Question

Find the ratio in which $AB$ divides $CD$ if given points are $A(0,1),B(15,2),C(1,2)$ and $D(4,-5)$.

• A. $\frac{87}{132}$
• B. $\frac{2}{3}$
• C. $\frac{3}{2}$
• D. $\frac{4}{3}$

Answer 1

The correct option is B $\frac{2}{3}$

A(0,1) B(15,2)

Equation of AB

y - 1 =$\frac{15}{1}(x-0)$

15x - y + 1 = 0

C(1,2) D(4,-5)

Equation of CD

y - 2 = $\frac{7}{-3}(x-1)$

$-3y+6=7x-7$

$7x+3y-13=0$

Intersection of AB and CD

45x + 3 = 13 - 7x

52x = 10

$x=\frac{5}{26}$

$y=15x+1$

$y=\frac{75}{26}+1=\frac{101}{26}$

Let ratio be $\lambda :1$

$\frac{5}{26}=\frac{(4\times 1)+(-5\times \lambda )}{\lambda +1}5\lambda +5=92-130\lambda 135\lambda =87\lambda =\frac{87}{132}Ratio=87:132$