Mathematics · Straight lines

JEE Main 2024 — 5 April, Shift 2 — Question 8

Let $A(-1,1)$ and $B(2,3)$ be two points and $P$ be a variable point above the line AB such that the area of $\triangle \mathrm{PAB}$ is 10 . If the locus of P is $\mathrm{ax}+\mathrm{by}=15$ , then $5 a+2 b$ is :

Option A:
$-\frac{12}{5}$
Correct
Option B:
$-\frac{6}{5}$
Option C:
$4$
Option D:
$6$

Answer: A

Step-by-step solution

$\frac12\begin{vmatrix}h&k&1\\-1&1&1\\2&3&1\end{vmatrix}=10$

$-2 x+3 y=25$

$-\frac{6}{5} x+\frac{9}{5} y=15$

$a=-\frac{6}{5}, b=\frac{9}{5}$

$5 a=-6,2 b=\frac{18}{5}$

$5a+2b=-\frac{12}{5}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Mathematics
Chapter: Straight lines
Topic: Locus