Question

A triangle $ABC$ is placed so that the midpoints of its sides are on the $x,y$ and $z$ axes respectively. Lengths of the intercepts made by the plane containing the triangle on these axes are respectively $\alpha ,\beta ,\gamma$, then the coordinates of the centroid of the triangle $ABC$ are

• A. $(-\frac{\alpha }{3},\frac{\beta }{3},\frac{\gamma }{3})$
• B. $(\frac{\alpha }{3},-\frac{\beta }{3},\frac{\gamma }{3})$
• C. $(\frac{\alpha }{3},\frac{\beta }{3},-\frac{\gamma }{3})$
• D. $(\frac{\alpha }{3},\frac{\beta }{3},\frac{\gamma }{3})$

Answer 1

The correct option is D $(\frac{\alpha }{3},\frac{\beta }{3},\frac{\gamma }{3})$

Equation of the plane containing the triangle
$ABC$ is $\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$
which meets the axes in $(\alpha ,0,0),(0,\beta ,0)$ and $(0,0,y)$.
Let the coordinates of $A$ be $(x_1,y_1,z_1)$
Since the middle point of $AB$ lies on the $z$-axis it is $(0,0,\gamma )$ and thus the coordinates of $B$ are $(-x_1,-y_1,2\gamma -z_1)$
Similarly the coordinates of $C$ are $(-x_1,2\beta -y_1,-z_1)$
So that the middle point of $BC=(-x_1,\beta -y_1,\gamma -z_1)=(\alpha ,0,0)$
$\Rightarrow x_1=-\alpha ,y_1=\beta ,z_1=\gamma$.
And thus the coordinates of $A$ are $(-\alpha ,\beta ,\gamma )$
Similarly the coordinates of $B$ are $(\alpha ,-\beta ,\gamma )$ and those of $C$ are $(\alpha ,\beta ,-\gamma )$.
Hence, the coordinates of the centroid of the triangle $ABC$ are $(\frac{\alpha }{3},\frac{\beta }{3},\frac{\gamma }{3})$