Question

Let A(-6, -5) and B(-6, 4) be the two points such that a point P on the line AB satisfies AP = (2/9) AB.

Answer 1

AP = (2/9) AB

9AP = 2(AP+PB)

9AP = 2AP + 2PB

9AP – 2AP = 2PB

7AP = 2PB

AP/AB = 2/7

AP:PB = 2:7

So P divides the line segment in the ratio 2:7

Section formula

= $\frac{l{x}_2+m{x}_1}{l+m}$, $\frac{l{y}_2+m{y}_1}{l+m}$

l = 2, m = 7

= [(2(-6)+7(-6)]/(2+7), [(2x(4)+7(-5)]/(2+7)

= $\frac{-12-42}{2+7}$, $\frac{8-35}{2+7}$

= $\frac{-12-42}{9}$, $\frac{8-35}{9}$

= -54/9, -27/9

= $\frac{-54}{9}$, $\frac{-27}{9}$

= (-6, -3)