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

Answer:

\displaystyle (6, -4) \text{and } (3, 2)  

\displaystyle \text{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 \displaystyle (-2, -2) \text{and } (1,0) correct to \displaystyle 3 significant figures.

Answer:

\displaystyle (-2, -2) \text{and } (1,0)  

\displaystyle \text{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 \displaystyle P(7, 3), Q(6, 3 + \sqrt{3}) \text{ and } R (5, 3) form an equilateral triangle.

Answer:

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

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

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

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

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

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

Answer:

Distance of the points from the center are equal. Therefore

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

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

\displaystyle 18x-6y=78  

\displaystyle 3x-y=13 … … … … i)

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

\displaystyle 9x^2-6x+121+y^2-22y=196+x^2-28x+y^2  

\displaystyle 22x-22y=66  

\displaystyle x-y=3 … … … … ii)

Solving i) and ii), we get \displaystyle x = 5 \text{and } y = 2 .

Therefore the center is \displaystyle (5, 2)  

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

Answer:

\displaystyle BA=BC  

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

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

\displaystyle 5+2x-4y=41-8x-10y  

\displaystyle 10x+6y=36  

\displaystyle 5x+3y=18  

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

Answer:

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

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

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

\displaystyle \text{Therefore } P = (3, 8)  

Therefore

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

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

Answer:

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

Therefore

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

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

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

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

Therefore

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

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

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

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

Answer:

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

Therefore

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

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

\displaystyle \text{Therefore the point } P= (-5,6)  

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

Therefore

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

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

\displaystyle \text{Therefore the point } Q= (-5,0)  

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

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

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

\displaystyle \text{Which proves that } PQ = \frac{3}{8} BC  

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Question 9: Find the coordinates of points of trisection of the line segment joining the point \displaystyle (6, 9) and the origin.

Answer:

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

Therefore \displaystyle \text{For } P  

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

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

\displaystyle \text{Hence } P(4, -6)  

Therefore \displaystyle \text{For } Q  

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

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

\displaystyle \text{Hence } Q(2, -3)  

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

(i) Calculate the value of \displaystyle a  

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

Answer:

Therefore \displaystyle \text{For } P (0,y)  

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

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

\displaystyle \text{Hence } P(0, \frac{5}{2} )  

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

Answer:

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

\displaystyle \text{Since } y = \frac{ky_2+y_1}{k+1}  

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

\displaystyle \Rightarrow k=3  

\displaystyle \Rightarrow m_1:m_2 = 3:1  

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

\displaystyle \text{Therefore } P(3,0)  

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

Answer:

\displaystyle \text{Given Midpoint of } AB = (4,-3)  

Therefore

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

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

\displaystyle \text{Therefore } A = (8, 0) \text{ and } B(0, -6)  

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

(i) the length of radius \displaystyle AC  

(ii) the coordinates of \displaystyle B [2013]

Answer:

\displaystyle \text{Given Midpoint of } AB = C(-2,5)  

Therefore

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

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

\displaystyle \text{Therefore } B = (-7, 17)  

\displaystyle 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 \displaystyle ABC whose vertices are : \displaystyle A(-1, 3), B(1, -1) \text{ and } C(5, 1)  [2006]

Answer:

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

Therefore

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

\displaystyle y = \frac{3-1+1}{3}=1  

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

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

Answer:

\displaystyle \text{Given Midpoint of } = C(2, -2a)  

Therefore

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

\displaystyle -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 \displaystyle (2a, 4) \text{ and } (-2, 2b) is (1, 2a+1) . Find the values of \displaystyle a \text{ and } b . [2007]

Answer:

\displaystyle \text{Given Midpoint of } = C(1, 2a+1)  

Therefore

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

\displaystyle 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 \displaystyle P that divides the line joining \displaystyle A(- 4, l) \text{ and } B(17, 10) in the ratio \displaystyle 1 : 2 .

(ii) Calculate the \displaystyle \text{Distance } OP , where \displaystyle O is the origin.

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

Answer:

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

Therefore

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

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

\displaystyle \text{Therefore the point } P= (3, 4)  

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

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

\displaystyle \text{Since } y = \frac{ky_2+y_1}{k+1}  

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

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

\displaystyle \Rightarrow m_1:m_2 = 4:17  

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

Answer:

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

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

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

\displaystyle \text{Since } AB=BC (two sides are equal). Hence triangle \displaystyle ABC is a isosceles triangle.

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

Answer:

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

Therefore

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

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

\displaystyle \text{Therefore the point } M= (3, 3)  

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

\displaystyle \text{Since } y = \frac{ky_2+y_1}{k+1}  

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

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

\displaystyle \Rightarrow p:q=1:2  

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

Answer:

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

\displaystyle \text{Since } y = \frac{ky_2+y_1}{k+1}  

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

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

\displaystyle \Rightarrow m_1:m_2=1:3  

\displaystyle \text{Since } x = \frac{kx_2+x_1}{k+1}  

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

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

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