The straight line $2 x + 3 y + 1 = 0$ bisects the angle between two straight lines one of which is $3 x + 2 y + 4 = 0$ . Determine the equation of the other line.

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Solution

The line $l_{2}$ will pass through the intersection of $l_{1} = 0$ and $B = 0$ . Its equation is
$3 x + 2 y + 4 + λ (2 x + 3 y +!) = 0.... (1)$
Now choose a point $(α, β)$ on the bisector so that $2 α + 3 β + 1 = 0$ . Apply the condition $p_{1} = p_{2}$ for the two lines
$\frac{3 α + 2 β + 4 + λ (2 α + 3 β + 1)}{{(3 + 2 λ)}^{2} + {(2 + 3 λ)}^{2}} = \frac{3 α + 2 β + 4}{\sqrt{9 + 4}}$
Put $2 α + 3 β + 1 = 0$ . Cancel $3 α + 2 β + 4$
$∴ 13 = 13 λ^{2} + 24 λ + 13 = 0$
$∴ λ = 0, - \frac{24}{13}$
But $λ = 0$ corresponds to given line hence $λ = - \frac{24}{13}$ will give the other line whose equation from (1) is $9 x + 46 y = 28$

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