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 (1) `-1/4` (2) `-4` (3) `-2` (4) `-1/2`

Options

A. `-1//4`
B. `-4`
C. `-2`
D. `-1//2`

Answer

C

Explanation

Let `m` be the slope of the line PQ then the equation of PQ is

`y-2=m(x-1) `
Now `PQ` meets X-axis at `P(1-2/m,0)` and Y-axis at `Q(0,2-m)`
`:.OP=1-2/m`
and `OQ=2-m`
Also area of `DeltaOPQ=1/2(OP)(OQ)=1/2|(1-2/m)(2-m)|`
`=1/2|2-m- 4/m+2|`
`=1/2|4-(m+4/m)|`
Let `f(m)=4-(m+4/m)`
`impliesf'(x)=1+4/(m^(2))`
Put `f'(m)=0impliesm^(2)=4`
`impliesm=+-2`
At `m-2, f(2)=0`
and at `m=-2,f(-2)=8`
Since the area cannot be zero, hence the required value `m` is `-2`.