Question

Let $A=(3,-4),B=(1,2)$ .Let $P=(2k-1,2k+1)$ be a variable point such that PA+PB is the minimum then $k$ is

• A. $\frac{7}{9}$
• B. $0$
• C. $\frac{7}{8}$
• D. none of these

Answer 1

The correct option is C $\frac{7}{8}$

Given, $A=(3,-4)$, $B=(1,2)$, $P=(2k-1,2k+1)$

By property of triangle, $PA+PB\ge AB$

For $PA+PB$ to be minimum, $PA+PB=AB$

This indicates that points A, B and C are collinear and $A-P-B$

Thus, area of triangle will be zero.

$\therefore \left|\begin{array}{ccc}{x}_A& y_A& 1\\ x_B& y_B& 1\\ x_P& y_P& 1\end{array}\right|=0$

$\therefore x_A\left|\begin{array}{cc}{y}_B& 1\\ y_P& 1\end{array}\right|-y_A\left|\begin{array}{cc}{x}_B& 1\\ x_P& 1\end{array}\right|+1\left|\begin{array}{cc}{x}_B& y_B\\ x_P& y_P\end{array}\right|=0$

$\therefore x_A[y_B-y_P]-y_A[x_B-x_P]+1[x_B{y}_P-x_P{y}_B]=0$

$\therefore x_A{y}_B-x_A{y}_P-y_A{x}_B+y_A{x}_P+x_B{y}_P-x_P{y}_B=0$

$\therefore (3\times 2)-(3\times (2k+1))-(-4\times 1)+(-4(2k-1))+\left(1(2k+1)\right)-\left((2k-1)2\right)=0$

$\therefore 6-(6k+3)+4-8k+4+(2k+1)-(4k-2)=0$

$\therefore 6-6k-3+4-8k+4+2k+1-4k+2=0$

$\therefore -16k+14=0$

$\therefore 16k=14$

$\therefore k=\frac{14}{16}$

$\therefore k=\frac{7}{8}$