# If the Point P (M, 3) Lies on the Line Segment Joining the Points a ( − 2 5 , 6 ) and B (2, 8), Find the Value of M. - Mathematics

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 .

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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.5 | Q 17 | Page 54