Let the point P(α, β) be at a unit distance from each of the two...

Given, the point P\left(&alpha;\right)$\left(, &beta;\right)$ be at a unit distance from each of the two lines ${L}_{1}: 3x - 4y + 12 = 0$, and ${L}_{2}: 8x + 6y + 11 = 0$, so on plotting the diagram we get,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/d44e0e3d-e905-4d01-a411-2ab8ff51914c-image.png" style="width: 244px; height: 300px;" /></p><p>Now, ${L}_{1}: 3x - 4y + 12 = 0$ and ${L}_{2}: 8x + 6y + 11 = 0$</p><p>Since ${L}_{1} \text{ and } {L}_{2}$ are perpendicular so it will form square of unit length, so equation of angle bisector of ${L}_{1}$ and ${L}_{2}$ of angle containing origin and will pass through \left(&alpha;, &beta;\right)$, so</p><p>Equation of angle bisector will be, $\dfrac{\left(3x - 4y + 12\right)}{\sqrt{{3}^{2} + {4}^{2}}} = \dfrac{8x + 6y + 11}{\sqrt{{8}^{2} + {6}^{2}}}$</p><p>$\to 2\left(3x - 4y + 12\right) = 8x + 6y + 11$</p><p>$\to 2x + 14y - 13 = 0$</p><p>$\to 2&alpha; + 14&beta; - 13 = 0 ⋯ \left(i\right)$</p><p>Also given perpendicular distance is one unit so, $\dfrac{3&alpha; - 4&beta; + 12}{5} = 1$</p><p>$\to 3&alpha; - 4&beta; + 7 = 0 ⋯ \left(ii\right)$</p><p>Solving equation \left(\text{i}\right) \text{ and } \left(\text{ii}\right)$</p><p> $2&alpha; + 14&beta; - 13 = 0$</p><p> $3&alpha; - 4&beta; + 7 = 0$ </p><p>$\toP\left(α\right)\left(, β\right), α = \dfrac{- 23}{25}, β = \dfrac{53}{50}$</p><p>So,$100\left(α + β\right) = 14$