Question

$A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ are the vertices of $△ABC$. $lx+my+n=0$ is an equation of the line $L$. If the centroid of $△ABC$ is at the origin and algebraic sum of the lengths of the perpendiculars from the vertices of $△ABC$ on the line $L$ is equal to $1$ then sum of the squares of the reciprocal of intercepts made by $L$ on the coordinate axes is equal to

• A. $0$
• B. $4$
• C. $9$
• D. $16$

Answer 1

The correct option is C $9$

Since, the centroid is at the origin,

$\frac{x_1+x_2+x_3}{3}=0,\frac{y_1+y_2+y_3}{3}=0$
Since, the sum of perpendiculars is 1

$\frac{l{x}_1+m{y}_1+n}{\sqrt{l^2+m^2}}+\frac{l{x}_2+m{y}_2+n}{\sqrt{l^2+m^2}}+\frac{l{x}_3+m{y}_3+n}{\sqrt{l^2+m^2}}=1$
$\Rightarrow l(x_1+x_2+x_3)+m(y_1+y_2+y_3)+3n=\sqrt{l^2+m^2}$
$\Rightarrow l^2+m^2=9n^2$ from (4)
Sum required is ${\left(\frac{l}{-n}\right)}^2+{\left(\frac{m}{-n}\right)}^2=\frac{l^2+m^2}{n^2}=9.$