Question

A triangle $ABC$ is placed so that the mid-points of the sides are on the $x,y,z$ axes. Lengths of the intercepts made by the plane containing the triangle on these axes are respectively $\alpha ,\beta ,\gamma$. Coordinates of the centroid of the triangle $ABC$ are

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

Answer 1

The correct option is D $(\alpha /3,\beta /3,\gamma /3)$

$△ABC$ has $P$ as mid-point of $AB$ on x-axis.

$Q$ as midpoint of $BC$ on y-axis.

$R$ as midpoint of $AC$ on z-axis.

$\therefore A=\left(x_1{y}_1{z}_1\right)$

$\therefore B=\left(x_2{y}_2{z}_2\right)$

$\therefore C=\left(x_3{y}_3{z}_3\right)$

$\Rightarrow P=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})Q=(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2},\frac{z_2+z_3}{2})R=(\frac{x_3+x_1}{2},\frac{y_3+y_1}{2},\frac{z_3+z_1}{2})P=(\alpha ,0,0)Q=(0,\beta ,0)R=(0,0,\gamma )\therefore \frac{x_1+x_2}{2}=\alpha ,\frac{y_1+y_2}{2}=0,\frac{z_1+z_2}{2}=0\frac{x_2+x_3}{2}=0,\frac{y_2+y_3}{2}=\beta ,\frac{z_2+z_3}{2}=0\frac{x_3+x_1}{2}=0,\frac{y_3+y_1}{2}=0,\frac{z_3+z_1}{2}=\gamma \therefore x_1+x_2+x_3=\alpha ,y_1+y_2+y_3=\beta ,z_1+z_2+z_3=\gamma$

$\therefore$ centroid $\left(G\right)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3})=(\frac{\alpha }{3},\frac{\beta }{3},\frac{\gamma }{3})$