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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Updated on: 10-Oct-2022

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