Question

P(x, y), A(3, 4) and B (5, -2) are the vertices of triangle PAB such that |PA|=|PB| and area of $\Delta PAB=10$ sq. units, then $PA=k\sqrt{5}$ units. Find the value of k.

• A. 2
• B. 1
• C. $\sqrt{5}$
• D. 3

Answer 1

The correct option is A 2


Given, |PA|=|PB|
$\therefore P{A}^2=P{B}^2\Rightarrow (x-3{)}^2+(y-4{)}^2=(x-5{)}^2+(y+2{)}^2\Rightarrow x^2-6x+9+y^2-8y+16=x^2-10x+25+y^2+4y+4\Rightarrow 4x-12y=4\Rightarrow x-3y=1\dots \left(i\right)$
area of $\Delta PAB=\frac{1}{2}|x(4+2)+3(-2-y)+5(y-4)|\Rightarrow 10=\frac{1}{2}|6x-6-3y+5y-20|\Rightarrow 20=6x+2y-26\Rightarrow 6x+2y-26=20or6x+2y-26=-20\Rightarrow 6x+2y=46or6x+2y=63x+y=23\dots \left(2\right)$

or $3x+y=3\dots \left(3\right)$
Solving (1) and (3), we get
$x=1,y=0$
Solving (1) and (2), we get
$x=7,y=2$
$\therefore$ Coordinates of P are (7, 2) and (1, 0)
Length of $PA=\sqrt{(7-3{)}^2+(2-4{)}^2}=\sqrt{16+4}=2\sqrt{5}\therefore 2\sqrt{5}=k\sqrt{5}\Rightarrow k=2$
Also, Length of $PA=\sqrt{(1-3{)}^2+(0-4{)}^2}=\sqrt{4+16}=2\sqrt{5}\therefore 2\sqrt{5}=k\sqrt{5}\Rightarrow k=2$