Question

$A(4,2),B(6,5)$ and $C(1,4)$ are the vertices of $△ABC$.

(i) The median from $A$ meets $BC$ at $D$. Find the coordinates of the point $D$.

(ii) Find the coordinates of point $P$ on $AD$ such that $AP:PD=2:1$

Answer 1

Given,

$A(4,2),B(6,5),C(1,4)$

(i)

Since AD is median, Point D becomes the mid-point of BC.

Coordinates of D$=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$=(\frac{6+1}{2},\frac{5+4}{2})$

$=(\frac{7}{2},\frac{9}{2})$

(ii)

Given,

$AP:PD=2:1$

As per section formula, the coordinates of point P $=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$

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

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