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

$\displaystyle \text{(i) } (-6, 7) \text{ and } (3, 5)$

$\displaystyle \text{(ii) } (5, -3) \text{ and } (-1, 7)$

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

Let the coordinates of the point $\displaystyle P \text{ be } (x, y)$

Therefore

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

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

$\displaystyle \text{Therefore } P = (-1.5, 6)$

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

Let the coordinates of the point $\displaystyle P \text{ be } (x, y)$

Therefore

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

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

$\displaystyle \text{Therefore } P = (2, 2)$

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

$\displaystyle \text{Given Midpoint of } AB = (2, 3)$

Therefore

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

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

$\displaystyle \text{Therefore } B = (1, 1)$

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

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

Let the coordinates of the point $\displaystyle L \text{ be } (x_1, y_1)$

Therefore

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

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

$\displaystyle \text{Therefore } L = (2, 2)$

Similarly

Let the coordinates of the point $\displaystyle M \text{ be } (x_2, y_2)$

Therefore

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

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

$\displaystyle \text{Therefore } M = (6, 0)$

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

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

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

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Question 4: $\displaystyle \text{Given } M$ is the mid-point of $\displaystyle AB$ , find the co-ordinates of :

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

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

i) $\displaystyle \text{Given Midpoint of } AB = (1, 7) \text{ and let } A=(x,y)$

Therefore

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

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

$\displaystyle \text{Therefore } A = (7, 4)$

ii) $\displaystyle \text{Given Midpoint of } AB = (-1, 3) \text{ and let } A=(x, y)$

Therefore

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

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

$\displaystyle \text{Therefore } B = (-5, 7)$

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

$\displaystyle \text{Given Midpoint of } AB = (-1, 3) \text{ and let } A=(0, y) \text{ and } B =(x, 0)$

Therefore

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

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

$\displaystyle \text{Therefore } B = (-6, 0) \text{ and } A=(0, 4)$

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

$\displaystyle \text{Given Midpoint of } AB = (4, 2) \text{ and let } A=(x,0) \text{ and } B =(0,y)$

Therefore

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

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

$\displaystyle \text{Therefore } A = (8, 0) \text{ and } B= (0, 4)$

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

$\displaystyle \text{Let } P(x, y)$.be the midpoint of $\displaystyle BC$ .

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

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

$\displaystyle \text{Therefore } P = (5, -1)$

Therefore

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

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

$\displaystyle \text{For } A(x_1, y_1)$

$\displaystyle \text{Ratio } =1:1$

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

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

$\displaystyle \text{Hence the coordinates of } A = (-1, -2)$

$\displaystyle \text{For } D(x_2, y_2)$

$\displaystyle \text{Ratio } =1:1$

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

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

$\displaystyle \text{Hence the coordinates of } D = (2, 13)$

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

$\displaystyle \text{Let } P(x, y)$ be the other end of the diameter

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

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

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

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

$\displaystyle \text{Let } P(x_1, y_1)$ be the midpoint of $\displaystyle AC$ .

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

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

$\displaystyle \text{Therefore } P = (-1, 4)$

$\displaystyle \text{Let } M(x_2, y_2)$.be the midpoint of $\displaystyle BD$ .

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

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

$\displaystyle \text{Therefore } M = (-1, 4)$

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

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

$\displaystyle \text{Given Midpoint of } PR = M(-3, 2) \text{ and let } S=(x_1, y_1)$

Therefore

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

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

$\displaystyle \text{Therefore } R = (-10, 2)$

$\displaystyle \text{Given Midpoint of } SQ = M(-3, 2) \text{ and let } S=(x_2, y_2)$

Therefore

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

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

$\displaystyle \text{Therefore } S = (-5, -1)$

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

$\displaystyle \text{Let } M(x, y)$ be the midpoint of $\displaystyle BD$

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

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

$\displaystyle \text{Therefore } M = (2, 4)$

$\displaystyle \text{Given Midpoint of } AC = M(2,4) \text{ and let } C=(x, y)$

Therefore

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

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

$\displaystyle \text{Therefore } C = (5, 8)$

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

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

$\displaystyle \text{Given Midpoint of } BC = P(2,-1)$

Therefore

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

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

$\displaystyle \text{Given Midpoint of } AC = Q(-1, 4)$

Therefore

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

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

$\displaystyle \text{Given Midpoint of } AB = R(-2, 2)$

Therefore

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

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

Adding i), iii) and v) we get

$\displaystyle x_1+x_2+x_3=-1 \ \ \cdots \ \cdots \ \cdots \ vii)$

Adding ii), iv) and vi) we get

$\displaystyle y_1+y_2+y_3=5 \ \ \cdots \ \cdots \ \cdots \ viii)$

Using i), iii), v) and vii) we get

$\displaystyle x_1=1, x_2=3 \text{ and } x_3=-5$

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

$\displaystyle y_1=-3, y_2=1 \text{ and } y_3=7$

Hence the vertices are $\displaystyle (1, -3), (3, 1) \text{ and } (-5, 7)$

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

$\displaystyle \text{Given } B$ is the midpoint of $\displaystyle AC$.Therefore

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

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

$\displaystyle \text{Therefore } A = (-5, 17) \text{ and } B=(-2, 7)$

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

$\displaystyle \text{Given } Q$ is the midpoint of $\displaystyle PR$.Therefore

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

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

$\displaystyle \text{Therefore } a=-4 \text{ and } b=-1$

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

$\displaystyle \text{Let } O$ be the centroid of triangle $\displaystyle ABC$ .

Therefore

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

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

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

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

$\displaystyle \text{Given } O(2, -5)$ be the centroid of triangle $\displaystyle PQR$

Therefore

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

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

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

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

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