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.

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

$AP+BP=AB$

⇒ $AP+BP=\frac{7AP}{3AB}(\because AP=\frac{3}{7}AB)$

⇒ $BP=\frac{7AP}{3}-AP=\frac{4AP}{3}$

⇒ $\frac{AP}{BP}=\frac{3}{4}$

Let (x,y) be the coordinates of P which divides AB in the ratio 3:4 internally. Then

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

Also, m = 3 and n = 4

Let the required point be P(x,y)

By section formula, we get

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

⇒$x=\frac{(3\times 2)+(4\times -2)}{3+4}$
⇒$x=\frac{6-8}{7}$
⇒$x=\frac{-2}{7}$
⇒$y=\frac{(3\times -4)+(4\times -2)}{3+4}$
⇒$y=\frac{-12-8}{7}$
⇒$y=\frac{-20}{7}$

Hence, the coordinates of the point P are $(\frac{-2}{7},\frac{-20}{7})$