Given:
Point P divides the line segment AB in 5:3 where A(-6,9) and B(2,1).
To find:
The coordinates of point P.
Solution:
Section formula: If a point divide a line segment in m:n, then the coordinates of point are
[tex]\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)[/tex]
Point P divides the line segment AB in 5:3 where A(-6,9) and B(2,1). Using section formula, we get
[tex]P=\left(\dfrac{5(2)+3(-6)}{5+3},\dfrac{5(1)+3(9)}{5+3}\right)[/tex]
[tex]P=\left(\dfrac{10-18}{8},\dfrac{5+27}{8}\right)[/tex]
[tex]P=\left(\dfrac{-8}{8},\dfrac{32}{8}\right)[/tex]
[tex]P=\left(-1,4\right)[/tex]
Therefore, the coordinates of point P are (-1,4).
