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Q)

For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.

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

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A)
Toolbox:
• 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$
But how the slopes are equal? And why the origin is taken in distance formula?