Mathematics · Straight lines

JEE Main 2024 — 4 April, Shift 1 — Question 3

The vertices of a triangle are $\mathrm{A}(-1,3), \mathrm{B}(-2,2)$ and $\mathrm{C}(3,-1)$ . A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

Option A:
$x-y-(2+\sqrt{2})=0$
Option B:
$-x+y-(2-\sqrt{2})=0$
Option C:
$x+y-(2-\sqrt{2})=0$
Correct
Option D:
$x+y+(2-\sqrt{2})=0$

Answer: C

Step-by-step solution

equation of $\mathrm{AC} \rightarrow \mathrm{x}+\mathrm{y}=2$

equation of line parallel to $AC$

$x+y=d$

$\left|\frac{\mathrm{d}-2}{\sqrt{2}}\right|=1$

$\mathrm{d}=2-\sqrt{2}$

eq ${ }^{\mathrm{n}}$ of new required line $x+y=2-\sqrt{2}$

$x+y-(2-\sqrt{2})=0$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Mathematics
Chapter: Straight lines
Topic: Angle between lines, perpendicular distance & distance between parallel lines, foot, image