Question

A (5,3) and B(3,-2) are two fixed points. Find the equation of the locus of P, so that the

area of triangle PAB is 9.​

Answer 1

`37=5x-2y,\,1=5x-2y`

Explanation:

If
`(x_1,y_1),\,(x_2,y_2),\,(x_3,y_3)` are coordinates of a triangle then area of a triangle is equal to
`(1)/(2)|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|`

Let point P be
`(x,y)`

Put
`(x_1,y_1)=(5,3),\,(x_2,y_2)=(3,-2),\,(x_3,y_3)=(x,y)`

Area of a triangle =
`(1)/(2)|5(-2-y)+3(y-3)+x(3+2)|`

`=(1)/(2)|-10-5y+3y-9+5x|\\\\=(1)/(2)|-19+5x-2y|`

Also,

Area of a triangle = 9 square units

`9=(1)/(2)|-19+5x-2y|\\\\18=|-19+5x-2y|`

±18 = -19 + 5x - 2y

`18=-19+5x-2y,\,-18=-19+5x-2y\\\\37=5x-2y,\,1=5x-2y`