Answer:
[tex]\frac{11}{3},4[/tex]
Explanation:
A line segment XY with points at X([tex]x_1,y_1[/tex]) and Y([tex]x_2,y_2[/tex]) divided by a point O(x, y) in the ratio n:m , the location of point O(x, y) is at:
[tex]x=\frac{n}{n+m}(x_2-x_1)+x_1\\ \\y=\frac{n}{n+m}(y_2-y_1)+y_1[/tex]
Given line segment AB with location A (3, 6) B(4, 3), point P is given as:
[tex]x=\frac{n}{n+m}(x_2-x_1)+x_1=\frac{2}{2+1}(4-3)+3=\frac{2}{3}(1)+3=\frac{11}{3} \\ \\y=\frac{n}{n+m}(y_2-y_1)+y_1=\frac{2}{2+1}(3-6)+6=\frac{2}{3}(-3)+6=-2+6=4[/tex]
The location of point P is at ([tex]\frac{11}{3},4[/tex])
