Question

Let $A(4,2),B(6,5)$ and $C(1,4)$ be the vertices of $\Delta ABC$.
(i) The median from $A$ meets $BC$ at $D$. Find the coordinates of the point $D$.
(ii) Find the coordinates of the point $P$ on $AD$ such that $AP:PD=2:1$
(iii) Find the coordinates of points $Q$ and $R$ on medians $BE$ and $CF$ respectively such that $BQ:QE=2:1$ and $CR:RF=2:1$.
(iv) What do yo observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio $2:1$.]
(v) If $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ are the vertices of $\Delta ABC$, find the coordinates of the centroid of the triangle.

Answer 1

i)

Median is the line joining the midpoint of one side of a triangle to the opposite vertex. So, the coordinates of $D$ would be$(\frac{6+1}{2},\frac{5+4}{2})$::$(\frac{7}{2},\frac{9}{2})$

ii)

$P$ divides $AD$ in the ratio $2:1$.

$A(x_1,y_1)=(4,2),D(x_2,y_2)=(\frac{7}{2},\frac{9}{2})$

$m:n=2:1$

Using section formula, we get the coordinates of $P$.

$P(x,y)=(\frac{n{x}_1+m{x}_2}{m+n},\frac{n{y}_1+m{y}_2}{m+n})$

$=(\frac{1\cdot 4+2\cdot \frac{7}{2}}{2+1},\frac{1\cdot 2+2\cdot \frac{9}{2}}{2+1})$

$=(\frac{11}{3},\frac{11}{3})$

iii)

Coordinates of $E$ will be $(\frac{5}{2},3)$ and the coordinates of $F=(5,\frac{7}{2})$.

Coordinates of $Q=(\frac{n{x}_1+m{x}_2}{m+n},\frac{n{y}_1+m{y}_2}{m+n})$

$=(\frac{1\cdot 6+2\cdot \frac{5}{2}}{2+1},\frac{1\cdot 5+2\cdot 3}{2+1})$

$=(\frac{11}{3},\frac{11}{3})$

Coordinates of $R=(\frac{n{x}_1+m{x}_2}{m+n},\frac{n{y}_1+m{y}_2}{m+n})$

$=(\frac{1\cdot 1+2\cdot 5}{2+1},\frac{1\cdot 4+2\cdot \frac{7}{2}}{2+1})$

$=(\frac{11}{3},\frac{11}{3})$

iv)

The coordinates of $P,Q$ and $R$ are the same which is $(\frac{11}{3},\frac{11}{3})$.

This point is called the centroid, denoted by $G$.

v)

Centroid of triangle $ABC=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$