Given:
Points A(-2,-1) and B(-6,5).
Point C which is 2/3 from point A to point B.
To find:
The y-coordinate of point C.
Solution:
Let the coordinates of point C are (a,b).
Point C which is 2/3 from point A to point B.
AC:AB = 2:3
AC:CB = AC:(AB-AC) = 2:(3-2) = 2:1
Section formula: If a point divides the line segment in m:n, then then
[tex]Point=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)[/tex]
Point C divides AB in 2:1, soby section formula
[tex]C=\left(\dfrac{2(-6)+1(-2)}{2+1},\dfrac{2(5)+1(-1)}{2+1}\right)[/tex]
[tex]C=\left(\dfrac{-12-2}{3},\dfrac{10-1}{3}\right)[/tex]
[tex]C=\left(\dfrac{-14}{3},\dfrac{9}{3}\right)[/tex]
[tex]C=\left(\dfrac{-14}{3},3\right)[/tex]
Therefore, the y-coordinated of point C is 3.
