Question

If the coordinates of points A and B are (−2, −2) and (2, −4) respectively, find the coordinates of the point P
such that $AP=\frac{3}{7}AB$, where P lies on the line segment AB. [CBSE 2015]

Answer 1

The coordinates of the points A and B are (−2, −2) and (2, −4) respectively, where $AP=\frac{3}{7}AB$ and P lies on the line segment AB. So
$$\begin{gathered} AP+BP=AB \\ \Rightarrow AP+BP=\frac{7AP}{3}\because AP=\frac{3}{7}AB \\ \Rightarrow BP=\frac{7AP}{3}-AP \\ \Rightarrow \frac{AP}{BP}=\frac{3}{4} \end{gathered}$$
Let (x, y) be the coordinates of P which divides AB in the ratio 3 : 4 internally. Then
$$\begin{gathered} x=\frac{3\times 2+4\times \left(-2\right)}{3+4}=\frac{6-8}{7}=-\frac{2}{7} \\ y=\frac{3\times \left(-4\right)+4\times \left(-2\right)}{3+4}=\frac{-12-8}{7}=-\frac{20}{7} \end{gathered}$$
Hence, the coordinates of point P are $\left(-\frac{2}{7},-\frac{20}{7}\right)$.