Question

Find P on AB that is 2/3 of the distance from A(-3,-5) to B(9,4).

Answer 1

A line segment can be divided into ratios.

The coordinate of P on line segment AB is (5,1)

Given

$A = (-3,-5)$

$B = (9,4)$

The position of P from A to B is:

$P_A = \frac 23$

So, from P to B would be:

$P_B =1 - \frac 23$

$P_B = \frac 13$

So, the ratio is:

$m : n = \frac 23 : \frac 13$

Simplify

$m : n = 2 :1$

The coordinate of point P is:

$P = (\frac{mx_2 + nx_1}{m + n},\frac{my_2 + ny_1}{m + n})$

So, we have:

$P = (\frac{2 \times 9 + 1 \times -3}{2 + 1},\frac{2 \times 4 + 1 \times -5}{2 + 1})$

$P = (\frac{15}{3},\frac{3}{3})$

$P = (5,1)$

Hence, the coordinate of P is (5,1)

Read more about line ratios at:

brainly.com/question/8847082