABCD is a square by plotting:

Plot the four points A$(-2, 2)$, B$(-2, -1)$, C$(1, -1)$, and D$(1, 2)$. Join them in order A-B-C-D-A.

On the graph:

• The segment AB is vertical, from $y=2$ to $y=-1$ at $x=-2$. Length = 3 units.
• The segment BC is horizontal, from $x=-2$ to $x=1$ at $y=-1$. Length = 3 units.
• The segment CD is vertical, from $y=-1$ to $y=2$ at $x=1$. Length = 3 units.
• The segment DA is horizontal, from $x=1$ to $x=-2$ at $y=2$. Length = $|-2 - 1| = 3$ units.

All four sides have length 3 units.

Since AB is vertical and BC is horizontal, they are perpendicular, forming a right angle at B. Similarly, BC is horizontal and CD is vertical (at $x=1$), forming a right angle at C. CD is vertical and DA is horizontal (at $y=2$), forming a right angle at D. DA is horizontal and AB is vertical (at $x=-2$), forming a right angle at A.

Since all four sides are equal in length and all four angles are right angles, the quadrilateral ABCD is indeed a square.

Using the Coordinate System for Mapping and Data Representation:

The coordinate system, particularly the Cartesian coordinate system, is a fundamental tool for precisely locating points in space and for visualizing relationships between sets of data.

In mapping locations, a coordinate system allows us to specify the exact position of any point on a map or in physical space using a pair or set of numbers. For example, on a geographical map, latitude and longitude form a type of coordinate system to pinpoint locations on Earth's surface. In architecture or engineering, a coordinate system is used to define the position of points in a building or structure. By assigning numerical coordinates to locations, we can accurately measure distances, calculate areas, and understand spatial relationships.

In representing data graphically, the coordinate system (usually the 2D Cartesian plane) is used to plot pairs of related data points. Each point on the graph represents a single data entry, where the x-coordinate represents one variable's value and the y-coordinate represents the corresponding value of another variable. For example, if we have data on temperature ($y$) at different times of the day ($x$), we can plot points $(time, temperature)$ on a graph. Joining these points or observing their pattern allows us to visualize trends, relationships, and distributions in the data.

Essentially, the coordinate system provides a universal framework for translating positions or data pairs into numerical form, which can then be analyzed, communicated, or visualized consistently.

Simple Example: Describing Positions of Objects in a Room:

Imagine a rectangular room. We can set up a simple 2D Cartesian coordinate system in this room to describe the position of objects on the floor.

1. Choose one corner of the room as the origin $(0, 0)$. Let's say the corner where two walls meet.

2. Designate one of the walls starting from the origin as the positive x-axis and the adjacent wall as the positive y-axis.

3. Choose a unit of measurement, such as meters or feet.

Now, we can describe the position of objects by measuring their distance from these two reference walls (axes).

For example:

• A small rug might have one corner at the origin, so its coordinates are $(0, 0)$.
• A lamp is placed 2 meters away from the 'y-axis' wall (along the x-direction) and 1 meter away from the 'x-axis' wall (along the y-direction). Its position can be described by the coordinates $(2, 1)$.
• A chair is located 4 meters from the 'y-axis' wall and 3 meters from the 'x-axis' wall. Its position is $(4, 3)$.
• If there's a dustbin right against the 'y-axis' wall, 0.5 meters along that wall from the origin, its position could be $(0, 0.5)$ (assuming its width is negligible in the x-direction).

Using this coordinate system, we can accurately tell someone where each object is located relative to our chosen origin and axes.

Plotting the points:

Let the given points be A$(0, 6)$, B$(3, 6)$, C$(3, 0)$, and D$(0, 0)$. To plot these points on a graph sheet:

A$(0, 6)$: Start at the origin $(0, 0)$. The x-coordinate is 0, so stay on the y-axis. The y-coordinate is 6 (positive), so move 6 units up along the y-axis. Mark this point as A.

B$(3, 6)$: Start at the origin $(0, 0)$. The x-coordinate is 3 (positive), so move 3 units right along the x-axis. The y-coordinate is 6 (positive), so from the position on the x-axis, move 6 units up parallel to the y-axis. Mark this point as B.

C$(3, 0)$: Start at the origin $(0, 0)$. The x-coordinate is 3 (positive), so move 3 units right along the x-axis. The y-coordinate is 0, so stay on the x-axis. Mark this point as C.

D$(0, 0)$: This is the origin itself. Mark this point as D.

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

Identifying the shape formed:

Let's look at the coordinates and the segments formed:

Segment AB connects A$(0, 6)$ and B$(3, 6)$. Both points have the same y-coordinate ($y=6$). This is a horizontal segment.

Segment BC connects B$(3, 6)$ and C$(3, 0)$. Both points have the same x-coordinate ($x=3$). This is a vertical segment.

Segment CD connects C$(3, 0)$ and D$(0, 0)$. Both points have the same y-coordinate ($y=0$). This is a horizontal segment (along the x-axis).

Segment DA connects D$(0, 0)$ and A$(0, 6)$. Both points have the same x-coordinate ($x=0$). This is a vertical segment (along the y-axis).

Since adjacent sides (AB and BC, BC and CD, CD and DA, DA and AB) are horizontal and vertical, they are perpendicular to each other. This means all interior angles of the quadrilateral ABCD are $90^\circ$.

Let's find the lengths of the sides:

Length of AB = $|x_B - x_A| = |3 - 0| = |3| = 3$ units.

Length of BC = $|y_B - y_C| = |6 - 0| = |6| = 6$ units.

Length of CD = $|x_C - x_D| = |3 - 0| = |3| = 3$ units.

Length of DA = $|y_A - y_D| = |6 - 0| = |6| = 6$ units.

We see that opposite sides have equal lengths: AB = CD = 3 units and BC = DA = 6 units.

A quadrilateral with four right angles and opposite sides equal in length is a rectangle.

The shape formed is a rectangle.

Finding the perimeter:

The perimeter of a rectangle is the sum of the lengths of all its sides, or $2 \times (\text{length} + \text{breadth})$.

Let the length be the longer side, 6 units, and the breadth be the shorter side, 3 units.

Perimeter = AB + BC + CD + DA

Perimeter = $3 + 6 + 3 + 6 = 18$ units.

Alternatively, Perimeter = $2 \times (6 + 3) = 2 \times 9 = 18$ units.

The perimeter of the rectangle is 18 units.

Finding the area:

The area of a rectangle is given by the formula: Area = Length $\times$ Breadth.

Area = $6 \times 3 = 18$ square units.

The area of the rectangle is 18 square units.