Question

If coordinate of two points $A$ and $B$ are $(3,4)$ and $(5,-2)$ find coordinates of $P$ if $PA$=$PB$ and area of $△PAB=10$

Answer 1

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

Since it is given that $PA=PB$

So, firstly we will calculate the distance $PA$.

$PA=(x,y)(3,4)$

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

$PB=(x,y)(5,-2)$

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

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

Squaring both the sides in the above equation,

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

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

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

$4x-12y=4$

$x-3y=1$ (Equation 1)

Now,it is given that Area of $△PAB=10$

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

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

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

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

$46=2y+6x$

$3x+y=23$ (Equation 2)

Now, solving equations 1 and 2.

Since $x-3y=1$

therefore, $x=3y+1$

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 of $P$ are $(7,2)$