JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Straight Lines

The vertices of a triangle are

A (- 1, 3)

B (- 2, 2)

and

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 :

(A) $x + y - (2 - 2) = 0$ ✓ Correct
(B) $x + y + (2 - 2) = 0$
(C) $- x + y - (2 - 2) = 0$
(D) $x - y - (2 + 2) = 0$

Answer: (A)

Solution

Side

A C

joins

A (- 1, 3)

and

C (3, - 1)

: slope

= \frac{- 1 - 3}{3 + 1} = - 1

, so

A C : x + y = 2

.

Shifting this side one unit inward (toward the interior/origin), the parallel line

x + y = d

satisfies

\frac{∣ d - 2∣}{2} = 1

, giving

d = 2 - 2

(the value closer to the origin).

The side of the new triangle nearest the origin is

x + y - (2 - 2) = 0.

Option A.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option

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