Q.Find the equation of the locus of a point $P$ which is equidistant from the points $A(1,2)$ and $B(3,-4)$.

Setting the area of $\triangle PAB$ equal to $9$ gives $|5x-2y-19|=18$.

Let $P(x,y)$, $A(5,3)$, $B(3,-2)$. The area of $\triangle PAB$ is

$\frac{1}{2}\big|\,x(3-(-2)) + 5(-2 - y) + 3(y - 3)\,\big| = 9.$

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

$\frac{1}{2}\big|\,5x - 2y - 19\,\big| = 9 \;\Rightarrow\; |5x - 2y - 19| = 18.$

Removing the modulus, $5x - 2y - 19 = \pm 18$, giving

$5x - 2y - 1 = 0 \quad\text{or}\quad 5x - 2y - 37 = 0.$

Write the area of triangle $PAB$ using the determinant formula in terms of $P(x,y)$, set it equal to $8.5$, and simplify to get the locus.

Let $P=(x,y)$, $A(2,3)$, $B(-3,4)$.

Area of $\triangle PAB = \dfrac{1}{2}\left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$

Using $A,B,P$:

$= \dfrac{1}{2}\left| 2(4-y) + (-3)(y-3) + x(3-4) \right|$

$= \dfrac{1}{2}\left| 8-2y -3y+9 -x \right|$

$= \dfrac{1}{2}\left| 17 - x - 5y \right|$

Given area $=8.5$:

$\dfrac{1}{2}|17-x-5y| = 8.5 \implies |17-x-5y| = 17$

So $17-x-5y = 17$ or $17-x-5y=-17$

Case 1: $x+5y = 0$

Case 2: $x+5y = 34$

Write $PA^2$, $PB^2$, $PC^2$ in terms of $(x,y)$ using the distance formula, impose the given condition, and simplify.

Let $P=(x,y)$. Given $A(1,2)$, $B(2,-3)$, $C(-2,3)$, and the condition $PA^{2}+PB^{2}=2PC^{2}$.

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

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

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

Left side:

$PA^{2}+PB^{2} = (x^{2}-2x+1)+(x^{2}-4x+4) + (y^{2}-4y+4)+(y^{2}+6y+9)$

$= 2x^{2}-6x+5 + 2y^{2}+2y+13 = 2x^{2}+2y^{2}-6x+2y+18$

Right side:

$2PC^{2} = 2\left[(x^{2}+4x+4)+(y^{2}-6y+9)\right] = 2x^{2}+2y^{2}+8x-12y+26$

Equating:

$2x^{2}+2y^{2}-6x+2y+18 = 2x^{2}+2y^{2}+8x-12y+26$

The $x^2$ and $y^2$ terms cancel:

$-6x+2y+18 = 8x-12y+26$

$-14x+14y-8 = 0$

Dividing by $-2$:

$7x - 7y + 4 = 0$

This is exactly the required equation of the locus.