Chapter 7 Coordinate Geometry

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

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(i)

Given points: $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (4, 1)$.

Using the distance formula:

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

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

$d = \sqrt{4 + 4}$

$d = \sqrt{8}$

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

The distance between the points (2, 3) and (4, 1) is $2\sqrt{2}$ units.

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(ii)

Given points: $(x_1, y_1) = (-5, 7)$ and $(x_2, y_2) = (-1, 3)$.

Using the distance formula:

$d = \sqrt{(-1 - (-5))^2 + (3 - 7)^2}$

$d = \sqrt{(-1 + 5)^2 + (-4)^2}$

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

$d = \sqrt{16 + 16}$

$d = \sqrt{32}$

$d = 4\sqrt{2}$ units.

The distance between the points (–5, 7) and (–1, 3) is $4\sqrt{2}$ units.

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(iii)

Given points: $(x_1, y_1) = (a, b)$ and $(x_2, y_2) = (-a, -b)$.

Using the distance formula:

$d = \sqrt{(-a - a)^2 + (-b - b)^2}$

$d = \sqrt{(-2a)^2 + (-2b)^2}$

$d = \sqrt{4a^2 + 4b^2}$

$d = \sqrt{4(a^2 + b^2)}$

$d = 2\sqrt{a^2 + b^2}$ units.

The distance between the points (a, b) and (–a, –b) is $2\sqrt{a^2 + b^2}$ units.

Part 1: Distance between points (0, 0) and (36, 15)

Let the points be $P(0, 0)$ and $Q(36, 15)$.

We use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Here, $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (36, 15)$.

Substituting the coordinates:

$d = \sqrt{(36 - 0)^2 + (15 - 0)^2}$

$d = \sqrt{(36)^2 + (15)^2}$

$d = \sqrt{1296 + 225}$

$d = \sqrt{1521}$

To find the square root of 1521:

We know $30^2 = 900$ and $40^2 = 1600$. The number ends in 1, so the square root must end in 1 or 9. Let's try 39.

$39 \times 39 = 1521$.

$d = 39$ units.

So, the distance between the points (0, 0) and (36, 15) is 39 units.

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Part 2: Distance between Town A and Town B

Yes, we can find the distance between the two towns A and B using the distance formula calculated above.

Let Town A be located at the origin (0, 0) on a coordinate plane.

Since Town B is located 36 km east and 15 km north of Town A, the coordinates of Town B can be represented as (36, 15).

The distance between Town A (0, 0) and Town B (36, 15) is the same as the distance calculated in Part 1.

Distance $AB = \sqrt{(36 - 0)^2 + (15 - 0)^2}$

Distance $AB = \sqrt{36^2 + 15^2}$

Distance $AB = \sqrt{1296 + 225}$

Distance $AB = \sqrt{1521}$

Distance $AB = 39$ km.

Therefore, the distance between Town A and Town B is 39 km.

Given:

Let the points be $A = (1, 5)$, $B = (2, 3)$, and $C = (– 2, – 11)$.

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To Find:

To determine if the points A, B, and C are collinear.

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Solution:

Three points are collinear if they lie on the same straight line. This condition can be checked using the distance formula: if the sum of the lengths of any two line segments connecting the points is equal to the length of the third line segment, then the points are collinear.

We calculate the distances between each pair of points using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Calculating the distance AB:

$AB = \sqrt{(2 - 1)^2 + (3 - 5)^2}$

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

$AB = \sqrt{1 + 4}$

$AB = \sqrt{5}$ units.

Calculating the distance BC:

$BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2}$

$BC = \sqrt{(-4)^2 + (-14)^2}$

$BC = \sqrt{16 + 196}$

$BC = \sqrt{212}$ units.

Calculating the distance AC:

$AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2}$

$AC = \sqrt{(-3)^2 + (-16)^2}$

$AC = \sqrt{9 + 256}$

$AC = \sqrt{265}$ units.

Now, we check if the sum of any two distances equals the third distance.

Check if $AB + BC = AC$.

$AB + BC = \sqrt{5} + \sqrt{212}$

$AC = \sqrt{265}$

Since $\sqrt{5} + \sqrt{212} \neq \sqrt{265}$, the condition $AB + BC = AC$ is not satisfied.

We can also check the other combinations, but it's clear that the sum of the two smaller distances ($\sqrt{5}$ and $\sqrt{212}$) does not equal the largest distance ($\sqrt{265}$).

Since the sum of the lengths of any two line segments is not equal to the length of the third segment, the points A, B, and C do not lie on the same line.

Therefore, the points (1, 5), (2, 3) and (– 2, – 11) are not collinear.

Given:

Let the points be $A = (5, -2)$, $B = (6, 4)$, and $C = (7, -2)$.

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To Check:

Whether the points A, B, and C are the vertices of an isosceles triangle.

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Solution:

An isosceles triangle is a triangle that has at least two sides of equal length. We will calculate the lengths of the sides AB, BC, and AC using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Calculating the length of side AB:

$AB = \sqrt{(6 - 5)^2 + (4 - (-2))^2}$

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

$AB = \sqrt{1^2 + 6^2}$

$AB = \sqrt{1 + 36}$

$AB = \sqrt{37}$ units.

Calculating the length of side BC:

$BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2}$

$BC = \sqrt{(1)^2 + (-6)^2}$

$BC = \sqrt{1 + 36}$

$BC = \sqrt{37}$ units.

Calculating the length of side AC:

$AC = \sqrt{(7 - 5)^2 + (-2 - (-2))^2}$

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

$AC = \sqrt{2^2 + 0^2}$

$AC = \sqrt{4 + 0}$

$AC = \sqrt{4} = 2$ units.

Comparing the lengths of the sides:

$AB = \sqrt{37}$

$BC = \sqrt{37}$

$AC = 2$

Since $AB = BC = \sqrt{37}$, two sides of the triangle formed by points A, B, and C are equal in length.

Therefore, the points (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Given:

From the figure (Fig. 7.8), the coordinates of the four friends are:

$A = (3, 4)$

$B = (6, 7)$

$C = (9, 4)$

$D = (6, 1)$

Champa believes ABCD is a square, while Chameli disagrees.

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To Find:

Using the distance formula, determine if ABCD is a square and state whether Champa or Chameli is correct.

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Solution:

To check if ABCD is a square, we need to calculate the lengths of all four sides (AB, BC, CD, DA) and the lengths of the two diagonals (AC, BD).

A quadrilateral is a square if:

1. All four sides are equal in length.

2. The two diagonals are equal in length.

We use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Calculating the lengths of the sides:

Length of AB:

$AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Length of BC:

$BC = \sqrt{(9 - 6)^2 + (4 - 7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Length of CD:

$CD = \sqrt{(6 - 9)^2 + (1 - 4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Length of DA:

$DA = \sqrt{(3 - 6)^2 + (4 - 1)^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ units.

Since $AB = BC = CD = DA = 3\sqrt{2}$, all four sides are equal.

Calculating the lengths of the diagonals:

Length of AC:

$AC = \sqrt{(9 - 3)^2 + (4 - 4)^2} = \sqrt{6^2 + 0^2} = \sqrt{36 + 0} = \sqrt{36} = 6$ units.

Length of BD:

$BD = \sqrt{(6 - 6)^2 + (1 - 7)^2} = \sqrt{0^2 + (-6)^2} = \sqrt{0 + 36} = \sqrt{36} = 6$ units.

Since $AC = BD = 6$, the two diagonals are equal.

Because all four sides are equal ($AB = BC = CD = DA = 3\sqrt{2}$) and both diagonals are equal ($AC = BD = 6$), the quadrilateral ABCD is a square.

Therefore, Champa's observation that ABCD is a square is correct.

Conclusion: Champa is correct.

We use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to find the lengths of the sides and diagonals of the quadrilateral formed by the given points.

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(i)

Let the points be $A(-1, -2)$, $B(1, 0)$, $C(-1, 2)$, and $D(-3, 0)$.

Calculating side lengths:

$AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} = \sqrt{(1+1)^2 + (0+2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$

$BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$

$CD = \sqrt{(-3 - (-1))^2 + (0 - 2)^2} = \sqrt{(-3+1)^2 + (-2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$

$DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} = \sqrt{(-1+3)^2 + (-2)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$

Since $AB = BC = CD = DA = 2\sqrt{2}$, all sides are equal.

Calculating diagonal lengths:

$AC = \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} = \sqrt{(-1+1)^2 + (2+2)^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4$

$BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$

Since all sides are equal ($AB=BC=CD=DA$) and the diagonals are equal ($AC=BD$), the quadrilateral formed by the points is a square.

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(ii)

Let the points be $A(-3, 5)$, $B(3, 1)$, $C(0, 3)$, and $D(-1, -4)$.

Calculating side lengths:

$AB = \sqrt{(3 - (-3))^2 + (1 - 5)^2} = \sqrt{(3+3)^2 + (-4)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}$

$BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$

$CD = \sqrt{(-1 - 0)^2 + (-4 - 3)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}$

$DA = \sqrt{(-3 - (-1))^2 + (5 - (-4))^2} = \sqrt{(-3+1)^2 + (5+4)^2} = \sqrt{(-2)^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85}$

Calculating diagonal lengths:

$AC = \sqrt{(0 - (-3))^2 + (3 - 5)^2} = \sqrt{(0+3)^2 + (-2)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$

$BD = \sqrt{(-1 - 3)^2 + (-4 - 1)^2} = \sqrt{(-4)^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41}$

Let's check for collinearity among points A, B, C.

$AC + BC = \sqrt{13} + \sqrt{13} = 2\sqrt{13}$

$AB = \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}$

Since $AC + BC = AB$, the points A, C, and B are collinear.

Therefore, the given points do not form a quadrilateral.

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(iii)

Let the points be $A(4, 5)$, $B(7, 6)$, $C(4, 3)$, and $D(1, 2)$.

Calculating side lengths:

$AB = \sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$

$BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$

$CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$

$DA = \sqrt{(4 - 1)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$

We observe that opposite sides are equal: $AB = CD = \sqrt{10}$ and $BC = DA = \sqrt{18}$.

Calculating diagonal lengths:

$AC = \sqrt{(4 - 4)^2 + (3 - 5)^2} = \sqrt{0^2 + (-2)^2} = \sqrt{4} = 2$

$BD = \sqrt{(1 - 7)^2 + (2 - 6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$

Since opposite sides are equal ($AB=CD$ and $BC=DA$) and the diagonals are unequal ($AC \neq BD$), the quadrilateral formed by the points is a parallelogram.

Given:

Two points $A(2, -5)$ and $B(-2, 9)$.

We need to find a point on the x-axis which is equidistant from A and B.

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To Find:

The coordinates of the point on the x-axis equidistant from A and B.

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Solution:

Let the point on the x-axis be $P(x, 0)$. Since any point on the x-axis has its y-coordinate as 0.

We are given that the point P is equidistant from points A and B. Therefore, $PA = PB$.

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

Calculating the distance PA:

$PA = \sqrt{(2 - x)^2 + (-5 - 0)^2}$

$PA = \sqrt{(2 - x)^2 + (-5)^2}$

$PA = \sqrt{(2 - x)^2 + 25}$

Calculating the distance PB:

$PB = \sqrt{(-2 - x)^2 + (9 - 0)^2}$

$PB = \sqrt{(-(2 + x))^2 + (9)^2}$

$PB = \sqrt{(2 + x)^2 + 81}$

Since $PA = PB$, we have $PA^2 = PB^2$.

$\left(\sqrt{(2 - x)^2 + 25}\right)^2 = \left(\sqrt{(2 + x)^2 + 81}\right)^2$

$(2 - x)^2 + 25 = (2 + x)^2 + 81$

Expanding the squared terms:

$(4 - 4x + x^2) + 25 = (4 + 4x + x^2) + 81$

$x^2 - 4x + 29 = x^2 + 4x + 85$

Subtract $x^2$ from both sides:

$-4x + 29 = 4x + 85$

Rearrange the terms to solve for $x$:

$29 - 85 = 4x + 4x$

$-56 = 8x$

$x = \frac{-56}{8}$

$x = -7$

So, the coordinates of the point P on the x-axis are $(x, 0) = (-7, 0)$.

Therefore, the required point on the x-axis is (-7, 0).

Given:

Two points $P(2, -3)$ and $Q(10, y)$.

The distance between P and Q is 10 units, i.e., $PQ = 10$.

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To Find:

The possible values of $y$.

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Solution:

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

Here, $(x_1, y_1) = (2, -3)$ and $(x_2, y_2) = (10, y)$, and $d = 10$.

Substituting these values into the distance formula:

$10 = \sqrt{(10 - 2)^2 + (y - (-3))^2}$

Simplify the expression inside the square root:

$10 = \sqrt{(8)^2 + (y + 3)^2}$

$10 = \sqrt{64 + (y + 3)^2}$

To eliminate the square root, square both sides of the equation:

$10^2 = \left(\sqrt{64 + (y + 3)^2}\right)^2$

$100 = 64 + (y + 3)^2$

Isolate the term $(y + 3)^2$:

$(y + 3)^2 = 100 - 64$

$(y + 3)^2 = 36$

Take the square root of both sides:

$\sqrt{(y + 3)^2} = \sqrt{36}$

$y + 3 = \pm 6$

We have two possible cases:

Case 1: $y + 3 = 6$

$y = 6 - 3$

$y = 3$

Case 2: $y + 3 = -6$

$y = -6 - 3$

$y = -9$

Therefore, the possible values of $y$ for which the distance between P(2, –3) and Q(10, y) is 10 units are 3 and –9.

Given:

Points $P(5, -3)$, $Q(0, 1)$, and $R(x, 6)$.

Point Q is equidistant from P and R, which means $QP = QR$.

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To Find:

1. The value(s) of $x$.

2. The distances QR and PR.

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Solution:

Since Q is equidistant from P and R, we have $QP = QR$.

This implies $QP^2 = QR^2$.

We use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, or $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.

Calculating $QP^2$:

Points are $Q(0, 1)$ and $P(5, -3)$.

$QP^2 = (5 - 0)^2 + (-3 - 1)^2$

$QP^2 = (5)^2 + (-4)^2$

$QP^2 = 25 + 16 = 41$

Calculating $QR^2$:

Points are $Q(0, 1)$ and $R(x, 6)$.

$QR^2 = (x - 0)^2 + (6 - 1)^2$

$QR^2 = x^2 + (5)^2$

$QR^2 = x^2 + 25$

Now, setting $QP^2 = QR^2$:

$41 = x^2 + 25$

$x^2 = 41 - 25$

$x^2 = 16$

$x = \pm \sqrt{16}$

$x = 4$ or $x = -4$.

So, the possible values of $x$ are 4 and -4.

The point R can be either $(4, 6)$ or $(-4, 6)$.

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Finding the distance QR:

We know $QR^2 = x^2 + 25$. Since $x^2 = 16$,

$QR^2 = 16 + 25 = 41$

$QR = \sqrt{41}$ units.

The distance QR is $\sqrt{41}$ units (this is true for both $x=4$ and $x=-4$).

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Finding the distance PR:

We need to calculate PR for both possible coordinates of R.

Case 1: R is $(4, 6)$

Points are $P(5, -3)$ and $R(4, 6)$.

$PR = \sqrt{(4 - 5)^2 + (6 - (-3))^2}$

$PR = \sqrt{(-1)^2 + (6 + 3)^2}$

$PR = \sqrt{1^2 + 9^2}$

$PR = \sqrt{1 + 81} = \sqrt{82}$ units.

Case 2: R is $(-4, 6)$

Points are $P(5, -3)$ and $R(-4, 6)$.

$PR = \sqrt{(-4 - 5)^2 + (6 - (-3))^2}$

$PR = \sqrt{(-9)^2 + (6 + 3)^2}$

$PR = \sqrt{(-9)^2 + 9^2}$

$PR = \sqrt{81 + 81} = \sqrt{162}$

$PR = \sqrt{81 \times 2} = 9\sqrt{2}$ units.

Thus, the distances PR are $\sqrt{82}$ units or $9\sqrt{2}$ units.

Given:

Let $P(x, y)$ be a point that is equidistant from the points $A(3, 6)$ and $B(-3, 4)$.

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To Find:

A relation between $x$ and $y$.

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Solution:

Since point $P(x, y)$ is equidistant from points $A(3, 6)$ and $B(-3, 4)$, the distance $PA$ must be equal to the distance $PB$.

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

Calculating the distance PA:

$PA = \sqrt{(3 - x)^2 + (6 - y)^2}$

Calculating the distance PB:

$PB = \sqrt{(-3 - x)^2 + (4 - y)^2}$

According to the given condition, $PA = PB$.

This implies $PA^2 = PB^2$. Squaring both sides eliminates the square roots:

$(3 - x)^2 + (6 - y)^2 = (-3 - x)^2 + (4 - y)^2$

Expand the terms:

$(9 - 6x + x^2) + (36 - 12y + y^2) = ((-1)(3 + x))^2 + (16 - 8y + y^2)$

$x^2 - 6x + y^2 - 12y + 9 + 36 = (3 + x)^2 + y^2 - 8y + 16$

$x^2 - 6x + y^2 - 12y + 45 = (9 + 6x + x^2) + y^2 - 8y + 16$

$x^2 - 6x + y^2 - 12y + 45 = x^2 + 6x + y^2 - 8y + 25$

Subtract $x^2$ and $y^2$ from both sides:

$-6x - 12y + 45 = 6x - 8y + 25$

Rearrange the terms to bring variables to one side and constants to the other:

$45 - 25 = 6x + 6x - 8y + 12y$

$20 = 12x + 4y$

Divide the entire equation by the greatest common divisor, which is 4:

$\frac{20}{4} = \frac{12x}{4} + \frac{4y}{4}$

$5 = 3x + y$

This can also be written as $3x + y - 5 = 0$.

Therefore, the required relation between $x$ and $y$ is $3x + y = 5$.

Given:

The line segment joins the points $A(4, -3)$ and $B(8, 5)$.

The ratio of internal division is $m_1 : m_2 = 3 : 1$.

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To Find:

The coordinates of the point which divides the line segment AB internally in the ratio 3 : 1.

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Solution:

Let $P(x, y)$ be the point that divides the line segment joining $A(x_1, y_1) = (4, -3)$ and $B(x_2, y_2) = (8, 5)$ internally in the ratio $m_1 : m_2 = 3 : 1$.

We use the section formula for internal division:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

Substitute the given values:

$x_1 = 4$, $y_1 = -3$

$x_2 = 8$, $y_2 = 5$

$m_1 = 3$, $m_2 = 1$

Calculating the x-coordinate:

$x = \frac{3(8) + 1(4)}{3 + 1}$

$x = \frac{24 + 4}{4}$

$x = \frac{28}{4}$

$x = 7$

Calculating the y-coordinate:

$y = \frac{3(5) + 1(-3)}{3 + 1}$

$y = \frac{15 - 3}{4}$

$y = \frac{12}{4}$

$y = 3$

So, the coordinates of the point P are $(7, 3)$.

Therefore, the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally are (7, 3).

Given:

The line segment is formed by joining points $A(-6, 10)$ and $B(3, -8)$.

The point $P(-4, 6)$ divides this line segment.

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To Find:

The ratio in which the point P divides the line segment AB.

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Solution:

Let the point $P(-4, 6)$ divide the line segment joining $A(x_1, y_1) = (-6, 10)$ and $B(x_2, y_2) = (3, -8)$ in the ratio $m_1 : m_2$.

Using the section formula for internal division, the coordinates of the point P are given by:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

We are given the coordinates of P as $(-4, 6)$. We can use either the x-coordinate or the y-coordinate formula to find the ratio.

Using the x-coordinate:

$x = -4$

$-4 = \frac{m_1(3) + m_2(-6)}{m_1 + m_2}$

Multiply both sides by $(m_1 + m_2)$:

$-4(m_1 + m_2) = 3m_1 - 6m_2$

$-4m_1 - 4m_2 = 3m_1 - 6m_2$

Rearrange the terms to group $m_1$ and $m_2$:

$-4m_2 + 6m_2 = 3m_1 + 4m_1$

$2m_2 = 7m_1$

To find the ratio $m_1 : m_2$, we rearrange the equation:

$\frac{m_1}{m_2} = \frac{2}{7}$

So, the ratio $m_1 : m_2$ is $2 : 7$.

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Alternate Solution (using the y-coordinate):

We use the y-coordinate formula:

$y = 6$

$6 = \frac{m_1(-8) + m_2(10)}{m_1 + m_2}$

Multiply both sides by $(m_1 + m_2)$:

$6(m_1 + m_2) = -8m_1 + 10m_2$

$6m_1 + 6m_2 = -8m_1 + 10m_2$

Rearrange the terms:

$6m_1 + 8m_1 = 10m_2 - 6m_2$

$14m_1 = 4m_2$

Rearrange to find the ratio $m_1 : m_2$:

$\frac{m_1}{m_2} = \frac{4}{14}$

Simplify the fraction:

$\frac{m_1}{m_2} = \frac{2}{7}$

The ratio $m_1 : m_2$ is $2 : 7$.

Both methods yield the same result.

Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.

Given:

The line segment joins the points $A(2, -2)$ and $B(-7, 4)$.

We need to find the points of trisection.

---

To Find:

The coordinates of the two points that divide the line segment AB into three equal parts.

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Solution:

Let P and Q be the points of trisection of the line segment joining A(2, -2) and B(-7, 4).

This means that AP = PQ = QB.

Point P divides the line segment AB internally in the ratio $AP : PB = 1 : (PQ + QB) = 1 : (1 + 1) = 1 : 2$.

Point Q divides the line segment AB internally in the ratio $AQ : QB = (AP + PQ) : 1 = (1 + 1) : 1 = 2 : 1$.

We use the section formula to find the coordinates of P and Q. The coordinates $(x, y)$ of a point dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m_1 : m_2$ are:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

Coordinates of P (dividing AB in the ratio 1 : 2):

Here, $(x_1, y_1) = (2, -2)$, $(x_2, y_2) = (-7, 4)$, $m_1 = 1$, $m_2 = 2$.

$x_P = \frac{1(-7) + 2(2)}{1 + 2} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1$

$y_P = \frac{1(4) + 2(-2)}{1 + 2} = \frac{4 - 4}{3} = \frac{0}{3} = 0$

So, the coordinates of point P are (-1, 0).

Coordinates of Q (dividing AB in the ratio 2 : 1):

Here, $(x_1, y_1) = (2, -2)$, $(x_2, y_2) = (-7, 4)$, $m_1 = 2$, $m_2 = 1$.

$x_Q = \frac{2(-7) + 1(2)}{2 + 1} = \frac{-14 + 2}{3} = \frac{-12}{3} = -4$

$y_Q = \frac{2(4) + 1(-2)}{2 + 1} = \frac{8 - 2}{3} = \frac{6}{3} = 2$

So, the coordinates of point Q are (-4, 2).

Therefore, the coordinates of the points of trisection of the line segment joining A(2, – 2) and B(– 7, 4) are (-1, 0) and (-4, 2).

Given:

The line segment joins the points $A(5, -6)$ and $B(-1, -4)$.

This line segment is divided by the y-axis.

---

To Find:

1. The ratio in which the y-axis divides the line segment AB.

2. The coordinates of the point of intersection.

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Solution:

Let the y-axis divide the line segment joining $A(5, -6)$ and $B(-1, -4)$ in the ratio $m_1 : m_2$.

Let the point of intersection on the y-axis be $P(x, y)$. Since the point lies on the y-axis, its x-coordinate must be 0. So, the point of intersection is $P(0, y)$.

Using the section formula, the coordinates of the point P dividing the line segment joining $(x_1, y_1) = (5, -6)$ and $(x_2, y_2) = (-1, -4)$ in the ratio $m_1 : m_2$ are given by:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

We know the x-coordinate of the intersection point P is 0. So, we use the formula for the x-coordinate:

$0 = \frac{m_1(-1) + m_2(5)}{m_1 + m_2}$

Multiply both sides by $(m_1 + m_2)$:

$0 \times (m_1 + m_2) = -m_1 + 5m_2$

$0 = -m_1 + 5m_2$

$m_1 = 5m_2$

Rearrange to find the ratio $m_1 : m_2$:

$\frac{m_1}{m_2} = \frac{5}{1}$

Therefore, the ratio in which the y-axis divides the line segment is $5 : 1$.

---

Now, we find the coordinates of the point of intersection P(0, y) using the formula for the y-coordinate and the ratio $m_1 = 5$, $m_2 = 1$.

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

$y = \frac{5(-4) + 1(-6)}{5 + 1}$

$y = \frac{-20 - 6}{6}$

$y = \frac{-26}{6}$

Simplify the fraction:

$y = -\frac{13}{3}$

So, the coordinates of the point of intersection are $(0, y) = (0, -13/3)$.

Therefore, the point of intersection is $(0, -13/3)$.

Given:

The points $A(6, 1)$, $B(8, 2)$, $C(9, 4)$, and $D(p, 3)$ are the vertices of a parallelogram ABCD, taken in order.

---

To Find:

The value of $p$.

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Solution:

We know that the diagonals of a parallelogram bisect each other. This means that the midpoint of the diagonal AC is the same as the midpoint of the diagonal BD.

The midpoint formula for a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Midpoint of diagonal AC:

Using points $A(6, 1)$ and $C(9, 4)$: $(x_1, y_1) = (6, 1)$, $(x_2, y_2) = (9, 4)$.

Midpoint of AC = $\left( \frac{6 + 9}{2}, \frac{1 + 4}{2} \right) = \left( \frac{15}{2}, \frac{5}{2} \right)$

Midpoint of diagonal BD:

Using points $B(8, 2)$ and $D(p, 3)$: $(x_1, y_1) = (8, 2)$, $(x_2, y_2) = (p, 3)$.

Midpoint of BD = $\left( \frac{8 + p}{2}, \frac{2 + 3}{2} \right) = \left( \frac{8 + p}{2}, \frac{5}{2} \right)$

Since the midpoints of AC and BD are the same point:

$\left( \frac{15}{2}, \frac{5}{2} \right) = \left( \frac{8 + p}{2}, \frac{5}{2} \right)$

Equating the x-coordinates:

$\frac{15}{2} = \frac{8 + p}{2}$

Multiply both sides by 2:

$15 = 8 + p$

Solve for $p$:

$p = 15 - 8$

$p = 7$

(Equating the y-coordinates, $\frac{5}{2} = \frac{5}{2}$, which is consistent.)

Therefore, the value of $p$ is 7.

Given:

The line segment joins the points $A(-1, 7)$ and $B(4, -3)$.

The ratio of internal division is $m_1 : m_2 = 2 : 3$.

---

To Find:

The coordinates of the point which divides the line segment AB internally in the ratio 2 : 3.

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Solution:

Let $P(x, y)$ be the point that divides the line segment joining $A(x_1, y_1) = (-1, 7)$ and $B(x_2, y_2) = (4, -3)$ internally in the ratio $m_1 : m_2 = 2 : 3$.

We use the section formula for internal division:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

Substitute the given values:

$x_1 = -1$, $y_1 = 7$

$x_2 = 4$, $y_2 = -3$

$m_1 = 2$, $m_2 = 3$

Calculating the x-coordinate:

$x = \frac{2(4) + 3(-1)}{2 + 3}$

$x = \frac{8 - 3}{5}$

$x = \frac{5}{5}$

$x = 1$

Calculating the y-coordinate:

$y = \frac{2(-3) + 3(7)}{2 + 3}$

$y = \frac{-6 + 21}{5}$

$y = \frac{15}{5}$

$y = 3$

So, the coordinates of the point P are $(1, 3)$.

Therefore, the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3 are (1, 3).

Given:

The line segment joins the points $A(4, -1)$ and $B(-2, -3)$.

We need to find the points of trisection.

---

To Find:

The coordinates of the two points that divide the line segment AB into three equal parts.

---

Solution:

Let P and Q be the points of trisection of the line segment joining $A(4, -1)$ and $B(-2, -3)$.

This means that AP = PQ = QB.

Point P divides the line segment AB internally in the ratio $AP : PB = 1 : (PQ + QB) = 1 : (1 + 1) = 1 : 2$.

Point Q divides the line segment AB internally in the ratio $AQ : QB = (AP + PQ) : 1 = (1 + 1) : 1 = 2 : 1$.

We use the section formula to find the coordinates of P and Q. The coordinates $(x, y)$ of a point dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m_1 : m_2$ are:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

Coordinates of P (dividing AB in the ratio 1 : 2):

Here, $(x_1, y_1) = (4, -1)$, $(x_2, y_2) = (-2, -3)$, $m_1 = 1$, $m_2 = 2$.

$x_P = \frac{1(-2) + 2(4)}{1 + 2} = \frac{-2 + 8}{3} = \frac{6}{3} = 2$

$y_P = \frac{1(-3) + 2(-1)}{1 + 2} = \frac{-3 - 2}{3} = \frac{-5}{3}$

So, the coordinates of point P are $(2, -5/3)$.

Coordinates of Q (dividing AB in the ratio 2 : 1):

Here, $(x_1, y_1) = (4, -1)$, $(x_2, y_2) = (-2, -3)$, $m_1 = 2$, $m_2 = 1$.

$x_Q = \frac{2(-2) + 1(4)}{2 + 1} = \frac{-4 + 4}{3} = \frac{0}{3} = 0$

$y_Q = \frac{2(-3) + 1(-1)}{2 + 1} = \frac{-6 - 1}{3} = \frac{-7}{3}$

So, the coordinates of point Q are $(0, -7/3)$.

Therefore, the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3) are $(2, -5/3)$ and $(0, -7/3)$.

Given:

Let the rectangular ground be represented on a coordinate plane with AD along the y-axis and AB along the x-axis. The lines are 1m apart, so the line numbers correspond to the x-coordinates.

The total distance along AD is 100 m (100 flower pots at 1m distance).

Niharika runs on the 2nd line (x = 2).

Distance covered by Niharika along AD = $\frac{1}{4} \times 100 = 25$ m.

Coordinates of the green flag posted by Niharika (let's call this point G) are $(2, 25)$.

Preet runs on the 8th line (x = 8).

Distance covered by Preet along AD = $\frac{1}{5} \times 100 = 20$ m.

Coordinates of the red flag posted by Preet (let's call this point R) are $(8, 20)$.

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To Find:

1. The distance between the green flag (G) and the red flag (R).

2. The coordinates where Rashmi should post her blue flag, which is exactly halfway between G and R.

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Solution:

1. Distance between the flags (GR):

We use the distance formula to find the distance between G(2, 25) and R(8, 20).

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$GR = \sqrt{(8 - 2)^2 + (20 - 25)^2}$

$GR = \sqrt{(6)^2 + (-5)^2}$

$GR = \sqrt{36 + 25}$

$GR = \sqrt{61}$ m.

The distance between both the flags is $\sqrt{61}$ m.

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2. Position of the blue flag (Midpoint of GR):

Rashmi posts her blue flag exactly halfway between the line segment joining G(2, 25) and R(8, 20). Let the coordinates of this point be M(x, y).

We use the midpoint formula:

$M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

$x = \frac{2 + 8}{2} = \frac{10}{2} = 5$

$y = \frac{25 + 20}{2} = \frac{45}{2} = 22.5$

The coordinates of the point where Rashmi should post her blue flag are $(5, 22.5)$.

This means Rashmi should post her flag on the 5th line at a distance of 22.5 m from the starting point along the line AD.

Given:

The line segment joins the points $A(-3, 10)$ and $B(6, -8)$.

The point $P(-1, 6)$ divides this line segment.

---

To Find:

The ratio in which the point P divides the line segment AB.

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Solution:

Let the point $P(-1, 6)$ divide the line segment joining $A(x_1, y_1) = (-3, 10)$ and $B(x_2, y_2) = (6, -8)$ in the ratio $m_1 : m_2$.

Using the section formula for internal division, the coordinates of the point P are given by:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

We are given the coordinates of P as $(-1, 6)$. We can use either the x-coordinate or the y-coordinate formula to find the ratio.

Using the x-coordinate:

$x = -1$

$-1 = \frac{m_1(6) + m_2(-3)}{m_1 + m_2}$

Multiply both sides by $(m_1 + m_2)$:

$-1(m_1 + m_2) = 6m_1 - 3m_2$

$-m_1 - m_2 = 6m_1 - 3m_2$

Rearrange the terms to group $m_1$ and $m_2$:

$-m_2 + 3m_2 = 6m_1 + m_1$

$2m_2 = 7m_1$

To find the ratio $m_1 : m_2$, we rearrange the equation:

$\frac{m_1}{m_2} = \frac{2}{7}$

So, the ratio $m_1 : m_2$ is $2 : 7$.

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Alternate Solution (using the y-coordinate):

Using the y-coordinate:

$y = 6$

$6 = \frac{m_1(-8) + m_2(10)}{m_1 + m_2}$

Multiply both sides by $(m_1 + m_2)$:

$6(m_1 + m_2) = -8m_1 + 10m_2$

$6m_1 + 6m_2 = -8m_1 + 10m_2$

Rearrange the terms:

$6m_1 + 8m_1 = 10m_2 - 6m_2$

$14m_1 = 4m_2$

Rearrange to find the ratio $m_1 : m_2$:

$\frac{m_1}{m_2} = \frac{4}{14}$

Simplify the fraction:

$\frac{m_1}{m_2} = \frac{2}{7}$

The ratio $m_1 : m_2$ is $2 : 7$.

Both methods yield the same result.

Therefore, the point (– 1, 6) divides the line segment joining the points (– 3, 10) and (6, – 8) in the ratio 2 : 7.

Given:

The line segment joins the points $A(1, -5)$ and $B(-4, 5)$.

This line segment is divided by the x-axis.

---

To Find:

1. The ratio in which the x-axis divides the line segment AB.

2. The coordinates of the point of division.

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Solution:

Let the x-axis divide the line segment joining $A(1, -5)$ and $B(-4, 5)$ in the ratio $m_1 : m_2$.

Let the point of intersection on the x-axis be $P(x, y)$. Since the point lies on the x-axis, its y-coordinate must be 0. So, the point of intersection is $P(x, 0)$.

Using the section formula, the coordinates of the point P dividing the line segment joining $(x_1, y_1) = (1, -5)$ and $(x_2, y_2) = (-4, 5)$ in the ratio $m_1 : m_2$ are given by:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

We know the y-coordinate of the intersection point P is 0. So, we use the formula for the y-coordinate:

$0 = \frac{m_1(5) + m_2(-5)}{m_1 + m_2}$

Multiply both sides by $(m_1 + m_2)$:

$0 \times (m_1 + m_2) = 5m_1 - 5m_2$

$0 = 5m_1 - 5m_2$

$5m_1 = 5m_2$

Rearrange to find the ratio $m_1 : m_2$:

$\frac{m_1}{m_2} = \frac{5}{5} = \frac{1}{1}$

Therefore, the ratio in which the x-axis divides the line segment is $1 : 1$.

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Now, we find the coordinates of the point of division P(x, 0) using the formula for the x-coordinate and the ratio $m_1 = 1$, $m_2 = 1$.

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$x = \frac{1(-4) + 1(1)}{1 + 1}$

$x = \frac{-4 + 1}{2}$

$x = \frac{-3}{2}$

So, the x-coordinate is $-3/2$, and the y-coordinate is 0.

Therefore, the coordinates of the point of division are $(-3/2, 0)$.

(Note: Since the ratio is 1:1, the point of division is the midpoint of the line segment AB.)

Given:

Let the vertices of the parallelogram, taken in order, be $A(1, 2)$, $B(4, y)$, $C(x, 6)$, and $D(3, 5)$.

---

To Find:

The values of $x$ and $y$.

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Solution:

We know that the diagonals of a parallelogram bisect each other. This means the midpoint of diagonal AC is the same as the midpoint of diagonal BD.

The midpoint formula for a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Midpoint of diagonal AC:

Using points $A(1, 2)$ and $C(x, 6)$:

Midpoint of AC = $\left( \frac{1 + x}{2}, \frac{2 + 6}{2} \right) = \left( \frac{1 + x}{2}, \frac{8}{2} \right) = \left( \frac{1 + x}{2}, 4 \right)$

Midpoint of diagonal BD:

Using points $B(4, y)$ and $D(3, 5)$:

Midpoint of BD = $\left( \frac{4 + 3}{2}, \frac{y + 5}{2} \right) = \left( \frac{7}{2}, \frac{y + 5}{2} \right)$

Since the midpoints of AC and BD must be the same point for a parallelogram:

Midpoint of AC = Midpoint of BD

$\left( \frac{1 + x}{2}, 4 \right) = \left( \frac{7}{2}, \frac{y + 5}{2} \right)$

Equating the corresponding coordinates:

Equating the x-coordinates:

$\frac{1 + x}{2} = \frac{7}{2}$

Multiply both sides by 2:

$1 + x = 7$

$x = 7 - 1$

$x = 6$

Equating the y-coordinates:

$4 = \frac{y + 5}{2}$

Multiply both sides by 2:

$8 = y + 5$

$y = 8 - 5$

$y = 3$

Therefore, the values are $x = 6$ and $y = 3$.

Given:

AB is the diameter of a circle.

The coordinates of the centre of the circle (let's call it C) are $(2, -3)$.

The coordinates of point B are $(1, 4)$.

---

To Find:

The coordinates of point A.

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Solution:

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

Since AB is the diameter of the circle, the centre C must be the midpoint of the diameter AB.

We use the midpoint formula. The midpoint of a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Here, the midpoint is the centre $C(2, -3)$, and the endpoints of the diameter are $A(x, y)$ and $B(1, 4)$.

Applying the midpoint formula to the segment AB:

Coordinates of C = $\left( \frac{x + 1}{2}, \frac{y + 4}{2} \right)$

We are given that the coordinates of the centre C are $(2, -3)$. Therefore, we can equate the coordinates:

Equating the x-coordinates:

$\frac{x + 1}{2} = 2$

Multiply both sides by 2:

$x + 1 = 4$

$x = 4 - 1$

$x = 3$

Equating the y-coordinates:

$\frac{y + 4}{2} = -3$

Multiply both sides by 2:

$y + 4 = -6$

$y = -6 - 4$

$y = -10$

So, the coordinates of point A are $(x, y) = (3, -10)$.

Therefore, the coordinates of point A are (3, -10).

Given:

The coordinates of points A and B are $A(-2, -2)$ and $B(2, -4)$.

Point P lies on the line segment AB.

The condition $AP = \frac{3}{7} AB$ holds.

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To Find:

The coordinates of point P.

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Solution:

Since P lies on the line segment AB, it divides AB internally.

We are given $AP = \frac{3}{7} AB$.

We can find the length of the segment PB:

$PB = AB - AP$

$PB = AB - \frac{3}{7} AB$

$PB = \left(1 - \frac{3}{7}\right) AB$

$PB = \frac{7 - 3}{7} AB = \frac{4}{7} AB$

Now, we find the ratio in which P divides AB, which is $AP : PB$.

$AP : PB = \frac{3}{7} AB : \frac{4}{7} AB$

Dividing both parts by $\frac{1}{7} AB$, we get the ratio:

$AP : PB = 3 : 4$

So, point P divides the line segment AB internally in the ratio $m_1 : m_2 = 3 : 4$.

Let the coordinates of A be $(x_1, y_1) = (-2, -2)$ and the coordinates of B be $(x_2, y_2) = (2, -4)$.

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

Using the section formula for internal division:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

Substitute the values $m_1 = 3$, $m_2 = 4$, $x_1 = -2$, $y_1 = -2$, $x_2 = 2$, $y_2 = -4$:

$x = \frac{3(2) + 4(-2)}{3 + 4}$

$x = \frac{6 - 8}{7}$

$x = \frac{-2}{7}$

$y = \frac{3(-4) + 4(-2)}{3 + 4}$

$y = \frac{-12 - 8}{7}$

$y = \frac{-20}{7}$

Thus, the coordinates of point P are $(-\frac{2}{7}, -\frac{20}{7})$.

Therefore, the coordinates of P are $(-\frac{2}{7}, -\frac{20}{7})$.

Given:

The line segment joins the points $A(-2, 2)$ and $B(2, 8)$.

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To Find:

The coordinates of the three points that divide the line segment AB into four equal parts.

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Solution:

Let P, Q, and R be the points that divide the line segment AB into four equal parts such that $AP = PQ = QR = RB$.

Point Q divides the line segment AB into two equal parts ($AQ = QB$), so Q is the midpoint of AB.

Point P divides the line segment AQ into two equal parts ($AP = PQ$), so P is the midpoint of AQ.

Point R divides the line segment QB into two equal parts ($QR = RB$), so R is the midpoint of QB.

We use the midpoint formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.

Coordinates of Q (Midpoint of AB):

Using $A(-2, 2)$ and $B(2, 8)$:

$x_Q = \frac{-2 + 2}{2} = \frac{0}{2} = 0$

$y_Q = \frac{2 + 8}{2} = \frac{10}{2} = 5$

So, the coordinates of point Q are (0, 5).

Coordinates of P (Midpoint of AQ):

Using $A(-2, 2)$ and $Q(0, 5)$:

$x_P = \frac{-2 + 0}{2} = \frac{-2}{2} = -1$

$y_P = \frac{2 + 5}{2} = \frac{7}{2}$

So, the coordinates of point P are $(-1, 7/2)$.

Coordinates of R (Midpoint of QB):

Using $Q(0, 5)$ and $B(2, 8)$:

$x_R = \frac{0 + 2}{2} = \frac{2}{2} = 1$

$y_R = \frac{5 + 8}{2} = \frac{13}{2}$

So, the coordinates of point R are $(1, 13/2)$.

Therefore, the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts are $(-1, 7/2)$, (0, 5), and $(1, 13/2)$.

Given:

Let the vertices of the rhombus, taken in order, be $A(3, 0)$, $B(4, 5)$, $C(-1, 4)$, and $D(-2, -1)$.

---

To Find:

The area of the rhombus ABCD.

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Solution:

The diagonals of the rhombus are AC and BD.

We need to find the lengths of the diagonals using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Length of diagonal AC:

Using points $A(3, 0)$ and $C(-1, 4)$:

$AC = \sqrt{(-1 - 3)^2 + (4 - 0)^2}$

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

$AC = \sqrt{16 + 16}$

$AC = \sqrt{32} = 4\sqrt{2}$ units.

Length of diagonal BD:

Using points $B(4, 5)$ and $D(-2, -1)$:

$BD = \sqrt{(-2 - 4)^2 + (-1 - 5)^2}$

$BD = \sqrt{(-6)^2 + (-6)^2}$

$BD = \sqrt{36 + 36}$

$BD = \sqrt{72} = 6\sqrt{2}$ units.

Now, we use the formula for the area of a rhombus:

Area of rhombus = $\frac{1}{2} \times (\text{product of diagonals})$

Area = $\frac{1}{2} \times AC \times BD$

Area = $\frac{1}{2} \times (4\sqrt{2}) \times (6\sqrt{2})$

Area = $\frac{1}{2} \times (4 \times 6) \times (\sqrt{2} \times \sqrt{2})$

Area = $\frac{1}{2} \times 24 \times 2$

Area = $\frac{1}{2} \times 48$

Area = 24 square units.

Therefore, the area of the rhombus is 24 square units.

Given:

The vertices of the triangle are $A(1, -1)$, $B(-4, 6)$, and $C(-3, -5)$.

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To Find:

The area of the triangle ABC.

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Solution:

Let the vertices be $(x_1, y_1) = (1, -1)$, $(x_2, y_2) = (-4, 6)$, and $(x_3, y_3) = (-3, -5)$.

The formula for the area of a triangle with given vertices is:

Area = $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substitute the coordinates of the vertices into the formula:

Area = $\frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)(-1 - 6)|$

Simplify the terms inside the absolute value:

Area = $\frac{1}{2} |1(6 + 5) + (-4)(-5 + 1) + (-3)(-7)|$

Area = $\frac{1}{2} |1(11) + (-4)(-4) + (-3)(-7)|$

Area = $\frac{1}{2} |11 + 16 + 21|$

Area = $\frac{1}{2} |48|$

Since the area must be positive, the absolute value of 48 is 48.

Area = $\frac{1}{2} \times 48$

Area = 24 square units.

Therefore, the area of the triangle is 24 square units.

Given:

The vertices of the triangle are $A(5, 2)$, $B(4, 7)$, and $C(7, -4)$.

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To Find:

The area of the triangle ABC.

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Solution:

Let the vertices be $(x_1, y_1) = (5, 2)$, $(x_2, y_2) = (4, 7)$, and $(x_3, y_3) = (7, -4)$.

The formula for the area of a triangle with given vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is:

$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substitute the coordinates of the vertices A, B, and C into the formula:

$Area = \frac{1}{2} |5(7 - (-4)) + 4(-4 - 2) + 7(2 - 7)|$

Simplify the terms inside the absolute value:

$Area = \frac{1}{2} |5(7 + 4) + 4(-6) + 7(-5)|$

$Area = \frac{1}{2} |5(11) + (-24) + (-35)|$

$Area = \frac{1}{2} |55 - 24 - 35|$

$Area = \frac{1}{2} |55 - 59|$

$Area = \frac{1}{2} |-4|$

Since the area must be positive, the absolute value of -4 is 4.

$Area = \frac{1}{2} \times 4$

$Area = 2$ square units.

Therefore, the area of the triangle is 2 square units.

Given:

The vertices of the triangle are $P(-1.5, 3)$, $Q(6, -2)$, and $R(-3, 4)$.

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To Find:

The area of the triangle PQR.

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Solution:

Let the vertices be $(x_1, y_1) = (-1.5, 3)$, $(x_2, y_2) = (6, -2)$, and $(x_3, y_3) = (-3, 4)$.

The formula for the area of a triangle with given vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is:

$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substitute the coordinates of the vertices P, Q, and R into the formula:

$Area = \frac{1}{2} |(-1.5)(-2 - 4) + (6)(4 - 3) + (-3)(3 - (-2))|$

Simplify the terms inside the absolute value:

$Area = \frac{1}{2} |(-1.5)(-6) + (6)(1) + (-3)(3 + 2)|$

$Area = \frac{1}{2} |(-1.5)(-6) + 6(1) + (-3)(5)|$

$Area = \frac{1}{2} |9 + 6 - 15|$

$Area = \frac{1}{2} |15 - 15|$

$Area = \frac{1}{2} |0|$

$Area = 0$ square units.

Since the area of the triangle is 0, the points P, Q, and R are collinear (they lie on the same straight line).

Therefore, the area of the triangle formed by the given points is 0 square units.

Given:

The points are $A(2, 3)$, $B(4, k)$, and $C(6, -3)$.

The points A, B, and C are collinear.

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To Find:

The value of $k$.

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Solution:

If three points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are collinear, then the area of the triangle formed by these points is zero.

The formula for the area of a triangle is:

$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

For collinear points, Area = 0. This implies:

$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$

Substitute the coordinates of A(2, 3), B(4, k), and C(6, -3) into the condition:

$(x_1, y_1) = (2, 3)$

$(x_2, y_2) = (4, k)$

$(x_3, y_3) = (6, -3)$

$2(k - (-3)) + 4(-3 - 3) + 6(3 - k) = 0$

Simplify the equation:

$2(k + 3) + 4(-6) + 6(3 - k) = 0$

$2k + 6 - 24 + 18 - 6k = 0$

Combine like terms:

$(2k - 6k) + (6 - 24 + 18) = 0$

$-4k + (24 - 24) = 0$

$-4k + 0 = 0$

$-4k = 0$

Divide by -4:

$k = \frac{0}{-4}$

$k = 0$

Therefore, the value of $k$ for which the points are collinear is 0.

p>The vertices of the quadrilateral are $A(-5, 7)$, $B(-4, -5)$, $C(-1, -6)$, and $D(4, 5)$.

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To Find:

The area of the quadrilateral ABCD.

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Solution:

To find the area of the quadrilateral ABCD, we can divide it into two triangles by drawing one of the diagonals, for example, diagonal AC.

The area of the quadrilateral ABCD will be the sum of the areas of triangle ABC and triangle ADC.

The formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is:

$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Area of Triangle ABC:

Vertices are $A(-5, 7)$, $B(-4, -5)$, $C(-1, -6)$.

$(x_1, y_1) = (-5, 7)$

$(x_2, y_2) = (-4, -5)$

$(x_3, y_3) = (-1, -6)$

$Area(\triangle ABC) = \frac{1}{2} |(-5)(-5 - (-6)) + (-4)(-6 - 7) + (-1)(7 - (-5))|$

$Area(\triangle ABC) = \frac{1}{2} |(-5)(-5 + 6) + (-4)(-13) + (-1)(7 + 5)|$

$Area(\triangle ABC) = \frac{1}{2} |(-5)(1) + (52) + (-1)(12)|$

$Area(\triangle ABC) = \frac{1}{2} |-5 + 52 - 12|$

$Area(\triangle ABC) = \frac{1}{2} |52 - 17|$

$Area(\triangle ABC) = \frac{1}{2} |35| = \frac{35}{2}$ square units.

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Area of Triangle ADC:

Vertices are $A(-5, 7)$, $D(4, 5)$, $C(-1, -6)$.

$(x_1, y_1) = (-5, 7)$

$(x_2, y_2) = (4, 5)$

$(x_3, y_3) = (-1, -6)$

$Area(\triangle ADC) = \frac{1}{2} |(-5)(5 - (-6)) + (4)(-6 - 7) + (-1)(7 - 5)|$

$Area(\triangle ADC) = \frac{1}{2} |(-5)(5 + 6) + (4)(-13) + (-1)(2)|$

$Area(\triangle ADC) = \frac{1}{2} |(-5)(11) + (-52) + (-2)|$

$Area(\triangle ADC) = \frac{1}{2} |-55 - 52 - 2|$

$Area(\triangle ADC) = \frac{1}{2} |-109|$

$Area(\triangle ADC) = \frac{109}{2}$ square units.

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Area of Quadrilateral ABCD:

Area(ABCD) = Area($\triangle ABC$) + Area($\triangle ADC$)

Area(ABCD) = $\frac{35}{2} + \frac{109}{2}$

Area(ABCD) = $\frac{35 + 109}{2}$

Area(ABCD) = $\frac{144}{2}$

Area(ABCD) = 72 square units.

Therefore, the area of the quadrilateral ABCD is 72 square units.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

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(i)

Given vertices are $A(2, 3)$, $B(-1, 0)$, and $C(2, -4)$.

Let $(x_1, y_1) = (2, 3)$, $(x_2, y_2) = (-1, 0)$, and $(x_3, y_3) = (2, -4)$.

Substitute these coordinates into the area formula:

$Area = \frac{1}{2} |2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)|$

$Area = \frac{1}{2} |2(0 + 4) + (-1)(-7) + 2(3)|$

$Area = \frac{1}{2} |2(4) + 7 + 6|$

$Area = \frac{1}{2} |8 + 7 + 6|$

$Area = \frac{1}{2} |21|$

$Area = \frac{21}{2}$ square units.

The area of the triangle is $\frac{21}{2}$ or 10.5 square units.

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(ii)

Given vertices are $P(-5, -1)$, $Q(3, -5)$, and $R(5, 2)$.

Let $(x_1, y_1) = (-5, -1)$, $(x_2, y_2) = (3, -5)$, and $(x_3, y_3) = (5, 2)$.

Substitute these coordinates into the area formula:

$Area = \frac{1}{2} |(-5)(-5 - 2) + (3)(2 - (-1)) + (5)(-1 - (-5))|$

$Area = \frac{1}{2} |(-5)(-7) + (3)(2 + 1) + (5)(-1 + 5)|$

$Area = \frac{1}{2} |35 + 3(3) + 5(4)|$

$Area = \frac{1}{2} |35 + 9 + 20|$

$Area = \frac{1}{2} |64|$

$Area = \frac{1}{2} \times 64$

$Area = 32$ square units.

The area of the triangle is 32 square units.

Concept:

For three points to be collinear, the area of the triangle formed by these points must be zero.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Therefore, for collinear points, the condition is:

$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$

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(i)

Given:

The points are $A(7, -2)$, $B(5, 1)$, and $C(3, k)$. The points are collinear.

To Find:

The value of $k$.

Solution:

Let $(x_1, y_1) = (7, -2)$, $(x_2, y_2) = (5, 1)$, and $(x_3, y_3) = (3, k)$.

Applying the condition for collinearity:

$7(1 - k) + 5(k - (-2)) + 3(-2 - 1) = 0$

$7(1 - k) + 5(k + 2) + 3(-3) = 0$

$7 - 7k + 5k + 10 - 9 = 0$

Combine the terms involving $k$ and the constant terms:

$(-7k + 5k) + (7 + 10 - 9) = 0$

$-2k + (17 - 9) = 0$

$-2k + 8 = 0$

$-2k = -8$

$k = \frac{-8}{-2}$

$k = 4$

Therefore, the value of $k$ is 4.

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(ii)

Given:

The points are $P(8, 1)$, $Q(k, -4)$, and $R(2, -5)$. The points are collinear.

To Find:

The value of $k$.

Solution:

Let $(x_1, y_1) = (8, 1)$, $(x_2, y_2) = (k, -4)$, and $(x_3, y_3) = (2, -5)$.

Applying the condition for collinearity:

$8(-4 - (-5)) + k(-5 - 1) + 2(1 - (-4)) = 0$

$8(-4 + 5) + k(-6) + 2(1 + 4) = 0$

$8(1) - 6k + 2(5) = 0$

$8 - 6k + 10 = 0$

Combine the constant terms:

$-6k + 18 = 0$

$-6k = -18$

$k = \frac{-18}{-6}$

$k = 3$

Therefore, the value of $k$ is 3.

Given:

The vertices of the given triangle (let's call it $\triangle ABC$) are $A(0, -1)$, $B(2, 1)$, and $C(0, 3)$.

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To Find:

1. The area of the triangle formed by joining the mid-points of the sides of $\triangle ABC$.

2. The ratio of the area of the smaller triangle (formed by midpoints) to the area of the given triangle ($\triangle ABC$).

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Solution:

Step 1: Find the coordinates of the midpoints of the sides of $\triangle ABC$.

Let D, E, and F be the midpoints of the sides BC, AC, and AB, respectively.

We use the midpoint formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.

Midpoint D of BC (B(2, 1), C(0, 3)):

$D = \left( \frac{2 + 0}{2}, \frac{1 + 3}{2} \right) = \left( \frac{2}{2}, \frac{4}{2} \right) = (1, 2)$

Midpoint E of AC (A(0, -1), C(0, 3)):

$E = \left( \frac{0 + 0}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{0}{2}, \frac{2}{2} \right) = (0, 1)$

Midpoint F of AB (A(0, -1), B(2, 1)):

$F = \left( \frac{0 + 2}{2}, \frac{-1 + 1}{2} \right) = \left( \frac{2}{2}, \frac{0}{2} \right) = (1, 0)$

So, the vertices of the triangle formed by joining the midpoints ($\triangle DEF$) are D(1, 2), E(0, 1), and F(1, 0).

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Step 2: Find the area of the triangle formed by the midpoints ($\triangle DEF$).

We use the area formula for a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

For $\triangle DEF$ with D(1, 2), E(0, 1), F(1, 0):

$Area(\triangle DEF) = \frac{1}{2} |1(1 - 0) + 0(0 - 2) + 1(2 - 1)|$

$Area(\triangle DEF) = \frac{1}{2} |1(1) + 0(-2) + 1(1)|$

$Area(\triangle DEF) = \frac{1}{2} |1 + 0 + 1|$

$Area(\triangle DEF) = \frac{1}{2} |2| = 1$ square unit.

The area of the triangle formed by joining the mid-points is 1 square unit.

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Step 3: Find the area of the given triangle ($\triangle ABC$).

Vertices are A(0, -1), B(2, 1), C(0, 3).

$Area(\triangle ABC) = \frac{1}{2} |0(1 - 3) + 2(3 - (-1)) + 0(-1 - 1)|$

$Area(\triangle ABC) = \frac{1}{2} |0(-2) + 2(3 + 1) + 0(-2)|$

$Area(\triangle ABC) = \frac{1}{2} |0 + 2(4) + 0|$

$Area(\triangle ABC) = \frac{1}{2} |8| = 4$ square units.

The area of the given triangle is 4 square units.

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Step 4: Find the ratio of the areas.

Ratio = $\frac{Area(\triangle DEF)}{Area(\triangle ABC)}$

Ratio = $\frac{1}{4}$

The ratio of the area of the triangle formed by the mid-points to the area of the given triangle is 1 : 4.

Given:

The vertices of the quadrilateral, taken in order, are $A(-4, -2)$, $B(-3, -5)$, $C(3, -2)$, and $D(2, 3)$.

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To Find:

The area of the quadrilateral ABCD.

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Solution:

To find the area of the quadrilateral ABCD, we can divide it into two triangles by joining the diagonal AC.

Area(Quadrilateral ABCD) = Area($\triangle ABC$) + Area($\triangle ADC$)

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Area of Triangle ABC:

Vertices are $A(-4, -2)$, $B(-3, -5)$, $C(3, -2)$.

Let $(x_1, y_1) = (-4, -2)$, $(x_2, y_2) = (-3, -5)$, $(x_3, y_3) = (3, -2)$.

$Area(\triangle ABC) = \frac{1}{2} |(-4)(-5 - (-2)) + (-3)(-2 - (-2)) + (3)(-2 - (-5))|$

$Area(\triangle ABC) = \frac{1}{2} |(-4)(-5 + 2) + (-3)(-2 + 2) + (3)(-2 + 5)|$

$Area(\triangle ABC) = \frac{1}{2} |(-4)(-3) + (-3)(0) + (3)(3)|$

$Area(\triangle ABC) = \frac{1}{2} |12 + 0 + 9|$

$Area(\triangle ABC) = \frac{1}{2} |21| = \frac{21}{2}$ square units.

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Area of Triangle ADC:

Vertices are $A(-4, -2)$, $D(2, 3)$, $C(3, -2)$.

Let $(x_1, y_1) = (-4, -2)$, $(x_2, y_2) = (2, 3)$, $(x_3, y_3) = (3, -2)$.

$Area(\triangle ADC) = \frac{1}{2} |(-4)(3 - (-2)) + (2)(-2 - (-2)) + (3)(-2 - 3)|$

$Area(\triangle ADC) = \frac{1}{2} |(-4)(3 + 2) + (2)(-2 + 2) + (3)(-5)|$

$Area(\triangle ADC) = \frac{1}{2} |(-4)(5) + (2)(0) + (3)(-5)|$

$Area(\triangle ADC) = \frac{1}{2} |-20 + 0 - 15|$

$Area(\triangle ADC) = \frac{1}{2} |-35| = \frac{35}{2}$ square units.

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Area of Quadrilateral ABCD:

Area(ABCD) = Area($\triangle ABC$) + Area($\triangle ADC$)

Area(ABCD) = $\frac{21}{2} + \frac{35}{2}$

Area(ABCD) = $\frac{21 + 35}{2}$

Area(ABCD) = $\frac{56}{2}$

Area(ABCD) = 28 square units.

Therefore, the area of the quadrilateral ABCD is 28 square units.

Given:

The vertices of the triangle $\triangle ABC$ are $A(4, -6)$, $B(3, -2)$, and $C(5, 2)$.

We need to verify that a median divides the triangle into two triangles of equal areas.

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To Verify:

If AD is a median of $\triangle ABC$, then Area($\triangle ABD$) = Area($\triangle ACD$).

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Verification:

Let AD be the median from vertex A to the side BC. By definition, the median connects a vertex to the midpoint of the opposite side. So, D is the midpoint of BC.

Step 1: Find the coordinates of the midpoint D of BC.

Using the midpoint formula $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$ for points $B(3, -2)$ and $C(5, 2)$.

$x_D = \frac{3 + 5}{2} = \frac{8}{2} = 4$

$y_D = \frac{-2 + 2}{2} = \frac{0}{2} = 0$

The coordinates of the midpoint D are (4, 0).

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Step 2: Calculate the area of $\triangle ABD$.

The vertices of $\triangle ABD$ are $A(4, -6)$, $B(3, -2)$, and $D(4, 0)$.

Using the area formula $Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$:

$Area(\triangle ABD) = \frac{1}{2} |4(-2 - 0) + 3(0 - (-6)) + 4(-6 - (-2))|$

$Area(\triangle ABD) = \frac{1}{2} |4(-2) + 3(6) + 4(-6 + 2)|$

$Area(\triangle ABD) = \frac{1}{2} |-8 + 18 + 4(-4)|$

$Area(\triangle ABD) = \frac{1}{2} |-8 + 18 - 16|$

$Area(\triangle ABD) = \frac{1}{2} |18 - 24|$

$Area(\triangle ABD) = \frac{1}{2} |-6| = \frac{1}{2} \times 6 = 3$ square units.

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Step 3: Calculate the area of $\triangle ACD$.

The vertices of $\triangle ACD$ are $A(4, -6)$, $C(5, 2)$, and $D(4, 0)$.

Using the area formula:

$Area(\triangle ACD) = \frac{1}{2} |4(2 - 0) + 5(0 - (-6)) + 4(-6 - 2)|$

$Area(\triangle ACD) = \frac{1}{2} |4(2) + 5(6) + 4(-8)|$

$Area(\triangle ACD) = \frac{1}{2} |8 + 30 - 32|$

$Area(\triangle ACD) = \frac{1}{2} |38 - 32|$

$Area(\triangle ACD) = \frac{1}{2} |6| = \frac{1}{2} \times 6 = 3$ square units.

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Step 4: Compare the areas.

We found that Area($\triangle ABD$) = 3 square units and Area($\triangle ACD$) = 3 square units.

Since Area($\triangle ABD$) = Area($\triangle ACD$), the median AD divides $\triangle ABC$ into two triangles of equal areas.

This verifies the result for the given $\triangle ABC$.

Given:

The line segment joins the points $A(2, -2)$ and $B(3, 7)$.

The line is given by the equation $2x + y - 4 = 0$.

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To Find:

The ratio in which the line $2x + y - 4 = 0$ divides the line segment joining A and B.

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Solution:

Let the line $2x + y - 4 = 0$ intersect the line segment joining $A(x_1, y_1) = (2, -2)$ and $B(x_2, y_2) = (3, 7)$ at point $P(x, y)$.

Let the ratio in which P divides AB be $m_1 : m_2$.

Assume the ratio is $k : 1$. Then $m_1 = k$ and $m_2 = 1$.

Using the section formula, the coordinates of the point P are:

$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{k(3) + 1(2)}{k + 1} = \frac{3k + 2}{k + 1}$

$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{k(7) + 1(-2)}{k + 1} = \frac{7k - 2}{k + 1}$

So, the coordinates of the point P are $\left( \frac{3k + 2}{k + 1}, \frac{7k - 2}{k + 1} \right)$.

Since the point P lies on the line $2x + y - 4 = 0$, its coordinates must satisfy the equation of the line.

Substitute the coordinates of P into the equation of the line:

$2\left( \frac{3k + 2}{k + 1} \right) + \left( \frac{7k - 2}{k + 1} \right) - 4 = 0$

Multiply the entire equation by $(k + 1)$ to eliminate the denominator:

$2(3k + 2) + (7k - 2) - 4(k + 1) = 0$

$6k + 4 + 7k - 2 - 4k - 4 = 0$

Combine the terms with $k$ and the constant terms:

$(6k + 7k - 4k) + (4 - 2 - 4) = 0$

$9k - 2 = 0$

$9k = 2$

$k = \frac{2}{9}$

The ratio $k : 1$ is $\frac{2}{9} : 1$.

Multiplying by 9, we get the ratio $2 : 9$.

Since $k = 2/9$ is positive, the division is internal.

Therefore, the line $2x + y - 4 = 0$ divides the line segment joining the points A(2, – 2) and B(3, 7) in the ratio 2 : 9.

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Alternate Solution:

The ratio in which the line $ax + by + c = 0$ divides the line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula:

$Ratio = -\frac{ax_1 + by_1 + c}{ax_2 + by_2 + c}$

Here, the line is $2x + y - 4 = 0$, so $a=2, b=1, c=-4$.

The points are $A(x_1, y_1) = (2, -2)$ and $B(x_2, y_2) = (3, 7)$.

Calculate the value of the expression $ax + by + c$ for point A:

$2(2) + 1(-2) - 4 = 4 - 2 - 4 = -2$

Calculate the value of the expression $ax + by + c$ for point B:

$2(3) + 1(7) - 4 = 6 + 7 - 4 = 13 - 4 = 9$

Substitute these values into the ratio formula:

$Ratio = -\frac{(-2)}{(9)} = \frac{2}{9}$

The ratio is $2 : 9$.

Since the ratio is positive, the division is internal.

Thus, the required ratio is 2 : 9.

Given:

The points $A(x, y)$, $B(1, 2)$, and $C(7, 0)$ are collinear.

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To Find:

A relation between $x$ and $y$.

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Solution:

If three points are collinear, the area of the triangle formed by these points is zero.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

$Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

For collinear points, Area = 0. Therefore, the condition for collinearity is:

$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$

Let the points be $(x_1, y_1) = (x, y)$, $(x_2, y_2) = (1, 2)$, and $(x_3, y_3) = (7, 0)$.

Substitute these coordinates into the collinearity condition:

$x(2 - 0) + 1(0 - y) + 7(y - 2) = 0$

Simplify the equation:

$x(2) + 1(-y) + 7(y - 2) = 0$

$2x - y + 7y - 14 = 0$

Combine like terms:

$2x + (-y + 7y) - 14 = 0$

$2x + 6y - 14 = 0$

Divide the entire equation by 2 to simplify:

$\frac{2x}{2} + \frac{6y}{2} - \frac{14}{2} = \frac{0}{2}$

$x + 3y - 7 = 0$

This can also be written as $x + 3y = 7$.

Therefore, the required relation between $x$ and $y$ is $x + 3y = 7$.

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Alternate Solution (Using Slope):

If points A(x, y), B(1, 2), and C(7, 0) are collinear, then the slope of AB must be equal to the slope of BC.

Slope of AB = $\frac{2 - y}{1 - x}$

Slope of BC = $\frac{0 - 2}{7 - 1} = \frac{-2}{6} = -\frac{1}{3}$

Equating the slopes:

$\frac{2 - y}{1 - x} = -\frac{1}{3}$

Cross-multiply:

$3(2 - y) = -1(1 - x)$

$6 - 3y = -1 + x$

Rearrange the terms:

$6 + 1 = x + 3y$

$7 = x + 3y$

The relation is $x + 3y = 7$.

Given:

A circle passes through the points $A(6, -6)$, $B(3, -7)$, and $C(3, 3)$.

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To Find:

The coordinates of the centre of the circle.

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Solution:

Let the centre of the circle be $O(x, y)$.

Since the points A, B, and C lie on the circle, the distance from the centre O to each of these points is equal (equal to the radius).

Therefore, $OA = OB = OC$.

This implies that the squares of the distances are also equal: $OA^2 = OB^2 = OC^2$.

We use the distance formula squared: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.

Calculating the squared distances:

$OA^2 = (x - 6)^2 + (y - (-6))^2 = (x - 6)^2 + (y + 6)^2$

$OA^2 = (x^2 - 12x + 36) + (y^2 + 12y + 36) = x^2 + y^2 - 12x + 12y + 72$

$OB^2 = (x - 3)^2 + (y - (-7))^2 = (x - 3)^2 + (y + 7)^2$

$OB^2 = (x^2 - 6x + 9) + (y^2 + 14y + 49) = x^2 + y^2 - 6x + 14y + 58$

$OC^2 = (x - 3)^2 + (y - 3)^2$

$OC^2 = (x^2 - 6x + 9) + (y^2 - 6y + 9) = x^2 + y^2 - 6x - 6y + 18$

Now, we set up equations using the equalities.

Equating $OA^2$ and $OB^2$:

$OA^2 = OB^2$

$x^2 + y^2 - 12x + 12y + 72 = x^2 + y^2 - 6x + 14y + 58$

Cancel $x^2$ and $y^2$ terms:

$-12x + 12y + 72 = -6x + 14y + 58$

Rearrange terms to form a linear equation in $x$ and $y$:

$72 - 58 = -6x + 12x + 14y - 12y$

$14 = 6x + 2y$

Divide the equation by 2:

Equating $OB^2$ and $OC^2$:

$OB^2 = OC^2$

$x^2 + y^2 - 6x + 14y + 58 = x^2 + y^2 - 6x - 6y + 18$

Cancel $x^2$, $y^2$, and $-6x$ terms:

$14y + 58 = -6y + 18$

Rearrange terms to solve for $y$:

$14y + 6y = 18 - 58$

$20y = -40$

$y = \frac{-40}{20}$

$y = -2$

Finding the value of x:

Substitute the value $y = -2$ into equation (i):

$3x + (-2) = 7$

$3x - 2 = 7$

$3x = 7 + 2$

$3x = 9$

$x = \frac{9}{3}$

$x = 3$

Thus, the coordinates of the centre $O(x, y)$ are $(3, -2)$.

Therefore, the centre of the circle passing through the given points is (3, –2).

Given:

Let the vertices of the square be ABCD taken in order.

Let the given opposite vertices be $A = (-1, 2)$ and $C = (3, 2)$.

Let the other two opposite vertices be $B = (x, y)$ and $D = (x', y')$.

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To Find:

The coordinates of the other two vertices, B and D.

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Solution:

We know the properties of a square:

1. Diagonals are equal in length ($AC = BD$).

2. Diagonals bisect each other at right angles.

Step 1: Find the midpoint and length of diagonal AC.

Let M be the midpoint of AC. Using the midpoint formula:

$M = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = \left( \frac{-1 + 3}{2}, \frac{2 + 2}{2} \right) = \left( \frac{2}{2}, \frac{4}{2} \right) = (1, 2)$

The center of the square is M(1, 2).

The length of diagonal AC using the distance formula:

$AC = \sqrt{(3 - (-1))^2 + (2 - 2)^2} = \sqrt{(3 + 1)^2 + 0^2} = \sqrt{4^2} = 4$ units.

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Step 2: Use properties of the other diagonal BD.

Since the diagonals of a square bisect each other, M(1, 2) is also the midpoint of the diagonal BD.

Since the diagonals are equal, the length of diagonal BD must also be 4 units.

The diagonals are perpendicular. Let's find the slope of AC:

Slope of AC = $\frac{2 - 2}{3 - (-1)} = \frac{0}{4} = 0$.

Since the slope of AC is 0, AC is a horizontal line ($y=2$).

Diagonal BD must be perpendicular to AC, so BD must be a vertical line.

A vertical line passing through the center M(1, 2) has the equation $x = 1$.

Therefore, the vertices B and D must lie on the line $x = 1$. Their coordinates will be of the form $(1, y)$ and $(1, y')$.

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Step 3: Find the coordinates of B and D.

Let $B = (1, y)$ and $D = (1, y')$.

M(1, 2) is the midpoint of BD:

$\left( \frac{1 + 1}{2}, \frac{y + y'}{2} \right) = (1, 2)$

The x-coordinate is correct. Equating the y-coordinates:

$\frac{y + y'}{2} = 2 \implies y + y' = 4$ ...(i)

The length of diagonal BD is 4:

$BD = \sqrt{(1 - 1)^2 + (y' - y)^2} = 4$

$\sqrt{0^2 + (y' - y)^2} = 4$

$\sqrt{(y' - y)^2} = 4$

$|y' - y| = 4$

This gives two possibilities: $y' - y = 4$ or $y' - y = -4$.

Case 1: $y' - y = 4$ ...(ii)

Add equation (i) and (ii):

$(y + y') + (y' - y) = 4 + 4$

$2y' = 8 \implies y' = 4$

Substitute $y' = 4$ into equation (i):

$y + 4 = 4 \implies y = 0$

So, the vertices are (1, 0) and (1, 4).

Case 2: $y' - y = -4$ ...(iii)

Add equation (i) and (iii):

$(y + y') + (y' - y) = 4 + (-4)$

$2y' = 0 \implies y' = 0$

Substitute $y' = 0$ into equation (i):

$y + 0 = 4 \implies y = 4$

So, the vertices are (1, 4) and (1, 0).

Both cases give the same pair of points.

Therefore, the coordinates of the other two vertices are (1, 0) and (1, 4).

Given:

A rectangular plot of land with saplings planted at a distance of 1m from each other on the boundary.

A triangular grassy lawn PQR inside the plot.

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To Find:

(i) The coordinates of the vertices of the triangle PQR taking A as origin and its area.

(ii) The coordinates of the vertices of $\triangle$ PQR taking C as origin and its area. Also, observe the result.

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Solution:

Let the distance between two consecutive saplings be 1 unit. This implies the grid lines in the figure represent units of 1m distance.

(i) Taking A as origin (0,0):

When A is taken as the origin, the x-axis is along AB and the y-axis is along AD.

By counting the saplings along AB, the length of AB is 12 units (from the 1st to the 13th sapling, 12 intervals).

By counting the saplings along AD, the length of AD is 10 units (from the 1st to the 11th sapling, 10 intervals).

The vertices of the rectangular plot are A, B, C, and D.

The coordinates of the vertices of the plot are:

A = $(0,0)$

B = $(12,0)$

D = $(0,10)$

C = $(12,10)$

Now, we find the coordinates of the vertices of $\triangle$ PQR by counting units from the origin A along the grid lines:

For point P, move 4 units right and 2 units up from A. So, P = $(4,2)$.

For point Q, move 3 units right and 6 units up from A. So, Q = $(3,6)$.

For point R, move 6 units right and 5 units up from A. So, R = $(6,5)$.

The coordinates of the vertices of $\triangle$ PQR are P$(4,2)$, Q$(3,6)$, and R$(6,5)$.

Now, we calculate the area of $\triangle$ PQR using the area formula for a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substitute the coordinates of P$(4,2)$, Q$(3,6)$, and R$(6,5)$ into the formula:

Area of $\triangle$ PQR $= \frac{1}{2} |4(6 - 5) + 3(5 - 2) + 6(2 - 6)|$

Area $= \frac{1}{2} |4(1) + 3(3) + 6(-4)|$

Area $= \frac{1}{2} |4 + 9 - 24|$

Area $= \frac{1}{2} |13 - 24|$

Area $= \frac{1}{2} |-11|$

Since area is always non-negative, we take the absolute value.

Area $= \frac{1}{2} \times 11 = 5.5$ square units.

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(ii) Taking C as origin (0,0):

When C is taken as the origin, we can align the axes along the sides CD and CB. Let CD be along the negative x-axis and CB be along the positive y-axis.

C = $(0,0)$

D is 12 units left of C (along the negative x-axis). So, D = $(-12,0)$.

B is 10 units up from C (along the positive y-axis). So, B = $(0,10)$.

A is 12 units left and 10 units up from C. So, A = $(-12,10)$.

Now, we find the coordinates of the vertices of $\triangle$ PQR by counting units from the origin C along the grid lines in this new coordinate system:

For point P, move 8 units left from C and 8 units up from C. So, P = $(-8,8)$.

For point Q, move 9 units left from C and 4 units up from C. So, Q = $(-9,4)$.

For point R, move 6 units left from C and 5 units up from C. So, R = $(-6,5)$.

The coordinates of the vertices of $\triangle$ PQR are P$(-8,8)$, Q$(-9,4)$, and R$(-6,5)$.

Now, we calculate the area of $\triangle$ PQR using the area formula:

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substitute the coordinates of P$(-8,8)$, Q$(-9,4)$, and R$(-6,5)$ into the formula:

Area of $\triangle$ PQR $= \frac{1}{2} |(-8)(4 - 5) + (-9)(5 - 8) + (-6)(8 - 4)|$

Area $= \frac{1}{2} |(-8)(-1) + (-9)(-3) + (-6)(4)|$

Area $= \frac{1}{2} |8 + 27 - 24|$

Area $= \frac{1}{2} |35 - 24|$

Area $= \frac{1}{2} |11|$

Area $= \frac{1}{2} \times 11 = 5.5$ square units.

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Observation:

The area of $\triangle$ PQR is 5.5 square units when calculated with A as the origin and also 5.5 square units when calculated with C as the origin.

We observe that the area of the triangle remains the same regardless of the position of the origin. This is because the area of a geometric figure is invariant under translation (change of origin) and rotation of the coordinate axes.

Given:

The vertices of $\triangle$ ABC are A(4, 6), B(1, 5), and C(7, 2).

A line intersects sides AB and AC at D and E respectively, such that $\frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{4}$.

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To Find:

The area of $\triangle$ ADE and compare it with the area of $\triangle$ ABC.

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Solution:

First, we calculate the area of $\triangle$ ABC using the formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Using vertices A(4, 6), B(1, 5), and C(7, 2):

Area of $\triangle$ ABC $= \frac{1}{2} |4(5 - 2) + 1(2 - 6) + 7(6 - 5)|$

Area of $\triangle$ ABC $= \frac{1}{2} |4(3) + 1(-4) + 7(1)|$

Area of $\triangle$ ABC $= \frac{1}{2} |12 - 4 + 7|$

Area of $\triangle$ ABC $= \frac{1}{2} |15|$

We are given that $\frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{4}$.

According to the Converse of Basic Proportionality Theorem (Theorem 6.2), if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Since $\frac{AD}{AB} = \frac{AE}{AC}$, the line segment DE is parallel to BC.

Now consider $\triangle$ ADE and $\triangle$ ABC:

$\angle$ DAE = $\angle$ BAC (Common angle)

$\angle$ ADE = $\angle$ ABC (Corresponding angles, since DE || BC)

Therefore, $\triangle$ ADE is similar to $\triangle$ ABC by AA similarity criterion.

According to Theorem 6.6, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

So,

We are given $\frac{AD}{AB} = \frac{1}{4}$. Substituting this value:

From equation (ii), we can find the Area of $\triangle$ ADE:

Area($\triangle$ ADE) $= \frac{1}{16} \times \text{Area}(\triangle ABC)$

Substitute the Area of $\triangle$ ABC from equation (i):

Area($\triangle$ ADE) $= \frac{1}{16} \times 7.5$

Area($\triangle$ ADE) $= \frac{1}{16} \times \frac{15}{2}$

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Comparison:

Comparing the area of $\triangle$ ADE (from iii) and the area of $\triangle$ ABC (from i), we have:

Area($\triangle$ ADE) $= \frac{15}{32}$ and Area($\triangle$ ABC) $= \frac{15}{2}$.

We can express the comparison as a ratio:

$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \frac{15/32}{15/2} = \frac{15}{32} \times \frac{2}{15} = \frac{\cancel{15}}{\cancel{32}^{16}} \times \frac{\cancel{2}^1}{\cancel{15}} = \frac{1}{16}$

So, Area($\triangle$ ADE) $= \frac{1}{16} \times \text{Area}(\triangle ABC)$.

The area of $\triangle$ ADE is $\frac{1}{16}$ times the area of $\triangle$ ABC.

Given:

The vertices of $\triangle$ ABC are A(4, 2), B(6, 5), and C(1, 4).

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To Find:

(i) The coordinates of point D, where the median from A meets BC.

(ii) The coordinates of point P on AD such that AP : PD = 2 : 1.

(iii) The coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.

(iv) Observation about points P, Q, and R.

(v) The coordinates of the centroid of a triangle with vertices A(x₁ , y₁ ), B(x₂ , y₂ ) and C(x₃ , y₃ ).

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Solution:

Vertices of $\triangle$ ABC are A(4, 2), B(6, 5), and C(1, 4).

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(i) Coordinates of point D:

The median from A meets BC at D. This means D is the midpoint of the side BC.

Using the midpoint formula for points B$(6, 5)$ and C$(1, 4)$:

$D = \left(\frac{x_B+x_C}{2}, \frac{y_B+y_C}{2}\right)$

$D = \left(\frac{6+1}{2}, \frac{5+4}{2}\right)$

$D = \left(\frac{7}{2}, \frac{9}{2}\right)$

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(ii) Coordinates of point P:

P is on AD such that AP : PD = 2 : 1. We use the section formula for a point dividing the line segment AD joining A$(4, 2)$ and D$(\frac{7}{2}, \frac{9}{2})$ in the ratio $m_1=2$ and $m_2=1$.

$P = \left(\frac{m_1 x_D + m_2 x_A}{m_1 + m_2}, \frac{m_1 y_D + m_2 y_A}{m_1 + m_2}\right)$

$x_P = \frac{2 \times \frac{7}{2} + 1 \times 4}{2 + 1} = \frac{7 + 4}{3} = \frac{11}{3}$

$y_P = \frac{2 \times \frac{9}{2} + 1 \times 2}{2 + 1} = \frac{9 + 2}{3} = \frac{11}{3}$

$P = \left(\frac{11}{3}, \frac{11}{3}\right)$

Wait, I used the coordinates of A from the question prompt (4,2) but when reading the image in the question, A seems to be at (4,6). Let me recheck the question text. The question text says A(4,6), B(1,5), C(7,2) in Question 6. For Question 7, it says A(4,2), B(6,5), C(1,4). I should follow the text for Question 7. So A is (4,2). Let me recalculate yP.

Correcting the calculation for yP using A(4,2) and D($\frac{7}{2}$, $\frac{9}{2}$):

$y_P = \frac{2 \times \frac{9}{2} + 1 \times 2}{2 + 1} = \frac{9 + 2}{3} = \frac{11}{3}$

So, P = $\left(\frac{11}{3}, \frac{11}{3}\right)$. Let me double check the calculation of D. B(6,5), C(1,4). $x_D = (6+1)/2 = 7/2$, $y_D = (5+4)/2 = 9/2$. This is correct.

Let me double check the calculation of P again. A(4,2), D(7/2, 9/2), ratio 2:1.

$x_P = \frac{2 \times (7/2) + 1 \times 4}{2 + 1} = \frac{7 + 4}{3} = \frac{11}{3}$ (Correct)

$y_P = \frac{2 \times (9/2) + 1 \times 2}{2 + 1} = \frac{9 + 2}{3} = \frac{11}{3}$ (Correct)

So P is $(\frac{11}{3}, \frac{11}{3})$. This seems correct based on the given points. Let me proceed to calculate Q and R.

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(iii) Coordinates of points Q and R:

To find Q on median BE, we first need the coordinates of E, the midpoint of AC. Using A$(4, 2)$ and C$(1, 4)$:

$E = \left(\frac{x_A+x_C}{2}, \frac{y_A+y_C}{2}\right)$

$E = \left(\frac{4+1}{2}, \frac{2+4}{2}\right)$

$E = \left(\frac{5}{2}, \frac{6}{2}\right) = \left(\frac{5}{2}, 3\right)$

Q is on BE such that BQ : QE = 2 : 1. Using the section formula with B$(6, 5)$ and E$(\frac{5}{2}, 3)$, and ratio $m_1=2$, $m_2=1$:

$Q = \left(\frac{m_1 x_E + m_2 x_B}{m_1 + m_2}, \frac{m_1 y_E + m_2 y_B}{m_1 + m_2}\right)$

$x_Q = \frac{2 \times \frac{5}{2} + 1 \times 6}{2 + 1} = \frac{5 + 6}{3} = \frac{11}{3}$

$y_Q = \frac{2 \times 3 + 1 \times 5}{2 + 1} = \frac{6 + 5}{3} = \frac{11}{3}$

$Q = \left(\frac{11}{3}, \frac{11}{3}\right)$

Next, to find R on median CF, we first need the coordinates of F, the midpoint of AB. Using A$(4, 2)$ and B$(6, 5)$:

$F = \left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}\right)$

$F = \left(\frac{4+6}{2}, \frac{2+5}{2}\right)$

$F = \left(\frac{10}{2}, \frac{7}{2}\right) = \left(5, \frac{7}{2}\right)$

R is on CF such that CR : RF = 2 : 1. Using the section formula with C$(1, 4)$ and F$(5, \frac{7}{2})$, and ratio $m_1=2$, $m_2=1$:

$R = \left(\frac{m_1 x_F + m_2 x_C}{m_1 + m_2}, \frac{m_1 y_F + m_2 y_C}{m_1 + m_2}\right)$

$x_R = \frac{2 \times 5 + 1 \times 1}{2 + 1} = \frac{10 + 1}{3} = \frac{11}{3}$

$y_R = \frac{2 \times \frac{7}{2} + 1 \times 4}{2 + 1} = \frac{7 + 4}{3} = \frac{11}{3}$

$R = \left(\frac{11}{3}, \frac{11}{3}\right)$

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(iv) Observation:

The coordinates of points P, Q, and R are all $\left(\frac{11}{3}, \frac{11}{3}\right)$.

This shows that the points P, Q, and R coincide. These points lie on the three medians of the triangle and divide each median in the ratio 2 : 1. This confirms that the three medians of a triangle are concurrent, and their point of intersection is called the centroid. The centroid divides each median in the ratio 2 : 1.

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(v) Coordinates of the centroid:

Let the vertices of $\triangle$ ABC be A$(x_1, y_1)$, B$(x_2, y_2)$, and C$(x_3, y_3)$.

Let G be the centroid. G is the point of intersection of the medians.

Consider the median from A to the midpoint of BC, let's call it D. The coordinates of D are:

$D = \left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}\right)$

The centroid G lies on AD and divides it in the ratio AG : GD = 2 : 1.

Using the section formula with A$(x_1, y_1)$ and D$(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2})$, and ratio $m_1=2$, $m_2=1$:

$x_G = \frac{m_1 x_D + m_2 x_A}{m_1 + m_2} = \frac{2 \times \left(\frac{x_2+x_3}{2}\right) + 1 \times x_1}{2 + 1}$

$x_G = \frac{(x_2+x_3) + x_1}{3} = \frac{x_1+x_2+x_3}{3}$

$y_G = \frac{m_1 y_D + m_2 y_A}{m_1 + m_2} = \frac{2 \times \left(\frac{y_2+y_3}{2}\right) + 1 \times y_1}{2 + 1}$

$y_G = \frac{(y_2+y_3) + y_1}{3} = \frac{y_1+y_2+y_3}{3}$

Thus, the coordinates of the centroid G of the triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are given by:

Answer: