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…
JEE Main 2021 — Mathematics Coordinate Geometry
2021mcqmedium
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 :
Official previous-year question
Held on 31 Aug 2021 · Verified 6 Jul 2026.
Options
A
58
B
512
C
516
D
54
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Solution
Given:Area of triangle formed by (5,6),(3,2)&(\alpha ,\beta )=12\mathrm{square}\mathrm{units}
∴21∣α53β62111∣=12
⇒∣α53β62111∣=24
⇒α(6−2)−β(5−3)+1(10−18)=±24
⇒4α−2β−8=±24
⇒4α−2β=32,4α−2β+16=0
⇒2α−β−16=0,2α−β+8=0
So point will be either (α,2α+8),(α,2α−16)
For the point (α,2α+8),
Distance from origin
D=α2+(2α+8)2=5α2+32α+64(∵β=2α+8)
⇒D2=5α2+32α+64
For maximum or minimum length,
dαd(D2)=10α+32=0
⇒α=−516
∴β=5−32+8=58
∴D=(−516)2+(58)2=585=58
Similarly if β=2α−16,D=α2+(2α−16)2=5α2−64α+256
∴D2=5α2−64α+256
For maximum or minimum length,
dαd(D2)=0
⇒10α−64=0
∴α=532
∴D=(532)2+(564−16)2=516at α=532
So, least possible length of line segment is 58
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