Given: A and b are two points (3,4) , (5,-2) and PA=PB and area of triangle PAB= 10 SQUARE units.

To find: The coordinates of P

Solution: 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 triangle PAB = 10

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

Area of triangle is given by the FORMULA= \frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]

Area of triangle PAB = \frac{1}{2}[3(-2-y)+5(y-4)+x(4+2)]=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 are (7,2).

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