Question

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

Answer 1

Answer:

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

Step-by-step 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 $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

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 = $\frac{1}{2}|5(-2-y)+3(y-3)+x(3+2)|$

$=\frac{1}{2}|-10-5y+3y-9+5x|\\\\=\frac{1}{2}|-19+5x-2y|$

Also,

Area of a triangle = 9 square units

$9=\frac{1}{2}|-19+5x-2y|\\\\18=|-19+5x-2y|$

±18 = -19 + 5x - 2y

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