If the line l _ 1 : 3y - 2x = 3 is the angular bisector of the...

Given,

${L}_{1}: 3y - 2x = 3$ is angular bisector of ${L}_{2}: x - y + 1 = 0$ and ${L}_{3}: \alphax + \betay + 17 = 0$

Now finding point of intersection of ${L}_{1} \text{ and } {L}_{2}$ we get, \left(0, 1\right)$</p><p>And point will lie on${L}_{3}$, so$\alpha \times 0 + \beta \times 1 + 17 = 0$</p><p>$⇒ \beta = - 17$<br />Any point, say \left(\dfrac{- 3}{2}, 0\right)$ on ${L}_{1}$ should be equidistant from lines ${L}_{2} \text{ and } {L}_{3}$

Now using the formula of distance of a point from the line we get,
$⇒ \left|\dfrac{\dfrac{- 3}{2} - 0 + 1}{\sqrt{{1}^{2} + {1}^{2}}}\right| = \left|\dfrac{\dfrac{- 3\alpha}{2} + 0 + 17}{\sqrt{{\alpha}^{2} + ( - 17{)}^{2}}}\right|$

$⇒ \left(\alpha - 7\right)\left(\alpha - 17\right) = 0$

Now, for $\alpha = 17, {L}_{2} \text{ and } {L}_{3}$ coincides

So, $\alpha = 7$

Now putting the value of $\alpha \text{ and } \beta$ in given expression we get,

${\alpha}^{2} + {\beta}^{2} - \alpha - \beta = {\left(- 17\right)}^{2} + {7}^{2} - 7 + 17$$= 348$