Question

If point $P(h,k)$ lies on the line $2x+3y=5$ such that $|PA.PB|$ is maximum where $A(2,3)$ and $B(1,2)$ then the value of $(3h+2k)$ is

• A. $3$
• B. $2$
• C. $6$
• D. $4$

Answer 1

Answer: (D)

$4$

$|PA-PB|<AB$
$|PAPB{|}_{max}=AB$
In this case P, A & B are collinear
So 'P' is intersecting point of these two lines as shown in figure
Equation of AB: $\begin{array}{cc}y-x& =1\\ 2x+3y& =5\end{array}\}$
$\Rightarrow$ Intersection point $P(\frac{2}{5},\frac{7}{5})$
$\therefore 3h+2k=4$