Question

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

Answer 1

The formula for the area ‘A’ encompassed by three points, and is given by the formula,

$∆=\frac{1}{2}\left|\left(x_1{y}_2+x_2{y}_3+x_3{y}_1\right)-\left(x_2{y}_1+x_3{y}_2+x_1{y}_3\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

It is said that the point P(m,3) lies on the line segment joining the points and B(2,8). Hence we understand that these three points are collinear. So the area enclosed by them should be 0.
$$\begin{gathered} ∆=\frac{1}{2}\left|\left(-\frac{2}{5}\times 3+m\times 8+2\times 6\right)-\left(m\times 6+2\times 3+\left(-\frac{2}{5}\right)\times 8\right)\right| \\ 0=\frac{1}{2}\left|\left(-\frac{6}{5}+8m+12\right)-\left(6m+6-\frac{16}{5}\right)\right| \\ 0=\frac{1}{2}\left|2m+8\right| \\ 0=2m+8 \\ m=-4 \end{gathered}$$

Hence the value of ‘m’ for which the given condition is satisfied is.