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)

Answer:

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 .

Answer:

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 .

Answer:

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)

Answer:

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 .

Answer:

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) .

Answer:

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) .

Answer:

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 .

Answer:

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) .

Answer:

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.

Answer:

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 .

Answer:

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 .

Answer:

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.

Answer:

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 .

Answer:

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 .

Answer:

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) .

Answer:

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 .

Answer:

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 .

Answer:

Given O(0, 0) be the centroid of triangle ABC .

Therefore

0=\frac{5+(-4)+y}{3} \Rightarrow y = -1

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

Hence x=-1 \ and \ y = -1

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