Question

$A(2,3)$ and $B(-3,4)$ be two given points. Find the equation of the locus of $P$ so that the area of triangle $PAB$ is $8.5$ square units.

Answer 1

Let coordinate of P are $(h,t)A(2,3),B(3,-4)$

point P will lies on the both side of AB and the

Area of triangle PAB $=\frac{1}{2}\left(h\right(3-4)+2(4-k)-3(k-3\left)\right)$

$\pm 8.5=\frac{1}{2}\left[h\right(-1)+8-2k-3k+9]$

$\pm 17=-h+8-5x+9$

taking ${}^{\prime }{+}^{\prime }$ sign

$17+h-17+5k=0$

$h+5k=0$

$\therefore$ locus of P is

$x+5y=0$

taking -ve sign

$-17+h+5k-17=0$

$h+5k=34$

$\therefore$ locus of P is

$x+5y=34$