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

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

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

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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.

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

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

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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.

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

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

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

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

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 $\displaystyle x \text{ and } y$

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 $\displaystyle x \text{ and } y$

$\displaystyle BA=BC$

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

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

$\displaystyle 5+2x-4y=41-8x-10y$

$\displaystyle 10x+6y=36$

$\displaystyle 5x+3y=18$

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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 $\displaystyle AP$

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 $\displaystyle AB=6AQ$

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: $\displaystyle PQ = \frac{3}{8} BC$

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

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]

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]

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]

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

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]

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

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]

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 $\displaystyle a \text{ and } b$

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

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]

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

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]

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]

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

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

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

$\displaystyle \text{Since } AB=BC$ (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 $\displaystyle p : q$

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]

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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