Question 1: Find the mid-point of the line segment joining the points :

(i) $(-6, 7)$ and $(3, 5)$

(ii) $(5, -3)$ and $(-1, 7)$

i)  Ratio for being a midpoint: $m_1:m_2 = 1:1$

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

Therefore

$x =$ $\frac{1 \times 3+1 \times (-6)}{1+1}$ $= -1.5$

$y =$ $\frac{1 \times 5+1 \times 7}{1+1}$ $= 6$

Therefore $P = (-1.5, 6)$

ii) Ratio for being a midpoint: $m_1:m_2 = 1:1$

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

Therefore

$x =$ $\frac{1 \times (-1)+1 \times (5)}{1+1}$ $= 2$

$y =$ $\frac{1 \times 7+1 \times (-3)}{1+1}$ $= 2$

Therefore $P = (2, 2)$

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Question 2: Points $A \ and \ B$ have co-ordinates $(3, 5) \ and \ (x, y)$ respectively. The mid-point of $AB$  is $(2, 3)$ . Find the values of $x \ and \ y$.

Given Midpoint of $AB = (2, 3)$

Therefore

$2 =$ $\frac{1 \times x+1 \times (3)}{1+1}$ $\Rightarrow x = 1$

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

Therefore $B = (1, 1)$

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Question 3: $A (5, 3), B (-1, 1) \ and \ C (7, -3)$ are the vertices of triangle $ABC$ . If $L$ is the mid-point of $AB \ and \ M$ is the mid-point of $AC$ , show that $LM =$ $\frac{1}{2}BC$ .

Ratio for being a midpoint: $m_1:m_2 = 1:1$

Let the coordinates of the point $L \ be \ (x_1, y_1)$

Therefore

$x_1 =$ $\frac{1 \times (-1)+1 \times (5)}{1+1}$ $= 2$

$y_1 =$ $\frac{1 \times 1+1 \times (3)}{1+1}$ $= 2$

Therefore $L = (2, 2)$

Similarly

Let the coordinates of the point $M \ be \ (x_2, y_2)$

Therefore

$x_2 =$ $\frac{1 \times (7)+1 \times (5)}{1+1}$ $= 6$

$y_2 =$ $\frac{1 \times (-3)+1 \times (3)}{1+1}$ $= 0$

Therefore $M = (6, 0)$

Length of $LM = \sqrt{(6-2)^2+(0-2)^2} = \sqrt{20}$

Length of $BC = \sqrt{(7-(-1))^2+(-3-1)^2} = \sqrt{80} = 2\sqrt{20}$

Hence $LM =$ $\frac{1}{2}$ $BC$

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Question 4: Given $M$ is the mid-point of $AB$ , find the co-ordinates of :

(i) $A; if M = (1, 7) \ and \ B = (-5, l0)$

(ii) $B; if A = (3, -1) \ and \ M = (-1, 3)$

i) Given Midpoint of $AB = (1, 7)$ and let $A=(x,y)$

Therefore

$1 =$ $\frac{1 \times (-5)+1 \times (x)}{1+1}$ $\Rightarrow x = 7$

$7 =$ $\frac{1 \times (10)+1 \times (y)}{1+1}$ $\Rightarrow y = 4$

Therefore $A = (7, 4)$

ii) Given Midpoint of $AB = (-1, 3)$  and let $A=(x, y)$

Therefore

$-1 =$ $\frac{1 \times x+1 \times (3)}{1+1}$ $\Rightarrow x = -5$

$3 =$ $\frac{1 \times y+1 \times (-1)}{1+1}$ $\Rightarrow y = 7$

Therefore $B = (-5, 7)$

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Question 5: $P (-3, 2)$ is the midpoint of line segment $AB$ as shown in the given figure. Find the co-ordinates of Points $A \ and \ B$ .

Given Midpoint of $AB = (-1, 3)$ and let $A=(0, y)$ and $B =(x, 0)$

Therefore

$-3 =$ $\frac{1 \times x+1 \times (0)}{1+1}$ $\Rightarrow x = -6$

$2 =$ $\frac{1 \times y+1 \times (0)}{1+1}$ $\Rightarrow y = 4$

Therefore $B = (-6, 0) \ and \ A=(0, 4)$

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Question 6: In the given figure, $P (4, 2)$ is mid-point of line segment $AB$. Find the co-ordinates of $A (x, 0) \ and \ B (0, y)$.

Given Midpoint of $AB = (4, 2)$ and let $A=(x,0)$ and $B =(0,y)$

Therefore

$4 =$ $\frac{1 \times x+1 \times (0)}{1+1}$ $\Rightarrow x = 8$

$2 =$ $\frac{1 \times (0)+1 \times y}{1+1}$ $\Rightarrow y = 4$

Therefore $A = (8, 0) \ and \ B= (0, 4)$

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Question 7:  $(-5,2), (3, -6) \ and \ (7,4)$ are the vertices of a triangle. Find the length of its median though the vertex $(3, -6)$.

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

$x =$ $\frac{1 \times (7)+1 \times (3)}{1+1}$ $= 5$

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

Therefore $P = (5, -1)$

Therefore

Length of $AP = \sqrt{(7-(-5))^2+(-1-2)^2} = \sqrt{100+9} = \sqrt{109}$

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Question 8: Given a line $ABCD$ in which $AB = BC = CD B=(0, 3) \ and \ C=(1, 8)$ . Find the co-ordinates of $A \ and \ D$ .

For $A(x_1, y_1)$

Ratio $=1:1$

$0 =$ $\frac{1 \times (1) +1 \times x_1}{1+1}$ $\Rightarrow x_1 = -1$

$3 =$ $\frac{1 \times (8) +1 \times (y_1)}{1+1}$ $\Rightarrow y_1 = -2$

Hence the coordinates of $A = (-1, -2)$

For $D(x_2, y_2)$

Ratio $=1:1$

$1 =$ $\frac{1 \times (x_2) +1 \times (0)}{1+1}$ $\Rightarrow x_2 = 2$

$8 =$ $\frac{1 \times (y_2) +1 \times (3)}{1+1}$ $\Rightarrow y_2 = 13$

Hence the coordinates of $D = (2, 13)$

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Question 9: One end of the diameter of a circle is $(-2, 5)$ Find the co-ordinates of the other end of it, if the center of the circle is $(2, -1)$.

Let $P(x, y)$ be the other end of the diameter

$2 =$ $\frac{1 \times (-2) +1 \times x}{1+1}$ $\Rightarrow x = 6$

$-1 =$ $\frac{1 \times (5) +1 \times (y)}{1+1}$ $\Rightarrow y = -7$

Hence the coordinates of the other point of the diameter is $(6, -7)$

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Question 10: $A (2,5), B (1,0), C (-4,3) \ and \ D (-3,8)$ are the vertices of quadrilateral $ABCD$ . Find the co-ordinates of the midpoint of $AC \ and \ BD$ . Give a special name to the quadrilateral.

Let $P(x_1, y_1)$ be the midpoint of $AC$.

$x_1 =$ $\frac{1 \times (-4)+1 \times (2)}{1+1}$ $= -1$

$y_1 =$ $\frac{1 \times (3)+1 \times (5)}{1+1}$ $= 4$

Therefore $P = (-1, 4)$

Let $M(x_2, y_2)$. be the midpoint of $BD$.

$x_2 =$ $\frac{1 \times (1)+1 \times (-3)}{1+1}$ $= -1$

$y_2 =$ $\frac{1 \times (0)+1 \times (8)}{1+1}$ $= 4$

Therefore $M = (-1, 4)$

Because the diagonals bisect each other, the quadrilateral is a Parallelogram.

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Question 11: $P (4, 2) \ and \ Q (-1, 5)$ are the vertices of parallelogram $PQRS \ and \ (-3, 2)$ are the co-ordinates of the point of intersection of its diagonals. Find co-ordinates of $R \ and \ S$ .

Given Midpoint of $PR = M(-3, 2)$  and let $S=(x_1, y_1)$

Therefore

$-3 =$ $\frac{1 \times x_1+1 \times (4)}{1+1}$ $\Rightarrow x_1 = -10$

$2 =$ $\frac{1 \times y_1+1 \times (2)}{1+1}$ $\Rightarrow y_1 = 2$

Therefore $R = (-10, 2)$

Given Midpoint of $SQ = M(-3, 2)$  and let $S=(x_2, y_2)$

Therefore

$-3 =$ $\frac{1 \times x_2+1 \times (-1)}{1+1}$ $\Rightarrow x_2 = -5$

$2 =$ $\frac{1 \times (5)+1 \times y_2}{1+1}$ $\Rightarrow y_2 = -1$

Therefore $S = (-5, -1)$

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Question 12: $A (-1, 0), B (1, 3) \ and \ D (3, 5)$ are the vertices of a parallelogram $ABCD$. Find the co-ordinates of vertex $C$.

Let $M(x, y)$ be the midpoint of $BD$.

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

$y =$ $\frac{1 \times (3)+1 \times (5)}{1+1}$ $= 4$

Therefore $M = (2, 4)$

Given Midpoint of $AC = M(2,4)$  and let $C=(x, y)$

Therefore

$2 =$ $\frac{1 \times x+1 \times (-1)}{1+1}$ $\Rightarrow x = 5$

$4 =$ $\frac{1 \times (y)+1 \times (0)}{1+1}$ $\Rightarrow y = 8$

Therefore $C = (5, 8)$

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Question 13: The points $(2, -1), (-1,4) \ and \ (-2,2)$ are midpoints of the sides of a triangle. Find its vertices.

Let the vertices of the triangle be $B(x_1, y_1), C(x_2, y_2) \ and \ A(x_3, y_3)$

Given Midpoint of $BC = P(2,-1)$

Therefore

$2 =$ $\frac{1 \times x_1+1 \times x_2}{1+1}$ $\Rightarrow x_1+x_2 = 4$ … … … … i)

$-1 =$ $\frac{1 \times (y_1)+1 \times (y_2)}{1+1}$ $\Rightarrow y_1+y_2 = -2$ … … … … ii)

Given Midpoint of $AC = Q(-1, 4)$

Therefore

$-1 =$ $\frac{1 \times x_2+1 \times x_3}{1+1}$ $\Rightarrow x_2+x_3 = -2$ … … … … iii)

$4 =$ $\frac{1 \times (y_2)+1 \times (y_3)}{1+1}$ $\Rightarrow y_2+y_3 = -4$ … … … … iv)

Given Midpoint of $AB = R(-2, 2)$

Therefore

$-2 =$ $\frac{1 \times x_3+1 \times x_1}{1+1}$ $\Rightarrow x_3+x_1 = -4$ … … … … v)

$2 =$ $\frac{1 \times (y_3)+1 \times (y_1)}{1+1}$ $\Rightarrow y_3+y_1 = 4$ … … … … vi)

Adding i), iii) and v) we get

$x_1+x_2+x_3=-1$ … … … … vii)

Adding ii), iv) and vi) we get

$y_1+y_2+y_3=5$ … … … … viii)

Using i), iii),  v)  and vii) we get

$x_1=1, x_2=3 \ and \ x_3=-5$

Similarly Using ii), iv),  vi)  and viii) we get

$y_1=-3, y_2=1 \ and \ y_3=7$

Hence the vertices are $(1, -3), (3, 1) \ and \ (-5, 7)$

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Question 14: Points $A (-5, x), B (y, 7) \ and \ C (1, -3)$ are collinear (i.e. lie on the same straight line) such that $AB = BC$ . Calculate the values of $x \ and \ y$.

Given $B$ is the midpoint of $AC$. Therefore

$y =$ $\frac{1 \times (1)+1 \times (-5)}{1+1}$ $= -2$

$7 =$ $\frac{1 \times (-3)+1 \times x}{1+1}$ $\Rightarrow x = 17$

Therefore $A = (-5, 17) \ and \ B=(-2, 7)$

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Question 15: Points $P (a, -4), Q (2, b) \ and \ R (0, 2)$ are collinear. If $Q$ lies between $P \ and \ R$ , such that $PR = 2QR$ , calculate the values of $a \ and \ b$ .

Given $Q$ is the midpoint of $PR$. Therefore

$-2 =$ $\frac{1 \times (0)+1 \times a}{1+1}$ $\Rightarrow a = -4$

$b =$ $\frac{1 \times (2)+1 \times (-4)}{1+1}$ $= -1$

Therefore $a=-4 \ and \ b=-1$

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Question 16: Calculate the co-ordinates of the centroid of the triangle $ABC$, if $A = (7, -2), B = (0, l) \ and \ C=(-1,4)$ .

Let $O$ be the centroid of triangle $ABC$.

Therefore

$x=$ $\frac{7+0+(-1)}{3}$ $= 2$

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

Hence the coordinates of the centroid are $(2, 1)$

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Question 17: The co-ordinates of the centroid of a triangle $PQR$ are $(2, -5)$. lf $Q = (-6, 5) \ and \ R = (11,8)$ ; calculate the co-ordinates of vertex $P$.

Given $O(2, -5)$ be the centroid of triangle $PQR$.

Therefore

$2=$ $\frac{1+x+(-6)}{3}$ $\Rightarrow x = 1$

$-5 =$ $\frac{y+8+5}{3}$ $\Rightarrow y = -28$

Hence the coordinates of P are $(1, -28)$

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Question 18: $A (5, x), B (-4, 3) \ and \ C (y, -2)$ are the vertices of the triangle $ABC$ whose centroid is the origin. Calculate the values of $x \ and \ y$.

Given $O(0, 0)$ be the centroid of triangle $ABC$.
$0=$ $\frac{5+(-4)+y}{3}$ $\Rightarrow y = -1$
$0 =$ $\frac{x+3+(-2)}{3}$ $\Rightarrow x = -1$
Hence $x=-1 \ and \ y = -1$