Question

$A(2,3)$ , $B(-3,4)$ are two points. lf a point $P$ moves such that the area of $\Delta$ $PAB$ is $8.5$ sq.units then the locus of $P$ is:

• A. $x^2+10xy+25y^2-34x-170y=0$
• B. $x^2+10xy-25y^2-34x-170y=0$
• C. $x^2-10xy+25y^2-34x+170y=0$
• D. $x^2-10xy-25y^2+34x-170y=0$

Answer 1

The correct option is A $x^2+10xy+25y^2-34x-170y=0$

Given $A(2,3),B(-3,4),ar\left(△PAB\right)=8.5sq.units$

Let $P=(h,k)$

Area of triangle

$=\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$

The points are $(2,3)(-3,4)(h,k)$

$\frac{17}{2}=\left|\frac{2(4-k)+(-3)(k-3)+h(3-4)}{2}\right|$

${17}^2=(8-2k-3k+9-h{)}^2$

${17}^2=(h+5k-17{)}^2$

${17}^2=h^2+25k^2+{17}^2+10hk-34h-170k$

$h^2+25k^2+10hk-34h-170k=0$

Therefore, the locus is

$x^2+25y^2+10yx-34x-170y=0$