Find one of the two points of trisection of the line segment joining the points $A (7,\ – 2)$ and $B (1,\ – 5)$ which divides the line in the ratio $1:2$.
Given: Trisection of the line segment joining the points $A (7,\ -2)$ and $B (1,\ -5)$ which divides the line in the ratio $1:2$.
To do: To find one of the two points.
Solution:
Let $( x,\ y)$ be the required point.
Using the division formula,
$( x,\ y)=( \frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n})$
Here $x_1=7,\ y_1=-2,\ x_2=1,\ y_2=-5,\ m=1$ and $n=2$
$( x,\ y)=( \frac{1\times1+2\times7}{1+2},\ \frac{1\times(-5)+2\times(-2)}{1+2})$
$( x,\ y)=( \frac{15}{3},\ \frac{-9}{3})$
$( x,\ y)=( 5,\ -3)$
Thus, the required point is $( 5,\ -3)$.
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