If the point $P (m, 3)$ lies on the line segment joining the points $A (−\frac{2}{5}, 6)$ and $B (2, 8)$, find the value of $m$.


Given:

The point $P (m, 3)$ lies on the line segment joining the points $A (−\frac{2}{5}, 6)$ and $B (2, 8)$.

To do:

We have to find the value of $m$.

Solution:

We know that,

If the points $A, B$ and $C$ are collinear then the area of $\triangle ABC$ is zero.

Let $A(-\frac{2}{5}, 6), P(m, 3)$ and $B(2, 8)$ be the vertices of $\triangle APB$.

Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by, 

Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Therefore,

Area of triangle \( APB=\frac{1}{2}[-\frac{2}{5}(3-8)+m(8-6)+2(6-3)] \)

\( 0=\frac{1}{2}[-\frac{2}{5}(-5)+m(2)+2(3)] \)

\( 0(2)=(2+2m+6) \)

\( 0=8+2m \)

\( 2m=-8 \)

\( m=-4 \)

The value of $m$ is $-4$.

Tutorialspoint
Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

39 Views

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements