Question

$A(3,4)$ and $B(5,-2)$ are two given points. If $AP=PB$ and area of $△PAB=10$, then $P$ is -

• A. $(7,1)$
• B. $(7,2)$
• C. $(-7,2)$
• D. $(-7,-1)$

Answer 1

Answer: (B)

$(7,2)$

Let the coordinate P be $(x,y)$

Since it is given that $PA=PB$

So, by using distance formula

$P(x,y)$ and $A(3,4)$

$PA=\sqrt{(3-x{)}^2+(4-y{)}^2}$

$P(x,y)$ and $B(5,-2)$

$PA=\sqrt{(5-x{)}^2+(-2-y{)}^2}$

then,

$\sqrt{(3-x{)}^2+(4-y{)}^2}=\sqrt{(5-x{)}^2+(-2-y{)}^2}$

Squaring both sides

$(3-x{)}^2+(4-y{)}^2=(5-x{)}^2+(-2-y{)}^2$

$9+x^2-6x+16+y^2-8y=25+x^2-10x+4+y^2+4y$

$-6x-8y=-10x+4+4y$

$4x-12y=4$

$x-3y=\mathrm{1.......}\left(1\right)$

Area of $△PAB=10$

Area of triangle of (3,4),(5,-2) and (x,y)

Area of triangle $=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

$10=\frac{1}{2}\left[3\right(-2-y)+5(y-4)+x(4+2\left)\right]$

$-6+2y-20+6x=20$

$6x+2y=46$

$3x+y=\mathrm{23......}\left(2\right)$

Since $x-3y=1\Rightarrow x=3y+1$

Equation (2) implies

$3(3y+1)+y=23$

$9y+3+y=23$

$10y=20$

$y=2$

$x=3y+1$

$x=3\times 2+1$

$x=7$

Therefore, the coordinates are $(7,2)$.