Question

State true or false

Using vector method the centroid of the triangle formed by joining the mid points of the sides of a given triangle coincides with the centroid of the given triangle.

• A. True
• B. False

Answer 1

The correct option is A True

R.E.F image

Let ABC be the given triangle Let P,Q,R be

midpoints of sides $AB,BC,AC$

Let $\overline{a},\overline{b},\overline{c},\overline{p},\overline{q},\overline{r}$ be the position vectors of

Points $A,B,C,P,Q,R$

By midpoint formula, $\overline{P}=\frac{\overline{a}+\overline{b}}{2};\overline{q}=\frac{\overline{b}+\overline{c}}{2};\overline{r}=\frac{\overline{a}+\overline{b}}{2}$

Let $\overline{G}$ is centriod of $\Delta ABC$ ie,$\overline{g}$ & ${\overline{g}}^1$ be controid

of $\Delta PQR$

By centroid formula $\overline{g}=\frac{\overline{a}+\overline{b}+\overline{c}}{3}$

${\overline{g}}^1=\frac{\overline{p}+\overline{q}+\overline{r}}{3}=\frac{\left(\frac{\overline{a}+\overline{b}}{2}\right)+\left(\frac{\overline{b}+\overline{c}}{2}\right)+\left(\frac{\overline{a}+\overline{c}}{2}\right)}{3}$

$=\frac{1}{6}(2\overline{a}+2\overline{b}+2\overline{c})=\frac{1}{3}(\overline{a}+\overline{b}+\overline{c})$

${\overline{g}}^1=\frac{1}{3}(\overline{a}+\overline{b}+\overline{c})\Rightarrow {\overline{g}}^1=\overline{g}$

$\therefore$ centroid of triangle ABC & PQR are

same & they coincide

The centroid of the $\Delta le$ formed by joining

midpoints of sides of $\Delta le$ coincides with

centroid of that $\Delta le$