Question

Find the equation of the line which passes through the point (– 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.

• A. $9x-20y+96=0$
• B. $3x+5y=3$
• C. $9x+20y+96=0$
• D. $9x-20y-96=0$

Answer 1

Answer: (A)

$9x-20y+96=0$

Let the equation of the line

$\frac{x}{a}+\frac{y}{b}=1.$

The coordinates of the point which divides the joining $A(a,0)andB(b,0)$ in the ratio $5:3$

So,

$[\frac{5\times 0+3a}{5+3},\frac{5\times b+3\times 0}{5+3}]=(\frac{3a}{8},\frac{5b}{8})$

It is given that,

The point $(-4,3)$ divides $AB$ in the ratio $5:3$

So,

$\frac{3a}{8}=-4\Rightarrow a=\frac{-32}{3}$

And

$\frac{5b}{8}=3$

$\Rightarrow b=\frac{24}{5}$

Hence, the equation of line is

$\frac{x}{a}+\frac{y}{b}=1$

$\Rightarrow \frac{x}{\frac{-32}{3}}+\frac{y}{\frac{24}{5}}=1$

$\Rightarrow \frac{-3x}{32}+\frac{5y}{24}=1$

$\Rightarrow -9x+20y=96$

$\Rightarrow 9x-20y+96=0$

Hence, this is the answer.