Question

Find the value of m for which the points (3,5), (m,6) and $(\frac{1}{2},\frac{15}{2})$ are collinear. [2 marks]

Answer 1

The points are collinear if the area of the triangle formed by these three points is zero.
Let the points be
A(3,5), B(m,6) and C$(\frac{1}{2},\frac{15}{2})$

$ar△ABC=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$ [1 mark]

Substituting the values

$\frac{1}{2}\left[3\right(6-\frac{15}{2})+m(\frac{15}{2}-5)+\frac{1}{2}(5-6\left)\right]=0$
$\Rightarrow 3\left(\frac{-3}{2}\right)+\frac{5m}{2}-\frac{1}{2}=0$
$\Rightarrow \frac{-9}{2}-\frac{1}{2}+\frac{5m}{2}=0$
$\Rightarrow \frac{-10}{2}+\frac{5m}{2}=0$
$\Rightarrow 5m=10\Rightarrow m=2$
[1 mark]