Question

The locus of a point which is collinear with the points $(3,4)$ and $(-4,3)$ is

• A. $2x+3y-12=0$
• B. $2x+3y+12=0$
• C. $x+y+12=0$
• D. $x-7y+25=0$

Answer 1

The correct option is D $x-7y+25=0$

Let $(h,k)$ be any arbitrary point on the locus.

It is given that $(3,4),(-4,3),(h,k)$ are collinear. Hence, area of the triangle formed by these points will be $0$
$\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=0$
Condition of col-linearity of three points $(x_1,y_1)(x_2,y_2),(x_3,y_3)$

$\left|\begin{array}{ccc}3& 4& 1\\ -4& 3& 1\\ h& k& 1\end{array}\right|=0$

$\therefore h(4-3)-k(4+3)+1(9+16)=0$
$⟹h-7k+25=0$
$⟹x-7y+25=0$ is the required equation of locus.

Hence, option D is correct.