Question

Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) in such a way that $AP=\frac{2}{5}AB$.

• A. $(-3,-8)$
• B. $(-2,5)$
• C. $(3,4)$
• D. $(5,-8)$

Answer 1

The correct option is C $(3,4)$

Given,
$AP=\left(\frac{2}{5}\right)AB$
$\frac{AP}{AB}=\frac{2}{5}$

$\left(\frac{AP}{AB}\right)+\left(\frac{PB}{AB}\right)=1$ $(\because AB=AP+PB)$

$\left(\frac{PB}{AB}\right)=1-\left(\frac{AP}{AB}\right)$
$=1-\frac{2}{5}$
$=\frac{3}{5}$

Hence,
$\frac{AP}{PB}=\left(\frac{2}{5}\right)AB÷\left(\frac{3}{5}\right)AB$
$=\left(\frac{2}{5}\right)÷\left(\frac{3}{5}\right)$
$=\frac{2}{3}$

A point divides line internally in the ratio m : n

Point $P=[\frac{(m{x}_2+m{x}_1)}{(m+n)},\left[\frac{(m{y}_2+m{y}_1)}{(m+n)}\right]]$

Here,
$m=2,n=3,x_1=1,x_2=6,y_1=2,y_2=7$

Hence,
$P=\left[\right(2\times 6+3\times 1\left)\right(2+3),[(2\times 7+3\times 2)/(2+3)]$
$=[\frac{(12+3)}{5},\frac{(14+6)}{5}]$
$=\frac{15}{5},\frac{20}{5}$
$=(3,4)$