Question

If $A\left(-3,4\right),$ $B\left(3,-1\right)$ and $C\left(-2,4\right)$ are the vertices of $▵ABC$. Find the length of line segment $AP$, where $P$ lies inside $BC$ such that $BP:PC=2:3$.

Answer 1

Step-1: Finding the coordinates of the point $P$.

Let the point $P$ be $\left(a,b\right)$.The given point $P$ divides the line segment $BC$ such that $BP:PC=2:3$.

We are given the coordinates of the endpoints of $BC$ as $B\left(3,-1\right)$and $C\left(-2,4\right)$ respectively.

Substituting these values in the section formula we get the coordinates of point $P$ as:

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

$\left(a,b\right)=\left(\frac{2\left(-2\right)+3\left(3\right)}{2+3},\frac{2\left(4\right)+3\left(-1\right)}{2+3}\right)$ [ since $m=2andn=3$]

Simplifying which, we get:

$\left(a,b\right)=\left(\frac{-4+9}{5},\frac{8-3}{5}\right)$

$\left(a,b\right)=\left(\frac{5}{5},\frac{5}{5}\right)$

Or we can say:

$\left(a,b\right)=\left(1,1\right)$

The coordinates of point $P$ are found to be $\left(1,1\right)$.

Step-2: Finding the distance of the side $AP$.

Now, we now know the coordinates of both points $A$ and $P$ as $A\left(-3,4\right),$$P\left(1,1\right)$ respectively.

Substituting these in the distance formula, we get:

$D=\sqrt{[{(x_2–x_1)}^2+(y_2-y_1}{)}^2$

$D_{AP}=\sqrt{{\left(1-\left(-3\right)\right)}^2+{\left(1-4\right)}^2}$

$D_{AP}=\sqrt{{\left(4\right)}^2+{\left(-3\right)}^2}$

Simplifying which we get:

$D_{AP}=\sqrt{16+9}$

$D_{AP}=\sqrt{25}$

Or we can say:

$D_{AP}=5$units as the length cannot be negative.

Therefore , the length of the segment $AP$ is found to be $5$ units.