Note: We will use the formula. If a point divides two points $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m_1:m_2$, then the coordinates of the point at

$x = \frac{m_1x_2+m_2x_1}{m_1+m_2}$

$y = \frac{m_1y_2+m_2y_1}{m_1+m_2}$

Question 1: Calculate the co-ordinates of the point $P$ which divides the line segment joining:

(i) $A (1, 3)$ and $B (5, 9)$ in the ratio $1: 2$

(ii) $A (-4, 6)$ and $B (3, -5)$ in the ratio $3: 2$

i)

Ratio: $m_1:m_2 = 1:2$

Let the coordinates of the point $P \ be \ (x, y)$

Therefore

$x = \frac{1 \times 5+2 \times 1}{1+2} = \frac{7}{3}$

$y = \frac{1 \times 9+2 \times 3}{1+2} = \frac{15}{3} = 5$

Therefore $P = (\frac{7}{3}, 5)$

ii)

Ratio: $m_1:m_2 = 3:2$

Let the coordinates of the point $P \ be \ (x, y)$

Therefore

$x = \frac{2 \times (-4)+3 \times 3}{3+2} = \frac{1}{5}$

$y = \frac{2 \times 6+3 \times (-5)}{3+2} = -\frac{3}{5}$

Therefore $P = (\frac{1}{5}, -\frac{3}{5})$

$\\$

Question 2: In what ratio is the line joining $(2, -3)$ and $(5, 6)$ divided by the $x-axis$.

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

Since  $y = \frac{ky_2+y_1}{k+1}$

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

$\Rightarrow 6k-3=0$

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

$\Rightarrow m_1:m_2 = 1:2$

$\\$

Question 3: In what ratio is the line joining $(2, -4)$ and $(-3, 6)$ divided by the $y-axis$ .

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

Since  $x = \frac{kx_2+x_1}{k+1}$

$\Rightarrow 0 = \frac{k \times (-3) +2}{k+1}$

$\Rightarrow 3k-2=0$

$\Rightarrow k = \frac{2}{3}$

$\Rightarrow m_1:m_2 = 2:3$

$\\$

Question 4: In what ratio does the point $(1, a)$ divide the join of $(-1, 4)$ and $(4, -1)$? Also, find the value of $a$.

Let the point $(1, a)$ divide the join of $(-1, 4)$ and $(4, -1)$ in the ratio $k: 1$

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

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

$\Rightarrow k+1=4k-1$

$\Rightarrow k = \frac{2}{3}$

$\Rightarrow m_1:m_2 = 2:3$

Now calculate $a$

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

$\Rightarrow a = \frac{10}{5}=2$

$\\$

Question 5: In what ratio does the point $(a, 6)$ divide the join of $(-4, 3)$ and $(2, 8)$? Also, find the value of $a$.

Let the point $(a, 6)$ divide the join of $(-4, 3)$ and $(2, 8)$ in the ratio $k: 1$

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

$\Rightarrow 6 = \frac{k \times (8) +3}{k+1}$

$\Rightarrow 6k+6=8k+3$

$\Rightarrow k = \frac{3}{2}$

$\Rightarrow m_1:m_2 = 3:2$

Now calculate $a$

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

$\Rightarrow 5a = 6-8=-2$

$\Rightarrow a = -\frac{2}{5}$

$\\$

Question 6: In what ratio is the join of $(4, 3)$ and $(2, -6)$ divided by the $x-axis$ . Also, find the co-ordinates of the point of intersection.

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

Since  $y = \frac{ky_2+y_1}{k+1}$

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

$\Rightarrow 6k-3=0$

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

$\Rightarrow m_1:m_2 = 1:2$

Now calculate the coordinate of the point of intersection

$x = \frac{1 \times (2)+2 \times (4)}{1+2} = \frac{10}{3}$

$y = \frac{1 \times (-6)+2 \times (3)}{1+2} = 0$

Co-ordinates of the point of intersection = $(\frac{10}{3}, 0)$

$\\$

Question 7: Find the ratio in which the join of $(-4, 7)$ and $(3,0)$ is divided by the $y-axis$ . Also, find the co-ordinates of the point of intersection.

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

Since  $x = \frac{kx_2+x_1}{k+1}$

$\Rightarrow 0 = \frac{k \times (3) -4}{k+1}$

$\Rightarrow 3k-4=0$

$\Rightarrow k = \frac{4}{3}$

$\Rightarrow m_1:m_2 = 4:3$

Now calculate the coordinate of the point of intersection

$y = \frac{4 \times (0)+3 \times (7)}{4+3} = 3$

Co-ordinates of the point of intersection = $(0, 3)$

$\\$

Question 8: Points $A, B, C \ and \ D$ divide the line segment joining the point $(5, -10)$ and the origin in five equal parts. Find the co-ordinates of $B \ and \ D$.

$PB: BQ=2:3$

$PD: DQ=4:1$

For $B$

Ratio $=2:3$

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

$y = \frac{2 \times (-10) +3 \times (0)}{2+3} = -4$

Hence the coordinates of $B = (2, -4)$

For $D$

Ratio $=4:1$

$x = \frac{4 \times (5) +1 \times (0)}{4+1} = 4$

$y = \frac{4 \times (-10) +1 \times (0)}{4+1} = -8$

Hence the coordinates of $D = (4, -8)$

$\\$

Question 9: The line joining the points $A (-3, -10)$ and $B (-2,6)$ is divided by the point $P$  such that $\frac{PB}{AB}=\frac{1}{5}$ Find the co-ordinates of $P$ .

Given $\frac{PB}{AB}=\frac{1}{5}$

This implies that $AP: PB=4:1$

$x = \frac{4 \times (-2) +1 \times (-3)}{4+1} = -\frac{11}{5}$

$y = \frac{4 \times (6) +1 \times (-10)}{4+1} = -\frac{14}{5}$

Hence the coordinates of the point $= (-\frac{11}{5} , -\frac{14}{5} )$

$\\$

Question 10: $P$ is a point on the line joining $A (4, 3)$ and $B (-2, 6)$ such that $5AP = 2BP$ Find the coordinates of $P$.

Given $5AP = 2BP$

This implies $\frac{AP}{BP}=\frac{2}{5}$

This implies that $AP: PB=2:5$

$x = \frac{2 \times (-2) +5 \times (4)}{2+5} = \frac{16}{7}$

$y = \frac{2 \times (6) +5 \times (3)}{2+5} = \frac{27}{7}$

Hence the coordinates of the point $= (\frac{16}{7}, \frac{27}{7})$

$\\$

Question 11: Calculate the ratio in which the line joining the points $(-3, -1)$ and $(5, 7)$ is divided by the line $x = 2$. Also, find the co-ordinates of the point of intersection.

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

Since  $x = \frac{kx_2+x_1}{k+1}$

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

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

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

$\Rightarrow m_1:m_2 = 5:3$

Now calculate the coordinate of the point of intersection

$y = \frac{5 \times (7)+3 \times (-1)}{5+3} = 4$

Co-ordinates of the point of intersection = $(2, 4)$

$\\$

Question 12: 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 = $(4.25, 2)$

$\\$

Question 13: The point $P (5, -4)$ divides the line segment $AB$, as shown in the figure, in the ratio $2: 5$. Find the co-ordinates of points $A \ and \ B$.

Ratio: $m_1:m_2 = 2:5$

Therefore

$5 = \frac{2 \times (0)+5 \times x}{2+5} \Rightarrow x = 7$

$-4 = \frac{2 \times y+5 \times (0)}{2+5} \Rightarrow y = -14$

Therefore $A = (7, 0) \ and \ B=(0, -14)$

$\\$

Question 14: Find the co-ordinates of the points of trisection of the line joining the points $(-3, 0)$ and $(6, 6)$.

When Ratio: $m_1:m_2 = 1:2$

Therefore

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

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

Therefore the point $= (0, 2)$

When Ratio: $m_1:m_2 = 2:1$

Therefore

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

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

Therefore the point $= (3, 4)$

$\\$

Question 15: Show that the Line segment joining the points $(-5, 8)$ and $(10, -4)$ is trisection by the co-ordinate axes.

Let the two points trisecting the points $(-5, 8)$ and $(10, -4)$ are $A (x, 0) \ and \ B (0, y)$.

When Ratio for B: $m_1:m_2 = 1:2$

Therefore

$y = \frac{1 \times (-4)+2 \times (8)}{1+2} = 4$

Therefore the point $= (0, 4)$

When Ratio for A: $m_1:m_2 = 2:1$

Therefore

$x = \frac{2 \times (10)+1 \times (-5)}{2+1} = \frac{5}{3}$

Therefore the point $= (\frac{5}{3}, 0)$

$\\$

Question 16: Show that $A (3, -2)$ is a point of trisection of the line-segment joining the points $(2, 1)$ and $(5, -8)$. Also find the co-ordinates of the other point of trisection.

Let the point $A (3, -2)$ divide the join of $(2, 1)$ and $(5, -8)$ in the ratio $k: 1$

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

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

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

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

$\Rightarrow m_1:m_2 = 1:2$

For the other point

When Ratio: $m_1:m_2 = 2:1$

Therefore

$x = \frac{2 \times (5) +1 \times (2)}{2+1}=4$

$y = \frac{2 \times (-8)+1 \times (1)}{2+1} = -5$

Therefore the point $= (4, -5)$

$\\$

Question 17: 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 $y = \frac{ky_2+y_1}{k+1}$

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

$\Rightarrow 6k-3=0$

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

$\Rightarrow m_1:m_2 = 1:2$

$\\$

Question 18: The line segment joining the points $M (5, 7)$ and $N (-3, 2)$ is intersected by the $y-axis$ at point $L$ . Write down the abscissa of $L$ . Hence, find the ratio in which $L$ divides $MN$ . Also, find the co-ordinates of $L$ .

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

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

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

$\Rightarrow 3k-5=0$

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

$\Rightarrow m_1:m_2 = 5:3$

$y = \frac{5 \times (2)+3 \times (7)}{5+3} = \frac{31}{8}$

Therefore the point $= (0, \frac{31}{8})$

$\\$

Question 19: $A (2, 5), B (-1, 2) \ and \ C (5, 8)$ are the co-ordinates of the vertices of the triangle $ABC$. Points $P \ and \ Q$ lie on $AB$ and $AC$ respectively, such that: $AP: PB = AQ: QC = 1 : 2$

(i) Calculate the co-ordinates of $P$ and $Q$ .

(ii) Show that $PQ = \frac{1}{3} BC$

For P When Ratio: $m_1:m_2 = 1:2$

Therefore

$x = \frac{1 \times (-1)+2 \times (2)}{1+2}= 1$

$y = \frac{1 \times (2)+2 \times (5)}{1+2} = 4$

Therefore the point $P= (1, 4)$

For Q When Ratio: $m_1:m_2 = 1:2$

Therefore

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

$y = \frac{1 \times 8+2 \times (5)}{1+2} = 6$

Therefore the point $= (3, 6)$

$\\$

Question 20: $A (-3,4), B (3, -1) \ and \ C (-2, 4)$ are the vertices of a triangle $ABC$ . Find the length of line segment $AP$ , where point $P$ lies inside $BC$ , such that $BP: PC = 2 : 3$

For P When Ratio: $m_1:m_2 = 2:3$

Therefore

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

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

Therefore the point $P= (1,1)$

$AP = \sqrt{(1-(-3))^2+(1-4)^2} = \sqrt{16+9} = \sqrt{25} = 5$

$\\$

Question 21: 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 $K= (\frac{14}{4}, 0)$

$\\$

Question 22: The line segment joining $A(4, 7)$ and $B(-6, -2)$ is intercepted by the $y-axis$ at the point $K$ . Write down the abscissa of the point $K$ . Hence, find the ratio in which $K$ divides $AB$ . Also, find the co-ordinates of the point $K$ .

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

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

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

$\Rightarrow 6k-4=0$

$\Rightarrow k = \frac{2}{3}$

$\Rightarrow m_1:m_2 = 2:3$

$y = \frac{2 \times (-2)+3 \times (7)}{2+3} = \frac{17}{5}$

Therefore the point $K= (0, \frac{17}{5})$

$\\$

Question 23: The line joining $P(-4, 5)$ and $Q(3, 2)$ intersects the $y-axis$ at point $R$ . $PM$ and $QN$ are perpendiculars from $P \ and \ Q$ on the $x-axis$. Find:

(i) The ratio $PR: RQ$.

(ii) The co-ordinates of $R$.

(iii) The area of the quadrilateral $PMNQ$.

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

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

$\Rightarrow 0 = \frac{k \times (3)-4}{k+1}$

$\Rightarrow 3k-4=0$

$\Rightarrow k = \frac{4}{3}$

$\Rightarrow m_1:m_2 = 4:3$

$y = \frac{\frac{4}{3} \times (2)+1 \times (5)}{ \frac{4}{3}+1} = \frac{23}{7}$

Therefore the point $R= (0, \frac{23}{7})$

Area of quadrilateral $PMNQ = 7 \times \frac{5+2}{2} = 24.5$ sq. units

$\\$

Question 24: 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 $A (-6, 0) \ and \ B(0,6)$.

$\\$

Question 25: Given a line segment $AB$ joining the points $A(-4,6)$ and $B(8,-3)$ . Find:

1. i) The ratio in which $AB$ is divided by $y-axis$ .
2. 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 $AB = \sqrt{(8-(-4))^2+(-3-6)^2} = \sqrt{144+81} = \sqrt{225} = 15 units$.

$\\$