# If the point $P(2,1)$ lies on the line segment joining points $A(4,2)$ and $B(8,4)$, then(A) $\mathrm{AP}=\frac{1}{3} \mathrm{AB}$(B) $\mathrm{AP}=\mathrm{PB}$(C) $\mathrm{PB}=\frac{1}{3} \mathrm{AB}$(D) $\mathrm{AP}=\frac{1}{2} \mathrm{AB}$

Given:

The point $P(2,1)$ lies on the line segment joining points $A(4,2)$ and $B(8,4)$.

To do:

We have to choose the correct option.

Solution:

We know that,

The distance between two points $\left(x_{1}, y_{1}\right)$ and $(x_{2}, y_{2})$, d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$The distance between$A(4,2)$and$P(2,1)$is,$A P=\sqrt{(2-4)^{2}+(1-2)^{2}}=\sqrt{(-2)^{2}+(-1)^{2}}=\sqrt{4+1}=\sqrt{5}$The distance between$A(4,2)$and$B(8,4)$is,$A B =\sqrt{(8-4)^{2}+(4-2)^{2}}=\sqrt{(4)^{2}+(2)^{2}}=\sqrt{16+4}=\sqrt{20}=2 \sqrt{5}$The distance between$B(8,4)$and$P(2,1)$is,$B P=\sqrt{(8-2)^{2}+(4-1)^{2}}=\sqrt{6^{2}+3^{2}}=\sqrt{36+9}=\sqrt{45}=3 \sqrt{5}A B =2 \sqrt{5}=2 A PA P=\frac{A B}{2}\$

Hence, the required condition is $\mathrm{AP}=\frac{A B}{2}$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

56 Views