JEE Main 2022 — Straight Lines Question with Solution

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,

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Now, $L_1:3x-4y+12=0$ and $L_2:8x+6y+11=0$

Since $L_1 \& L_2$ 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

Equation of angle bisector will be, $\frac{\left(3x-4y+12\right)}{\sqrt{3^2+4^2}}=\frac{8x+6y+11}{\sqrt{8^2+6^2}}$

$\Rightarrow 2\left(3x-4y+12\right)=8x+6y+11$

$\Rightarrow 2x+14y-13=0$

$\Rightarrow 2\alpha +14\beta -13=0 \cdots \left(\mathrm{i}\right)$

Also given perpendicular distance is one unit so, $\frac{3\alpha -4\beta +12}{5}=1$

$\Rightarrow 3\alpha -4\beta +7=0 \cdots \left(\mathrm{ii}\right)$

Solving equation $\left(\text{i}\right) \& \left(\text{ii}\right)$

 $2\alpha +14\beta -13=0$

 $3\alpha -4\beta +7=0$ 

$\Rightarrow$$P\left(\alpha \right)\left(,\beta \right),\alpha =\frac{-23}{25},\beta =\frac{53}{50}$

So, $100\left(\alpha +\beta \right)=14$