Question

If $A$ and $B$ are $(-2,-2)$ and $(2,-4)$, respectively, find the coordinates of $P$ such that $AP=\frac{3}{7}AB$ and $P$ lies on the line segment $AB$.

Answer 1

Step 1: Finding the ratio in which the point divides the line

Given, coordinates of point $A$ is $(-2,-2)$ and that of $B$ is $(2,-4)$.

Also given,$AP=\frac{3}{7}AB$.

This means that, $AP$ is $3$ parts out of $7$ parts of $AB$.

Hence, the remaining $4$ parts is $PB$.

Therefore, $AP:PB=3:4$

Step 2 - Apply section formula

Let $m:n$ be the ratio in which point $P$ divides a line segment. Then coordinates of $P$ is given as,

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

Where, $x_1$, $x_2$ and $y_1$, $y_2$ are the $x$ and $y$ coordinates of the start and end point of the line segment.

Here, $(x_1,y_1)=(-2,-2)$ and $(x_2,y_2)=(2,-4)$.

Thus, coordinates of $P$ are,

$\begin{array}{rcl}P& =& \left(\frac{3\times 2+4\times \left(-2\right)}{3+4},\frac{3\times (-4)+4\times (-2)}{3+4}\right)\\ & =& \left(\frac{-2}{7},-\frac{20}{7}\right)\end{array}$

Hence, the coordinates of $P$ are $\left(\frac{-2}{7},-\frac{22}{7}\right)$.