Mathematics · Straight lines

JEE Main 2024 — 29 January, Shift 2 — Question 14

Let A be the point of intersection of the lines $3 x+$ $2 \mathrm{y}=14,5 \mathrm{x}-\mathrm{y}=6$ and $B$ be the point of intersection of the lines $4 x+3 y=8,6 x+y=5$ . The distance of the point $\mathrm{P}(5,-2)$ from the line AB is

Option A:
$\frac{13}{2}$
Option B:
8
Option C:
$\frac{5}{2}$
Option D:
6
Correct

Answer: D

Step-by-step solution

Sol. Solving lines $\mathrm{L}_{1}(3 x+2 y=14)$ and $L_{2}(5 x-y=6)$

to get $A(2,4)$ and solving lines $L_{3}(4 x+3 y=8)$ and $\mathrm{L}_{4}(6 \mathrm{x}+\mathrm{y}=5)$

to get $\mathrm{B}\left(\frac{1}{2}, 2\right)$ .

Finding Equation of $\mathrm{AB}: 4 \mathrm{x}-3 \mathrm{y}+4=0$

Calculate distance PM $\Rightarrow\left|\frac{4(5)-3(-2)+4}{5}\right|=6$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Mathematics
Chapter: Straight lines
Topic: Various forms of lines