Need help! Find the point, Q, along the directed line segment BA that divides BA into the ratio 2:3.

Answer:
[tex](-\frac{2}{5},1)[/tex]
Step-by-step explanation:
The problem is based on Section Formula which is given as under
[tex]x=\frac{mx_2+nx_1}{m+n}[/tex]
[tex]y=\frac{my_2+ny_1}{m+n}[/tex]
Where the m and n are the ratio in which the point [tex]P(x,y) bisect the line joining [tex]A(x_1,y_1)[/tex] and [tex]B(x_2,y_2) [/tex] internally
Now from the attached image we can guess the values of [tex](x_1,y_1) [/tex] and [tex](x_2,y_2)[/tex] and the values of m and n are 3 and 2 respectively
Substituting the values and simplifying we get
[tex]x=\frac{3 \times 2+2 \times -4}{3+2}[/tex]
[tex]x=\frac{6-8}{5}[/tex]
[tex]x=\frac{-2}{5}[/tex]
[tex]x=-\frac{2}{5}[/tex]
Now solving for y coordinate
[tex]y=\frac{3 \times -3+2 \times 7}{3+2}[/tex]
[tex]y=\frac{-9+14}{3+2}[/tex]
[tex]y=\frac{5}{5}[/tex]
[tex]y=1[/tex]
Hence the coordinates of point p which divides the line BA in ration 3:2 will be
[tex](-\frac{2}{5},1)[/tex]
Answer:
see explanation
Step-by-step explanation:
Using the section formula
[tex]x_{Q}[/tex] = [tex]\frac{2(-4)+3(2)}{2+3}[/tex] = [tex]\frac{-8+6}{5}[/tex] = - [tex]\frac{2}{5}[/tex]
[tex]y_{Q}[/tex] = [tex]\frac{2(7)+3(-3)}{2+3}[/tex] = [tex]\frac{14-9}{5}[/tex] = 1
Hence Q = (- [tex]\frac{2}{5}[/tex], 1 )