JEE Main · 2012 · 29 Apr 2012 · Offline · Application of Derivatives

A line is drawn through the point

(1, 2)

to meet the coordinate axes at P and Q such that it forms a triangle OPQ , where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is

(A) $- \frac{1}{4}$
(B)-4
(C)-2✓ Correct
(D) $- \frac{1}{2}$

Answer: (C)

Solution

Take the line through

(1, 2)

with slope

m < 0

so that it meets both positive axes:

y - 2 = m (x - 1)

. Its intercepts are

P (1 - \frac{2}{m}, 0)

and

Q (0, 2 - m)

, so the area of triangle

O P Q

Δ = \frac{( m - 2 ) ^{2}}{- 2 m} = - \frac{m}{2} + 2 - \frac{2}{m}

. Setting

\frac{d Δ}{d m} = - \frac{1}{2} + \frac{2}{m ^{2}} = 0

gives

m^{2} = 4

, and since the slope must be negative,

m = - 2

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option

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