Question

Let $A(2,-3)$ and $B(-2,1)$ be two angular points of $△ABC$. If the centroid of the triangle moves on the line $2x+3y=1$, then the locus of the angular point $C$ is given by

• A. $2x+3y=9$
• B. $2x-3y=9$
• C. $3x+2y=5$
• D. $3x-2y=3$

Answer 1

The correct option is A $2x+3y=9$

$A(2,-3),B(-2,1)$ are two angular points of $△ABC$

Let $C(\alpha ,\beta )$ be the third angular point of $△ABC$

Let $G$ be the centroid of $△ABC$

Then, $G=(\frac{2-2+\alpha }{3},\frac{-3+1+\beta }{3})$

$=(\frac{\alpha }{3},\frac{\beta -2}{3})$ .... $\left(i\right)$

Since, $G$ moves on the line $2x+3y=1$

Points on this line are of the form $(x,\frac{1-2x}{3})$

$⟹G=(t,\frac{1-2t}{3})$

From $\left(i\right)$

$\frac{\alpha }{3}=t$ and $\frac{\beta -2}{3}=\frac{1-2t}{3}$

$⟹\alpha =3t$ and $\beta =3-2t$

$\therefore C\equiv (3t,3-2t)$

Locus of point $C$ is given by

$(3t-2{)}^2+(3-2t+3{)}^3=(3t+2{)}^2+(3-2t-1{)}^2$

$⟹9t^2+4-12t+36+4t^2-24t=9t^2+4+12t+4+4t^2-8t$

$⟹40t=32⟹t=\frac{4}{5}$

$\therefore C\equiv (\frac{12}{5},\frac{7}{5})$

The equation of locus of point C will be parallel to line $2x+3y=1$, sice centroid $G$ lie on this line.

$\therefore$ Equation of locus of $C$ is given by $2x+3y=d$

Substituting $C$ in the given equation $2x+3y=d$ we get $d=9$

Thus, the locus of $C$ is $2x+3y=9$.