Question

A and B are the two points (3,4) and (5,-2). Find the point P such that PA=PB and area of triangle PAB =10 square units.

Answer 1

SOLUTION

Here, P=P(x,y) be the required point.

Then,

=) PA= PB= PA^2= PB^2

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

=) x-3y-1= 0.............(1)

Again,

Area of ∆PAB=10

=)1/2[x(4+2)+3(-2-y)+5(y-4)]=10

=) 3x+y-23=0.............(2)

On solving (1) &(2), we get

=) x= 7 and y= 2

Hence, the required points (7,2).

hope it helps ✔️

Answer 2

Step-by-step explanation:

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