Question

Question 7 (ii)
Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of triangle ABC.
(b) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.

Answer 1

(b) Coordinates of P can be calculated as follows using section formula:

$x=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}$,$y=\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}$
In this case $(m_1=1,m_2=2and{x}_1=4,x_2=\frac{7}{2},y_1=2,y_2=\frac{9}{2})$
$x=\frac{1\times 4+2\times \frac{7}{2}}{3}$
$=\frac{4+7}{3}=\frac{11}{3}$
$y=\frac{1\times 2+2\times \frac{9}{2}}{3}$
$=\frac{2+9}{3}=\frac{11}{3}$
Therefore, coordinates of point P are $(\frac{11}{3},\frac{11}{3})$