Question

The locus of P such that the area of $\Delta PAB$ is $12sq.$units where $A=\left(2,3\right)$ $B=\left(-4,5\right)$ is

• A. $x^2+6xy+9y^2-22x-66y-23=0$
• B. $x^2-6xy+9y^2+22x+66y+23=0$
• C. $x^2+5xy+9y^2+20x+62y+22=0$
• D. none

Answer 1

Answer: (A)

$x^2+6xy+9y^2-22x-66y-23=0$

Given, A(2,3) and B(-4,5)

Let the coordinates of P be (x,y)

${Area}^2=\frac{1}{4}{\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|}^2{12}^2=\frac{1}{4}{\left|\begin{array}{ccc}2& 3& 1\\ -4& 5& 1\\ x& y& 1\end{array}\right|}^2$

$144\times 4=\left[2\right(5-y)-3(-4-x)+1(-4y-5x){]}^{2}576=[10-2y+12+3x-4y-5x{]}^2$

$576={[22-6y-2x]}^{2}144={[11-3y-x]}^{2}144=121+9y^2+x^2-66y-22x+6xy9y^2+x^2-22x-66y+6xy-23=0$