Question

Consider the points $A(0,1)$ and $B(2,0)$, and $P$ be a point on the line $4x+3y+9=0$. The coordinates of $P$ such that $|PA-PB|$ is maximum are

• A. $(\frac{12}{5},\frac{-1}{5})$
• B. $(\frac{-84}{5},\frac{13}{5})$
• C. $(\frac{-24}{5},\frac{17}{5})$
• D. $(0,-3)$

Answer 1

Answer: (C)

$(\frac{-24}{5},\frac{17}{5})$

Consider $P,A,B$

If three points $P,A,B$ form a triangle, then according to triangle inequalities, $|PA-PB|<AB$

If they do not form a triangle, i.e they are collinear, $|PA-PB|=AB$

If they are collinear, then P is the point of intersection of lines $AB,4x+3y+9=0$ (given P is on this line)

equation of line $AB=x+2y=2$

Solving the equations $4x+3y=-9$ and $x+2y=2$

$\Rightarrow 4x+3y-4(x+2y)=-9-8$

$\Rightarrow 4x+3y-4x-8y=-17$

$\Rightarrow -5y=-17$ or $y=\frac{17}{5}$

Put $y=\frac{17}{5}$ in $x+2y=2$

$\Rightarrow x=2-2y=2-2\times \frac{17}{5}=2-\frac{34}{5}=\frac{10-34}{5}=\frac{-24}{5}$

point of intersection of the lines is $(\frac{-24}{5},\frac{17}{5})$

therefore $P$ is $(\frac{-24}{5},\frac{17}{5})$.