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 $x \ and \ y$.

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:

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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