Class 10: Distance and Section Formula – Exercise 13(b) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

$\displaystyle x = \frac{m_1x_2+m_2x_1}{m_1+m_2} y = \frac{m_1y_2+m_2y_1}{m_1+m_2}$

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

$\displaystyle \text{i) } A (1, 3) \text{ and } B (5, 9) \text{ in the ratio } 1: 2$

$\displaystyle \text{ii) } A (-4, 6) \text{ and } B (3, -5) \text{ in the ratio } 3: 2$

Answer:

$\displaystyle \text{i) } \text{Ratio: } m_1:m_2 = 1:2$

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

Therefore

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

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

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

$\displaystyle \text{ii) } \text{Ratio: } m_1:m_2 = 3:2$

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

Therefore

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

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

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

$\displaystyle \\$

Question 2: In what ratio is the line joining $\displaystyle (2, -3) \text{ and } (5, 6)$ divided by the $\displaystyle x-axis$

Answer:

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

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

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

$\displaystyle \Rightarrow 6k-3=0$

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

$\displaystyle \Rightarrow m_1:m_2 = 1:2$

$\displaystyle \\$

Question 3: In what ratio is the line joining $\displaystyle (2, -4) \text{ and } (-3, 6)$ divided by the $\displaystyle y-axis$

Answer:

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

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

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

$\displaystyle \Rightarrow 3k-2=0$

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

$\displaystyle \Rightarrow m_1:m_2 = 2:3$

$\displaystyle \\$

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

Answer:

Let the point $\displaystyle (1, a)$ divide the join of $\displaystyle (-1, 4) \text{ and } (4, -1) \text{ in the ratio } k: 1$

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

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

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

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

$\displaystyle \Rightarrow m_1:m_2 = 2:3$

Now calculate $\displaystyle a$

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

$\displaystyle \Rightarrow m_1:m_2 = 3:2$

Now calculate $\displaystyle a$

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

$\displaystyle \Rightarrow 6k-3=0$

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

$\displaystyle \Rightarrow m_1:m_2 = 1:2$

Now calculate the coordinate of the point of intersection

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

$\displaystyle \Rightarrow 3k-4=0$

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

$\displaystyle \Rightarrow m_1:m_2 = 4:3$

Now calculate the coordinate of the point of intersection

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle PB: BQ=2:3$

$\displaystyle PD: DQ=4:1$

For $\displaystyle B$

Ratio $\displaystyle =2:3$

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

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

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

For $\displaystyle D$

Ratio $\displaystyle =4:1$

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{ Given } 5AP = 2BP$

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

$\displaystyle \Rightarrow m_1:m_2 = 5:3$

Now calculate the coordinate of the point of intersection

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

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

$\displaystyle \\$

Question 12: Calculate the ratio in which the line joining $\displaystyle A (6, 5) \text{ and } B (4, -3)$ is divided by the line $\displaystyle y=2$ . [2006]

Answer:

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

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

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

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

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

$\displaystyle \Rightarrow m_1:m_2 = 3:5$

Now calculate the coordinate of the point of intersection

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Ratio: } m_1:m_2 = 2:5$

Therefore

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

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

$\displaystyle \text{Therefore } A = (7, 0) \ and \ B=(0, -14)$

$\displaystyle \\$

Question 14: Find the co-ordinates of the points of trisection of the line joining the points $\displaystyle (-3, 0) \text{ and } (6, 6)$

Answer:

When $\displaystyle \text{Ratio: } m_1:m_2 = 1:2$

Therefore

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

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

$\displaystyle \text{Therefore the point } = (0, 2)$

When $\displaystyle \text{Ratio: } m_1:m_2 = 2:1$

Therefore

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

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

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

$\displaystyle \\$

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

Answer:

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

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

Therefore

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

$\displaystyle \text{Therefore the point } = (0, 4)$

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

Therefore

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

$\displaystyle \Rightarrow m_1:m_2 = 1:2$

For the other point

When $\displaystyle \text{Ratio: } m_1:m_2 = 2:1$

Therefore

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

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

$\displaystyle \text{Therefore the point } = (4, -5)$

$\displaystyle \\$

Question 17: lf $\displaystyle A = (-4, 3) \text{ and } B = (8, -6)$

(i) find the length of $\displaystyle AB$

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

Answer:

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

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

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

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

$\displaystyle \Rightarrow 6k-3=0$

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

$\displaystyle \Rightarrow m_1:m_2 = 1:2$

$\displaystyle \\$

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

Answer:

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

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

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

$\displaystyle \Rightarrow 3k-5=0$

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

$\displaystyle \Rightarrow m_1:m_2 = 5:3$

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

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

$\displaystyle \\$

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

$\displaystyle \text{(i) Calculate the co-ordinates of } P \text{ and } Q$ .

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

Answer:

$\displaystyle \text{For P When Ratio: } m_1:m_2 = 1:2$

Therefore

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

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

$\displaystyle \text{Therefore the point } P= (1, 4)$

$\displaystyle \text{For Q When Ratio: } m_1:m_2 = 1:2$

Therefore

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

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

$\displaystyle \text{Therefore the point } = (3, 6)$

$\displaystyle \\$

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

Answer:

For P When $\displaystyle \text{Ratio: } m_1:m_2 = 2:3$

Therefore

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

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

$\displaystyle \text{Therefore the point } P= (1,1)$

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

$\displaystyle \\$

Question 21: The line segment joining $\displaystyle A(2, 3) \text{ and } 8(6, -5)$ is intercepted by $\displaystyle x-axis$ at the point $\displaystyle K$ . Write down the ordinate of the point $\displaystyle K$ . Hence, find the ratio in which $\displaystyle K$ divides $\displaystyle AB$ . Also, find the co-ordinates of the point $\displaystyle K$ . [1990, 2006]

Answer:

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

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

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

$\displaystyle \Rightarrow 5k-3=0$

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

$\displaystyle \Rightarrow m_1:m_2 = 3:5$

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

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

$\displaystyle \\$

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

Answer:

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

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

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

$\displaystyle \Rightarrow 6k-4=0$

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

$\displaystyle \Rightarrow m_1:m_2 = 2:3$

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

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

$\displaystyle \\$

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

$\displaystyle \text{(i) The ratio } PR: RQ$

$\displaystyle \text{(ii) The co-ordinates of } R$

$\displaystyle \text{(iii) The area of the quadrilateral } PMNQ$

Answer:

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

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

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

$\displaystyle \Rightarrow 3k-4=0$

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

$\displaystyle \Rightarrow m_1:m_2 = 4:3$

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

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

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

$\displaystyle \\$

Question 24: In the given figure, line $\displaystyle APB$ meets the $\displaystyle x-axis$ at point $\displaystyle A \text{ and } y-axis$ at point $\displaystyle B$ . $\displaystyle P$ is the point $\displaystyle (-4,2) \text{ and } AP: PB = 1 :2$ Find the co-ordinates of $\displaystyle A \ and \ B$ . [1999, 2013]