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If $G$ be the centroid of a triangle ABC, prove that:$AB^2 + BC^2 + CA^2 = 3 (GA^2 + GB^2 + GC^2)$
Given:
$G$ is the centroid of a triangle ABC.
To do:
We have to prove that $AB^2 + BC^2 + CA^2 = 3 (GA^2 + GB^2 + GC^2)$.
Solution:
Let the coordinates of the vertices of $\triangle ABC$ be $A (x_1, y_1), B (x_2, y_2), C (x_3, y_3)$ and $G$ be the centroid of the triangle.
Therefore,
The coordinates of \( \mathrm{G} \) are
Let us consider LHS,
\( A B^{2}+B C^{2}+C A^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\left(y_{2}-y_{3}\right)^{2}+\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2} \)
\( =\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\left(x_{3}-x_{1}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(y_{2}-y_{3}\right)^{2}+\left(y_{3}-y_{1}\right)^{2} \)
\( =x_{1}^{2}+x_{2}^{2}-2 x_{1} x_{2}+x_{2}^{2}+x_{3}^{2}-2 x_{2} x_{3}+x_{3}^{2}+x_{1}^{2}-2 x_{3} x_{1}+y_{1}^{2}+y_{2}^{2}-2 y y_{2}+y_{2}^{2}+y_{3}^{2}-2 y_{2} y_{3}+y_{3}^{2} +y_{1}^{2}-2 y y \)
\( =2\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)-2\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right)+2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)-2\left(y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{1}\right) \)
Now, let us consider RHS,
\( 3\left[\mathrm{GA}^{2}+\mathrm{GB}^{2}+\mathrm{GC}^{2}\right] \)
\( =3\left[\begin{array}{l}\left.\left(x_{1}-\frac{x_{1}+x_{2}+x_{3}}{3}\right)^{2}+\left(y_{1}-\frac{y_{1}+y_{2}+y_{3}}{3}\right)^{2}+\left(x_{2}-\frac{x_{1}+x_{2}+x_{3}}{3}\right)^{2}+\left(y_{2}-\frac{y_{1}+y_{2}+y_{3}}{3}\right)^{2}\right] \\ +\left(x_{3}-\frac{x_{1}+x_{2}+x_{3}}{3}\right)^{2}+\left(y_{3}-\frac{y_{1}+y_{2}+y_{3}}{3}\right)^{2}\end{array}\right] \)
\( =3\left[\begin{array}{l}\left(\frac{3 x_{1}-x_{1}-x_{2}-x_{3}}{3}\right)^{2}+\left(\frac{3 y_{1}-y_{1}-y_{2}-y_{3}}{3}\right)^{2}+\left(\frac{3 x_{2}-x_{1}-x_{2}-x_{3}}{3}\right)^{2} \\ +\left(\frac{3 y_{2}-y_{1}-y_{2}-y_{3}}{3}\right)^{2}+\left(\frac{3 x_{3}-x_{1}-x_{2}-x_{3}}{-3}\right)^{2}+\left(\frac{3 y_{3}-y_{1}-y_{2}-y_{3}}{3}\right)^{2}\end{array}\right] \)
\( =3\left[\begin{array}{l}\left(\frac{2 x_{1}-x_{2}-x_{3}}{3}\right)^{2}+\left(\frac{2 y_{1}+y_{2}-y_{3}}{3}\right)^{2}+\left(\frac{2 x_{2}-x_{1}-x_{3}}{3}\right)^{2} \\ +\left(\frac{2 y_{2}-y_{1}-y_{3}}{3}\right)^{2}+\left(\frac{2 x_{3}-x_{1}-x_{2}}{3}\right)^{2}+\left(\frac{2 y_{3}-y_{1}-y_{3}}{3}\right)^{2}\end{array}\right] . \)
\( =3 \times \frac{1}{9}\left[4 x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-4 x_{1} x_{2}+2 x_{z} x_{3}-4 x_{3} x_{1}+4 y_{1}^{2}+y_{2}^{2}+y_{3}^{2}-4 y_{1} y_{2}+2 y_{2} y_{3}-4 y_{3} y_{1}+4 x_{2}^{2}+\right. \)
\( x_{1}^{2}+x_{3}^{2}-4 x_{2} x_{1}+2 x_{1} x_{3}-4 x_{3} x_{1}+4 y_{2}^{2}+y_{3}^{2}+y_{1}^{2}-4 y_{2} y_{3}+y_{3} y_{1}-4 y_{1} y_{2}+4 x_{3}^{2}+x_{1}^{2}+x_{2}^{2}-4 x_{3} x_{1} \)
\( \left.+2 x_{1} x_{2}-4 x_{2} x_{3}+4 y_{3}^{2}+y_{1}^{2}+y_{2}^{2}-4 y_{3} y_{1}+2 y_{1} y_{2}-4 y_{2} y_{2}\right] \)
\( =\frac{1}{3}\left[6 x_{1}^{2}+6 x_{2}^{2}-6 x_{1}^{2}-6 x_{1} x_{2}-6 x_{2} x_{3}-6 x_{3} x_{1}+6 y_{1}^{2}+6 y_{2}^{2}+6 y_{3}^{2}-6 y_{1} y_{2}-6 y_{2} y_{3}-6 y_{3} y_{1}\right] \)
\( =\frac{1}{3}\left[6\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)-6\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right)+6\left(y_{1}^{2}+y^{2}+y_{3}^{2}\right)-6\left(y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{1}\right)\right] \)
\( =2\left[\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)-2\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right)+2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)-2\left(y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{1}\right)\right] \)
Therefore,
LHS $=$ RHS
Hence proved.
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