Let A be the set of all points $(\alpha, \beta)$ such that the area of triangle formed by the points $(5,6),(3,2)$ and $(\alpha, \beta)$ is 12 square units. Then the least possible length of a line segment joining the origin to a point in A , is :

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Straight Lines

Let A be the set of all points

(α, β)

such that the area of triangle formed by the points

(5, 6), (3, 2)

and

(α, β)

is 12 square units. Then the least possible length of a line segment joining the origin to a point in A , is :

(A) $\frac{16}{5}$
(B) $\frac{4}{5}$
(C) $\frac{12}{5}$
(D) $\frac{8}{5}$ ✓ Correct

Answer: (D)

Solution

The line through

(5, 6)

and

(3, 2)

2 x - y - 4 = 0

, with those points a distance

(5 - 3)^{2} + (6 - 2)^{2} = 25

apart. The area condition

12 = \frac{1}{2} \cdot 25 \cdot h

gives perpendicular distance

h = \frac{12}{5}

, so

(α, β)

lies on

2 x - y - 4 = \pm 12

, i.e.

2 x - y = 16

2 x - y = - 8

. The shortest distances from the origin to these lines are

\frac{16}{5}

and

\frac{8}{5}

; the least is

\frac{8}{5}

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option

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