Question

If the coordinates of two points $A,B$ are $(1,2)(3,8)$ respectively, find a point $P$ such that$|PA|=|PB|$ and area of $△PAB=10$.

Answer 1

Given that : $A$ and $B$ are two points

Let $A(x_1,y_1)=(1,2)$

$B(x_2,y_2)=(3,8)$

and $|PA|=|PB|$

Lt the point $P(x,y)$

Find $P(x,y)=?$

Since, It is given that,

$|PA|=|PB|$

$\sqrt{(x-x_1{)}^2+(y-y_1{)}^2}-\sqrt{(x-x_2{)}^2+(y-y_2{)}^2}$

$\sqrt{(x-1{)}^2+(y-2{)}^2}=\sqrt{(x-3{)}^2+(y-8{)}^2}$

Squaring both side and we get,

$(x-1{)}^2+(y-2{)}^2=(x-3{)}^2+(y-8{)}^2$

$x^2+1^2-2x+y^2+4-4y=x^2+9-6x+y^2+64-16y$

$-2x-4y+5+6x+16y=73$

$4x+12y=73-5$

$4x+12y=68$

$4(x+3y)=68$

$x+3y=\frac{68}{4}$

$x+3y=17.....\left(1\right)$

Area of $△PAB=10$

$\frac{1}{2}\left[x\right(y_1-y_2)+x_1(y_2-y)+x_2(y-y_1\left)\right]=10$

$x(2-8)+1(8-y)+3(y-2)=10$

$x(-6)+(8-y)+3y-6=10$

$-6x+8-y+3y-6-10=0$

$-6x+2y-8=0$

$6x-2y+8=0$

$3x-y+4=0$

$3x-y=-4....\left(2\right)$

From equation (1) and (2) to

$(x+3y=17)\times 3$

$3x-y=-4$

$3x+9y=51$

$3x-y=-9$

$3x+9y=51$

$\underset{–}{\underset{-}{3}x\underset{+}{-}y\underset{+}{=}-4}$

$\log =55$

$y=\frac{55}{10}$

$y=5.5$

put in equation (1)

$x+3y=17$

$x+3\times 5.5=17$

$x+16.5=17$

$x=17-16.5$

$x=0.5$

Hence, the point $P(x,y)=(0.5,5.5)$

Hence, this is the answer.