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Question 1: Calculate the distance between the points (6, -4) and (3,2) correct to 2 decimal places.

Answer:

(6, -4) and (3, 2)

Distance = \sqrt{(3-6)^2+(2-(-4))^2} = \sqrt{9+36} = \sqrt{45} = 6.71

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Question 2: Find the distance between the points (-2, -2) and (1,0) correct to 3 significant figures.

Answer:

(-2, -2) and (1,0)

Distance = \sqrt{(1-(-2))^2+(0-(-2))^2} = \sqrt{9+4} = \sqrt{13} =3.61

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Question 3: Show that the points P(7, 3), Q(6, 3 + \sqrt{3}) \ and \  R (5, 3)  form an equilateral triangle.

Answer:

P(7, 3), Q(6, 3 + \sqrt{3}) \ and \  R (5, 3)

PQ =   \sqrt{(6-7)^2+(3 + \sqrt{3}-3)^2} = \sqrt{1+3} = \sqrt{4} = 2  

PR =   \sqrt{(5-7)^2+(3-3)^2} = \sqrt{4+0} = \sqrt{4} = 2  

QR =   \sqrt{(5-6)^2+(3-(3 + \sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2

Therefore three sides PQ, PR \ and \ QR are equal which makes it an equilateral triangle.

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Question 4: The circle with center (x, y) passes though the points (3, 11), (14, 0) \ and \ (12, 8) . Find the values of x \ and \ y .21

Answer:

Distance of the points from the center are equal. Therefore

\sqrt{(3-x)^2+(11-y)^2}=\sqrt{(12-x)^2+(8-y)^2}

9x^2-6x+121+y^2-22y=144+x^2-24x+64+y^2-16y 

18x-6y=78  

3x-y=13 … … … … i)

\sqrt{(3-x)^2+(11-y)^2}=\sqrt{(14-x)^2+(0-y)^2} 

9x^2-6x+121+y^2-22y=196+x^2-28x+y^2 

22x-22y=66  

x-y=3  … … … … ii)

Solving i) and ii), we get x = 5 and  y = 2.

Therefore the center is  (5, 2) 

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Question 5: The points A(-1, 2), B(x,  y) \ and \ C = (4, 5) are such that BA = BC . Find a linear relation between x \ and \ y .

Answer:

 BA=BC 

\sqrt{(x-(-1))^2+(y-2)^2}=\sqrt{(4-x)^2+(5-y)^2} 

x^2+1+2x+y^2+4-4y=16+x^2-8x+25+y^2-10y 

5+2x-4y=41-8x-10y 

10x+6y=36 

5x+3y=18 

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Question 6: Given a triangle ABC in which A = (4, 4), B = (0, 5) and C = (5, 10) . A point P lies on BC such that BP : PC = 3 : 2 . Find the length of line segment AP .

Answer:

A point P (x,y) lies on BC such that BP : PC = 3 : 2 .

x = \frac{2 \times (0)+3 \times (5)}{2+3} = 3 

y = \frac{2 \times (5)+3 \times (10)}{2+3}  = 8

Therefore P = (3, 8)

Therefore

Length of AP = \sqrt{(3-4)^2+(8-(-4))^2} = \sqrt{1+144} = \sqrt{145}

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Question 7: A(20, 0) \ and \  B(10, -20) are two fixed points. Find the co-ordinates of the point P in AB such that: 3PB = AB . Also, find the co-ordinates of some other point Q \ in \ AB such that AB=6AQ .

Answer:

For P When Ratio: m_1:m_2 = 2:1  A(20, 0) \ and \  B(10, -20)

Therefore

x = \frac{2 \times (10)+1 \times (20)}{2+1}= \frac{40}{3}  

y = \frac{2 \times (-20)+1 \times (0)}{1+2} =\frac{-40}{3}

Therefore the point P= (\frac{40}{3},\frac{-40}{3})

For Q When Ratio: m_1:m_2 = 1:5  A(20, 0) \ and \  B(10, -20)

Therefore

x = \frac{1 \times (10)+5 \times (20)}{1+5}= \frac{55}{3}  

y = \frac{1 \times (-20)+5 \times (0)}{1+5} =\frac{-10}{3}

Therefore the point P= (\frac{55}{3},\frac{-10}{3})

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Question 8: A(-8, 0), B(0, 16) \ and \  C(0, 0) are the vertices of a triangle ABC . Point P lies on AB \ and \  Q lies on AC  such that AP: PB = 3: 5 \ and \ AQ: QC = 3:5 . Show that: PQ = \frac{3}{8}BC .

Answer:

For P(x_1,y_1) When Ratio: m_1:m_2 = 3:5  A(-8, 0) \ and \  B(0, 16)

Therefore

x_1 = \frac{3 \times (0)+5 \times (-8)}{3+5}= -5

y_1 = \frac{3 \times (16)+5 \times (0)}{3+5} =6

Therefore the point P= (-5,6)

For Q(x_2,y_2) When Ratio: m_1:m_2 = 3:5  A(-8, 0) \ and \  C(0, 0)

Therefore

x_2 = \frac{3 \times (0)+5 \times (-8)}{3+5}= -5

y_2 = \frac{3 \times (0)+5 \times (0)}{3+5} =0

Therefore the point Q= (-5,0)

PQ = \sqrt{(-5-(-5))^2+(0-6)^2} = \sqrt{0+36} = 6

BC = \sqrt{(0-0)^2+(16-0)^2} = \sqrt{256} = 16

Therefore PQ = \frac{3}{8} \times 16=6

Which proves that PQ = \frac{3}{8}BC

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Question 9: Find the co-ordinates of points of trisection of the line segment joining the point (6, 9) and the origin.

Answer:

Let P(x_1,y_1) \ and \  Q(x_2,y_2) be the two points dividing the points  (6, 9) and the origin in the ratio 1:2 and 2:1 respectively.

Therefore for  P 

x_1 = \frac{1 \times (0)+2 \times (6)}{1+2}= 4

y_1 = \frac{1 \times (0)+2 \times (-9)}{1+2} =-6

Hence P(4, -6) 

Therefore for  Q 

x_1 = \frac{2 \times (0)+1 \times (6)}{2+1}= 2

y_1 = \frac{2 \times (0)+1 \times (-9)}{2+1} =-3

Hence  Q(2, -3) 

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Question 10: A line segment joining A(-1, \frac{5}{3}) \ and \ B(a, 5) is divided in the ratio 1 : 3 \ at \ P , the point where the line segment AB   intersects the y-axis .

(i) Calculate the value of  a

(ii) Calculate the co-ordinates of P .      [1994]

Answer:

Therefore for  P (0,y)

0 = \frac{1 \times (a)+3 \times (-1)}{1+3} \Rightarrow a = 3

y = \frac{1 \times (5)+3 \times (\frac{5}{3})}{1+3} =\frac{5}{2}

Hence P(0, \frac{5}{2}) 

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Question 11: In what ratio is the line joining A(0, 3) \ and \ B (4, -1) divided by the  x-axis ? Write the co-ordinates of the point where AB intersects the x-axis .     [1993]

Answer:

Let the required ratio be  k:1   and the point of  x-axis    be  (x,0)

Since  y = \frac{ky_2+y_1}{k+1}

\Rightarrow 0 = \frac{k \times (-1) +3}{k+1}

\Rightarrow k=3

\Rightarrow  m_1:m_2 = 3:1

Therefore  x = \frac{3(4)+1(0)}{3+1} = 3

Therefore P(3,0)

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Question 12: The mid-point of the segment AB , as shown in diagram, is C(4, -3) . Write down the coordinates of A \ and \  B .     [1996]

Answer:

Given Midpoint of AB = (4,-3)

Therefore

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

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

Therefore A = (8, 0) \ and \  B(0, -6)

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Question 13: AB is a diameter of a circle with center C = (-2, 5) . If A = (3, -7) , find

(i) the length of radius AC 

(ii) the coordinates of B . [2013]

Answer:11

Given Midpoint of AB = C(-2,5)

Therefore

-2 = \frac{1 \times x+1 \times (3)}{1+1} \Rightarrow x = -7  

5 = \frac{1 \times (y)+1 \times (-7)}{1+1}  \Rightarrow y = 17

Therefore B = (-7, 17)

AC = \sqrt{(-7-3)^2+(17-(-7))^2} = \sqrt{676} = 26

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Question 14: Find the co-ordinates of the centroid of a triangle ABC whose vertices are : A(-1, 3), B(1, -1) \ and \ C(5, 1) .     [2006]

Answer:

Let O(x, y) be the centroid of triangle ABC .

Therefore

x=\frac{-1+1+5}{3}= \frac{5}{3}

y = \frac{3-1+1}{3}=1

Hence the coordinates of the centroid are (\frac{5}{3}, 1)

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Question 15: The mid-point of the line segment joining (4a, 2b-3) \ and \ (-4, 3b) \ is \  (2, -2a) . Find the values of a \ and \  b .

Answer:

Given Midpoint of  = C(2, -2a)

Therefore

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

-2(2) = \frac{1 \times (2b-3)+1 \times (3b)}{1+1}  \Rightarrow b = -1

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Question 16: The mid-point of the line segment joining (2a, 4) \ and \ (-2, 2b) \ is \ (1, 2a+1) . Find the values of a \ and \ b .      [2007]

Answer:

Given Midpoint of  = C(1, 2a+1)

Therefore

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

2(2)+1 = \frac{1 \times (4)+1 \times (2b)}{1+1}  \Rightarrow b = 3

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Question 17: (i) Write down the co-ordinates of the point P that divides the line joining A(- 4, l) \ and \ B(17, 10) in the ratio 1 : 2 .

(ii) Calculate the distance OP , where O is the origin.

(iii) In what ratio does the y-axis divide the line AB ? [1995]

Answer:

i) For P When Ratio: m_1:m_2 = 1:2 A(- 4, l) \ and \ B(17, 10)

Therefore

x = \frac{1 \times (17)+2 \times (-4)}{1+2}= 3  

y = \frac{1 \times (10)+2 \times (1)}{1+2} = 4

Therefore the point P= (3, 4)

ii) OP = \sqrt{(3-0)^2+(4-0)^2} = \sqrt{25} = 5

iii)  Let the required ratio be  k:1    and the point be Q(0,y)

Since  y = \frac{ky_2+y_1}{k+1}

\Rightarrow 0 = \frac{k \times (17) -4}{k+1}

\Rightarrow k=\frac{4}{17}

\Rightarrow  m_1:m_2 = 4:17

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Question 18: Prove that the points A(-5,4); B(-1, -2) \ and \ C(5, 2) are the vertices of an isosceles right-angled triangle. Find the co-ordinates of D so that ABCD is a square.      [1992]

Answer:

AC = \sqrt{(5-(-5))^2+(2-4)^2} = \sqrt{104}  

AB= \sqrt{(-1-(-5))^2+(-2-(-4))^2} = \sqrt{52}  

BC = \sqrt{(-1-(-5))^2+(-2-2)^2} = \sqrt{52}  

Since AB=BC (two sides are equal). Hence triangle ABC is a isosceles triangle.

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Question 19: M is the mid-point of the line segment joining the points A(-3, 7) \ and \ B(9, -1) . Find the coordinates of point M . Further, if R(2, 2) divides the line segment joining M and the origin in the ratio p : q , find the ratio p : q .

Answer:

For M When Ratio: m_1:m_2 = 1:1  for A(-3, 7) \ and \ B(9, -1)

Therefore

x = \frac{1 \times (9)+1 \times (-3)}{1+1}= 3  

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

Therefore the point M= (3, 3)

Let R(2,2) divide MO in the ratio  k:1  

Since  y = \frac{ky_2+y_1}{k+1}

\Rightarrow 0 = \frac{k \times (0) +3}{k+1}

\Rightarrow k=\frac{1}{2}

\Rightarrow  p:q=1:2

 

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Question 20: Calculate the ratio in which the line joining A(-4, 2) \ and \ B(3, 6) is divided by point P(x, 3) . Also, find (i) x (ii) length of AP .   [2014]

Answer:

Let P(x,3) divide MO in the ratio  k:1  

Since  y = \frac{ky_2+y_1}{k+1}

\Rightarrow 3 = \frac{k \times (6) +2}{k+1}

\Rightarrow k=\frac{1}{3}

\Rightarrow  m_1:m_2=1:3

Since  x = \frac{kx_2+x_1}{k+1}

\Rightarrow x = \frac{1 \times (3) +3 \times (-4)}{1+3} = \frac{-9}{4}

AP = \sqrt{(-\frac{9}{4}-(-4))^2+(3-2)^2} = \sqrt{(\frac{7}{4})^2+1)} = \sqrt{\frac{65}{16}}  

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