JEE Main 2022 · 25 Jul · Shift 2 · Mathematics Q10

← JEE Main 2022 · 25 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 10 of 90

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

-14

-22

Answer and worked solution

Official answer: D

Distances. Below $L_1$ and above $L_2$ : $3\alpha-4\beta+12=5$ and $8\alpha+6\beta+11=10$ .

Solve. $\beta=\frac{53}{50}$ , $\alpha=-\frac{46}{50}$ : $100(\alpha+\beta)=14$ .

(D)

Topics:Distance of a point from a line

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JEE Main 2022 · 25 Jul · Shift 2, Mathematics question 10