Question

If point $P(3,8)$ lies on the line segment joining $A(2,10)$ and $B(6,2),$ then:

[1 mark]

• A. $AP=\frac{AB}{4}$
• B. $AP=BP$
• C. $AB=4BP$
• D. $AP=\frac{AB}{3}$

Answer 1

The correct option is A $AP=\frac{AB}{4}$


Using the section formula,
$P(3,8)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})(3,8)=(\frac{m\left(6\right)+n\left(2\right)}{m+n},\frac{m\left(2\right)+n\left(10\right)}{m+n})3=\frac{6m+2n}{m+n}$
$3(m+n)=6m+2n$
$n=3m$
$m:n=1:3$
Option A : $Ap=\frac{AB}{4}$
$AP=m=1$
$AB=m+n=1+\frac{AB}{4}$
$\therefore AP=\frac{AB}{4}$