Given:
A line segment AB.
A point is [tex]\dfrac{3}{10}[/tex] of the way from A to B.
To find:
The coordinates of that point.
Solution:
Section formula: If a point divide a line segment in m:n, then
[tex]Point=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)[/tex]
From the given figure, it is clear that the two end points of the line segment are A(-4,-5) and B(12,4).
Let the unknown point be P. The point is [tex]\dfrac{3}{10}[/tex] of the way from A to B. It means,
[tex]\dfrac{AP}{AB}=\dfrac{3}{10}[/tex]
Let AP and AB are 3x and 10x, then
[tex]\dfrac{AP}{PB}=\dfrac{AP}{AB-AP}[/tex]
[tex]\dfrac{AP}{PB}=\dfrac{3x}{10x-3x}[/tex]
[tex]\dfrac{AP}{PB}=\dfrac{3x}{7x}[/tex]
[tex]\dfrac{AP}{PB}=\dfrac{3}{7}[/tex]
It means, point P divided the line segment in 3:7.
Using section formula, we get
[tex]P=\left(\dfrac{3(12)+7(-4)}{3+7},\dfrac{3(4)+7(-5)}{3+7}\right)[/tex]
[tex]P=\left(\dfrac{36-28}{10},\dfrac{12-35}{10}\right)[/tex]
[tex]P=\left(\dfrac{8}{10},\dfrac{-23}{10}\right)[/tex]
[tex]P=\left(0.8,-2.3\right)[/tex]
Therefore, the coordinates of the required point are (0.8, -2.3).
