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.
Solution
The formula for the area ‘A’ encompassed by three points ( x1 , y1) , (x2 , y2) and (x3 , y3) 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 A`(-2/5,6)` and B(2,8). Hence we understand that these three points are collinear. So the area enclosed by them should be 0.
\[∆ = \frac{1}{2}\left| \left( - \frac{2}{5} \times 3 + m \times 8 + 2 \times 6 \right) - \left( m \times 6 + 2 \times 3 + \left( - \frac{2}{5} \right) \times 8 \right) \right|\]
\[ 0 = \frac{1}{2}\left| \left( - \frac{6}{5} + 8m + 12 \right) - \left( 6m + 6 - \frac{16}{5} \right) \right|\]
\[ 0 = \frac{1}{2}\left| 2m + 8 \right|\]
\[ 0 = 2m + 8\]
\[ m = - 4\]
Hence the value of ‘m’ for which the given condition is satisfied is m = - 4 .
