Question

Find the equation of the straight lines passing through $(-2,-7)$ and having an intercept of length $3$ between the straight lines $4x+3y=12,4x+3y=3$

Answer 1

By geometry (Image)$=\frac{12-3}{\sqrt{4^2+3^2}}=\frac{9}{5}$
Let another intercept is $3$
There can be two line AB and AC but AB will meet in third quadrant $(-2,-7)$
$\tan \theta =\frac{12/5}{9/5}=\frac{12}{9}=\frac{4}{3}$
$m=\frac{4}{3}$
Equation of line passing through $(-2,-7)$
$(y-y_1)=m(x-x_1)$
$(y+7)=\frac{4}{3}(x+2)$
$3y+21=4x+8$
$4x-3y=13$