Question

If $A\equiv (1,3)$ , $B\equiv (5,6)$ , then a point $M$ on $x-$axis, such that $AM+MB$ is minimum is

• A. $(\frac{2}{3},0)$
• B. $(\frac{4}{3},0)$
• C. $(\frac{5}{3},0)$
• D. $(\frac{7}{3},0)$

Answer 1

The correct option is D $(\frac{7}{3},0)$

Given points $A\equiv (1,3),B\equiv (5,6)\in Q_1$ both lying on same side of $x$ axis.
Let $M\equiv (\alpha ,0)$ be a point on $x$ axis.



Now, $AM+MB=AM+M{B}^{\prime }$ will be minimum when $A,M,B^{\prime }$ are collinear
$\Rightarrow$ slope of $AM$ = slope of $A{B}^{\prime }$
$(\because$Image of $B$ w.r.t $x$ axis is $B^{\prime }\equiv (5,-6))$
$\Rightarrow \frac{3-0}{1-\alpha }=\frac{3+6}{1-5}$
$\Rightarrow \alpha =\frac{7}{3}$
$\therefore M\equiv (\frac{7}{3},0)$​​​​​​