For what values of a and b the intercepts

Given equation is $a x+b y+8=0$

It can also be re-written as $a x+b y=-8$

Now, dividing by $-8$ to both the sides, we get

$\frac{a}{-8} x+\frac{b}{-8} y=1$

$\Rightarrow \frac{x}{-\frac{8}{a}}+\frac{y}{-\frac{8}{b}}=1$

So, the intercepts on the axes are $-\frac{8}{a}$ and $-\frac{8}{b}$

Now, the second equation which is given is $2 x-3 y+6=0$ It can also be re-written as $2 x-3 y=-6$

Now, dividing by $-6$ to both the sides, we get

$\frac{2}{-6} x-\frac{3}{-6} y=1$

$\Rightarrow \frac{x}{-3}+\frac{y}{2}=1$

So, the intercepts are $-3$ and 2

Now, according to the question intercepts cut off on the coordinate axes by the line $a x+b y+8=0$ are equal in length but opposite in signs to those cut off by the line $2 x-3 y+6=0$ on the axes

Therefore, $-\frac{8}{a}=3$ and $-\frac{8}{b}=-2$

$\Rightarrow \mathrm{a}=-\frac{8}{3}$ And $\Rightarrow \mathrm{b}=4$