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 $-4$. 
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