Given:
The coordinates of point P(-4,0).
The coordinates of point A(2,-6).
Explanation:
The distance of AP is 3/4 times of AE. So,
[tex]\begin{gathered} AP=\frac{3}{4}AE \\ \frac{AE}{AP}=\frac{4}{3} \\ \frac{AP+PE}{AP}=\frac{4}{3} \\ \frac{PE}{AP}=\frac{4}{3}-1 \\ \frac{PE}{AP}=\frac{1}{3} \end{gathered}[/tex]
The formula for the coordinates of the point lying between endpoint (x_1,y_1) and (x_2,y_2) such that divide the line in m:n is,
[tex](x,y)=(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})[/tex]
So coordinates for point P lying on segment AE is,
[tex]\begin{gathered} (-4,0)=(\frac{3\cdot x+1\cdot2}{1+3},\frac{3\cdot y+1\cdot(-6)}{1+3}) \\ =(\frac{3x+2}{4},\frac{3y-6}{4}) \end{gathered}[/tex]
Solve the equations for x and y.
[tex]\begin{gathered} \frac{3x+2}{4}=-4 \\ 3x+2=-16 \\ x=-\frac{18}{3} \\ =-6 \end{gathered}[/tex]
For y,
[tex]\begin{gathered} \frac{3y-6}{4}=0 \\ 3y=6 \\ y=\frac{6}{3} \\ =2 \end{gathered}[/tex]
So coordinates of point E is (-6,2).
