Question

AB has coordinates A(-5,9) and B(7,- 7). Points P, Q, and I are collinear
points in AB with coordinates P(-2,5), Q(1, 1), and T(4, -3).
Part A: Which of the following line segments would contain the point that
partitions AB into a ratio of 3: 2?

Answer 1

Answer:

$\overline{QT}$

Step-by-step explanation:

We want to find the coordinates of a certain point C(x,y) such that C divides $A(x_1,y_1)$ and $B(x_2,y_2)$ in the ratio m:n=3:2

The x-coordinate is given by:

$x=\frac{mx_2+nx_1}{m+n}$

The y-coordinate is given by:

$y=\frac{my_2+ny_1}{m+n}$

AB has coordinates A(-5,9) and B(7,- 7)

We substitute the values to get:

$x=\frac{3*7+2*-5}{3+2}$

$x=\frac{21-10}{5}$

$x=\frac{11}{5}$

and

$y=\frac{3*-7+2*9}{3+2}$

$y=\frac{-21+18}{5}$

$y=-\frac{3}{5}$

Therefore C has coordinates  $(\frac{11}{5},-\frac{3}{5})$

The line segment that contains C is $\overline{QT}$

See attachment.