JEE MAIN - Mathematics (2022 - 25th July Evening Shift - No. 8)
Let the point $$P(\alpha, \beta)$$ be at a unit distance from each of the two lines $$L_{1}: 3 x-4 y+12=0$$, and $$L_{2}: 8 x+6 y+11=0$$. If $$P$$ lies below $$L_{1}$$ and above $${ }{L_{2}}$$, then $$100(\alpha+\beta)$$ is equal to :
$$-$$14
42
$$-$$22
14
Explanation
_25th_July_Evening_Shift_en_8_1.png)
$${L_1}:3x - 4y + 12 = 0$$
$${L_2}:8x + 6y + 11 = 0$$
Equation of angle bisector of L1 and L2 of angle containing origin
$$2(3x - 4y + 12) = 8x + 6y + 11$$
$$2x + 14y - 13 = 0$$ ...... (i)
$${{3\alpha - 4\beta + 12} \over 5} = 1$$
$$ \Rightarrow 3\alpha - 4\beta + 7 = 0$$ ...... (ii)
Solution of $$2x + 14y - 13 = 0$$ and $$3x - 4y + 7 = 0$$ gives the required point $$P(\alpha ,\beta ),\,\alpha = {{ - 23} \over {25}},\beta = {{53} \over {50}}$$
$$100(\alpha + \beta ) = 14$$
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