Question

Find the equation of the line cutting off intercepts $\frac{-4}{3}$ and $\frac{3}{4}$ on the $X$ and $Y$ axes respectively.

Answer 1

$X$-intercept$=a=\frac{-4}{3}$

$Y$-intercept$=b=\frac{3}{4}$

Equation of the line is $\frac{x}{a}+\frac{y}{b}=1$

$\frac{x}{\frac{-4}{3}}+\frac{y}{\frac{3}{4}}=1$

$\Rightarrow \frac{-3x}{4}+\frac{4y}{3}=1$

$\Rightarrow -9x+16y=12$

$\Rightarrow 9x-16y=-12$

$\therefore$ Equation of the line is $9x-16y+12=0$