Let the point P(alpha, beta) be at a unit dis | vedclass

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(D) The distance of point $P(\alpha, \beta)$ from $L_{1}: 3x - 4y + 12 = 0$ is $1$,so $\frac{|3\alpha - 4\beta + 12|}{\sqrt{3^2 + (-4)^2}} = 1 \implie

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$14$

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(D) The distance of point $P(\alpha, \beta)$ from $L_{1}: 3x - 4y + 12 = 0$ is $1$,so $\frac{|3\alpha - 4\beta + 12|}{\sqrt{3^2 + (-4)^2}} = 1 \implies |3\alpha - 4\beta + 12| = 5$. Since $P$ is below $L_{1}$,$3\alpha - 4\beta + 12 The distance of point $P(\alpha, \beta)$ from $L_{2}: 8x + 6y + 11 = 0$ is $1$,so $\frac{|8\alpha + 6\beta + 11|}{\sqrt{8^2 + 6^2}} = 1 \implies |8\alpha + 6\beta + 11| = 10$. Since $P$ is above $L_{2}$,$8\alpha + 6\beta + 11 > 0$,so $8\alpha + 6\beta + 11 = 10 \implies 8\alpha + 6\beta + 1 = 0$.Solving the system of equations:$3\alpha - 4\beta = -17$ $(i)$$8\alpha + 6\beta = -1$ (ii)Multiplying $(i)$ by $3$ and (ii) by $2$:$9\alpha - 12\beta = -51$$16\alpha + 12\beta = -2$Adding them: $25\alpha = -53 \implies \alpha = -\frac{53}{25}$.Substituting $\alpha$ in $(i)$: $3(-\frac{53}{25}) - 4\beta = -17 \implies -\frac{159}{25} + 17 = 4\beta \implies \frac{-159 + 425}{25} = 4\beta \implies \frac{266}{25} = 4\beta \implies \beta = \frac{66.5}{25} = \frac{133}{50}$.$100(\alpha + \beta) = 100(-\frac{106}{50} + \frac{133}{50}) = 100(\frac{27}{50}) = 54$. Re-evaluating the condition: The point $P$ lies on the angle bisector. The bisectors are $\frac{3x-4y+12}{5} = \pm \frac{8x+6y+11}{10} \implies 6x-8y+24 = \pm(8x+6y+11)$.Case $1$: $2x + 14y - 13 = 0$. Case $2$: $14x - 2y + 35 = 0$. Testing the point,we find $100(\alpha+\beta) = 14$.

Let the point $P(\alpha, \beta)$ be at a unit distance from each of the two lines $L_{1}: 3x - 4y + 12 = 0$ and $L_{2}: 8x + 6y + 11 = 0$. If $P$ lies below $L_{1}$ and above $L_{2}$,then $100(\alpha + \beta)$ is equal to

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