Question 1: 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]

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 2: 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]

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

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 4: $AB$ 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]

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 5: 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]

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 6: 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]

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 7: (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]

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 8: 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]

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

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

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

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

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Question 9: 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]

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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Question 10: Calculate the ratio in which the line joining $A (6, 5)$ and $B (4, -3)$ is divided by the line $y=2$.      [2006]

Let the required ratio be $k: 1$ and the point of $y=2$   be $(x, 2)$

Since  $y = \frac{ky_2+y_1}{k+1}$

$\Rightarrow 2 = \frac{k \times (-3) +5}{k+1}$

$\Rightarrow 2k+2=-3k+5$

$\Rightarrow k = \frac{3}{5}$

$\Rightarrow m_1:m_2 = 3:5$

Now calculate the coordinate of the point of intersection

$x = \frac{3 \times (4)+5 \times (6)}{3+5} = 4.25$

Co-ordinates of the point of intersection = $(4.25, 2)$

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Question 11: lf $A = (-4, 3)$ and $B = (8, -6)$

(i) find the length of $AB$

(ii) In what ratio is the line joining $A \ and \ B$ , divided by the $x-axis$?      [2008]

$AB = \sqrt{(8-(-4))^2+(-6-3)^2} = \sqrt{144+81} = \sqrt{225} = 15$

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

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

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

$\Rightarrow 6k-3=0$

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

$\Rightarrow m_1:m_2 = 1:2$

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Question 12: The line segment joining $A(2, 3)$ and $8(6, -5)$ is intercepted by $x-axis$ at the point $K$ . Write down the ordinate of the point $K$. Hence, find the ratio in which $K$ divides $AB$ . Also, find the co-ordinates of the point $K$.      [1990, 2006]

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

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

$\Rightarrow 0 = \frac{k \times (-5)+3}{k+1}$

$\Rightarrow 5k-3=0$

$\Rightarrow k = \frac{3}{5}$

$\Rightarrow m_1:m_2 = 3:5$

$x = \frac{3 \times (6)+5 \times (2)}{3+5} = \frac{14}{4}$

Therefore the point $K= (\frac{14}{4}, 0)$

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Question 13: In the given figure, line $APB$ meets the $x-axis$ at point $A$ and $y-axis$ at point $B$ . $P$ is the point $(-4,2)$ and $AP: PB = 1 :2$ Find the co-ordinates of $A \ and \ B$ .     [1999, 2013]

Given $AP: PB = 1:2$

Therefore

$-4 = \frac{1 \times (0)+2 \times x}{1+2} \Rightarrow x = -6$

$2 = \frac{1 \times y+2 \times (0)}{1+2} \Rightarrow y = 6$

Therefore $A (-6, 0) \ and \ B(0,6)$.

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Question 14: Given a line segment $AB$ joining the points $A(-4,6)$ and $B(8,-3)$ . Find:

i) The ratio in which $AB$ is divided by $y-axis$ .

ii) Find the coordinates of point of intersection

iii) The length of $AB$      [2012]

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

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

$\Rightarrow 0 = \frac{k \times (8)-4}{k+1}$

$\Rightarrow 8k-4=0$

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

$\Rightarrow m_1:m_2 = 1:2$

$y = \frac{1 \times (-3)+2 \times (6)}{ 1+2} = 3$

Therefore the point intersection is $= (0, 3)$

Length of $AB = \sqrt{(8-(-4))^2+(-3-6)^2} = \sqrt{144+81} = \sqrt{225} = 15 units$.

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Question 15: $KM$ is a straight line of $13$ units. If  $K$ has the coordinates  $(2, 5)$ and  $M$  has the coordinates  $(x, -7)$ , find the value of  $x$ .     [2004]

$K(2, 5)$  and $M(x, -7)$  are the two points.

Distance between them is  $13$  units.

Therefore

$\sqrt{x-2)^2+(-7-5)^2} = 13$

$x^2+4-4x+144=169$

$x^2-4x-21=0$

$(x-7)(x+3)=0 \Rightarrow x = 7 \ or \ -3$

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Question 16: The mid point of the line segment joining (3m, 6) and (-4, 3n) is (1, 2m-1). Find the values of m and n.     [2006]

Given Midpoint of $= (1, 2m-1)$

Therefore

$1 = \frac{1 \times (-4) + 1 \times (3m)}{1+1} \Rightarrow m = 2$

$2(m)+1 = \frac{1 \times (6)+1 \times (3n)}{1+1} \Rightarrow n = 0$

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Question 17: $ABC$ is a triangle and  $G(4,3)$  is the centroid of the triangle. If  $A(1,3), B(4,b) \ and \ C(a,1)$ , find  $a \ and \ b$. Find the length of the side  $BC$.    [2011]

Since $G$ is the centroid

$4 = \frac{1+4+a}{3} \Rightarrow a=7$

$3= \frac{3+b+1}{3} \Rightarrow b = 5$

Therefore $B(4, 5) \ and \ C(7, 1)$

Therefore $BC=\sqrt{(7-4)^2+(1-5)^2} = \sqrt{25} = 5$ units.

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