Question

If M be a point on the line L = 0 such that |AM - BM| is minimum, then the area of $△AMB$ equals

• A. $\frac{13}{4}$
• B. $\frac{13}{2}$
• C. $\frac{13}{6}$
• D. $\frac{13}{8}$

Answer 1

The correct option is A $\frac{13}{4}$

$x+y=0y=-x$
When,
$x=2,y=-2x=1,y=-1x=-1,y=1x=0,y=0$
M lies on $x=-y$
for M, let's say if $x=a$ then $y=-a$
$M(a,-a)$
Midpoint of $AB=(2,\frac{1}{2})$
As the point lies on the $\perp$ of AB, the equation of the perpendicular bisector is
$(y-\frac{1}{2})=\frac{2}{3}(x-2)4x-6y-5=0$
The point of intersection if the lines
$x+y=0$ and $4x-6y-5=0$
Showing the equation, we get the point $(\frac{1}{2},-\frac{1}{2})$
So that area of the triangle formed at this point A and B.
$=\left|\frac{1}{2}\left[\frac{1}{2}\right(2+1)+1(-1+\frac{1}{2})+3(-\frac{1}{2}-2)]\right|=\frac{13}{4}$