Question

Consider the points $A(3,4)$ and $B(7,13)$. If $P$ is a point on the line $y=x$ such that $PA+PB$ is minimum, then the coordinates of $P$ are

• A. $(12/7,12/7)$
• B. $(13/7,13/7)$
• C. $31/7,31/7$ )
• D. $(0,0)$

Answer 1

Answer: (C)

Consider a point $A$', the image of $A$ in $y=x$.
Therefore, the coordinates of $A$' are $(4,3)$

or
(Notice that $A$ and $B$ lie on the same side with respect to $y=x$ ).
Then $PA=PA$'.
Thus, $PA+PB$ is minimum, If $P{A}^{\prime }+PB$ is minimum,
i.e., if $P,A^{\prime },B$ are collinear. Now, the equation of $AB$ is
$y-3=\frac{13-3}{7-4}(x-4)$
or $3y-10x+31=0$
It intersects $y=x$ at $(31/7,31/7)$, which is the required point $P$.