Consider two points $A(1,2)$ and $B(3,-1)$.Let $M$ be a point on the straight line $L\equiv x+y=0$ such that $|AM-BM|$ is minimum,then the area of triangle AMB equals
$(A)\frac{13}{4}\hspace{1cm}(B)\frac{13}{2}\hspace{1cm}(C)\frac{13}{6}\hspace{1cm}(D)\frac{13}{8}$
I asked a similar problem What is the coordinate of a point $P$ on the line $2x-y+5=0$ such that $|PA-PB|$ is maximum where $A=(4,-2)$ and $B=(2,-4)$ and i was answered and i understood the concept and now i am applying the same thing here(the method suggested by Juantheron),but i am not getting the desired answer.
$A(1,2),B(3,-1)$,let angle $AMB=\theta,M$ lies on the line $x+y=0$.Using the cosine law and the concept $\cos\theta\geq-1,$ we get $(AM+BM)^2\geq AB^2\Rightarrow|AM+BM|\geq AB$,
but this is not helpful as this gives the minimum value of $|AM+BM|$.
Then i used the cosine law and the fact that $\cos\theta\leq 1$,we get $(AM-BM)^2\leq AB^2\Rightarrow|AM-BM|\leq AB$ but this is also not helpful as it gives the maximum value of $|AM-BM|$.
What should i do to find $M$ so that $|AM-BM|$ is minimum.Then i could find the area of triangle $AMB$.Please help me.
It can be clearly observed that $|PA - PB| \geq 0$. So, the least value is zero. But when is this achieved?
We know that the locus of points which are equidistant from two points is the perpendicular bisector of the line segment joining that two points. So, Let $P$ be any point on the perpendicular bisector of $AB$. So, from the definition of perpendicular bisector, we get that $PA = PB$, which implies that $PA - PB = 0$. So, the required point i.e. the point that satisfies the given condition lies on the perpendicular bisector $AB$. But the equation of $\perp$ bisector of $AB$ is
$$\left( y-\dfrac{1}{2} \right) = \dfrac{2}{3} \left(x-2 \right) \Leftrightarrow 4x-6y-5=0 $$
But this point lies on $L \equiv x+y=0$. So, the required point is actually the point of intersection of the lines $x+y=0$ and $ 4x-6y-5=0$. Therefore, the required point is $ \left( \dfrac{1}{2} , -\dfrac{1}{2} \right)$. So, the area of the triangle formed this point, $A$ and $B$ is equal to
$$\dfrac{1}{2} \left| \dfrac{1}{2} \left( 2+1 \right) + 1 \left(-1+\dfrac{1}{2}\right) +3 \left(-\dfrac{1}{2} - 2 \right) \right| = \dfrac{13}{4} $$