$\begin {array} {1 1} (A)\; \large\frac{8}{3}, 4 & \quad (B)\;\large\frac{-8}{3},4 \\ (C)\;\large\frac{3}{8},4 & \quad (D)\;\large\frac{-3}{8},4 \end {array}$

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- If two lines are parallel then their slopes are equal.
- Distance of the perpendicular from a point $(x_1, y_1)$ to the line $ax+by+c=0$ is $ d = \bigg| \large\frac{a(x_1)+b(y_1)+c}{\sqrt{a^2+b^2}} \bigg|$

Step 1 :

It is given that the intercepts cut off on the coordinate axes by the line $ ax+by+8=0$------(1) are equal in length but opposite in signs to those cut off by the line. $2x-3y+6=0$-------(2)

This implies that the slope of the two lines are equal.

Slope of line (1) is $ -\large\frac{a}{b}$

Slope of line (2) is $ \large\frac{2}{3}$

$ \therefore -\large\frac{a}{b} = \large\frac{2}{3}$

$ \Rightarrow a = \large\frac{-2b}{3}$

Step 2 :

The length of the perpendicular from the origin to the line (1) is

$ d_1 = \bigg| \large\frac{a(0)+b(0)+8}{\sqrt{a^2+b^2}} \bigg|$

Substituting for $a$ we get,

$ d_1 = \large\frac{8 \times 3 }{ \sqrt{13b^2}}$

Length of the perpendicular from the origin to line $(l_2)$ is

$ d_2 = \bigg| \large\frac{2(0)-3(0)+6}{\sqrt{2^2+3^2}} \bigg|$

But $d_1=d_2$

$ \therefore \large\frac{8 \times 3 }{\sqrt{13b^2}} $$ = \large\frac{6}{13}$

$ \Rightarrow b = 4$

$ \therefore a = \large\frac{-2b}{3}$$ = \large\frac{-2 \times 4}{3}$$ = \large\frac{-8}{3}$

hence the value of $a$ and $b$ are $ \large\frac{-8}{3}$$, 4$

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