Question

Find the value of $m$ for which the points with coordinates $(3,5),(m,6)and\left(\frac{1}{2},\frac{15}{2}\right)$ are colinear.

Answer 1

Step 1: Given data:

Points $(3,5),(m,6)and\left(\frac{1}{2},\frac{15}{2}\right)$ are colinear.

Then, the area from these points is equal to zero.

We know that the area of the triangle of vertices $\left(x_1,y_1\right),\left(x_2,y_2\right),and\left(x_3,y_3\right)$ is given by the formula

$∆=\frac{1}{2}\left|\left\{x_1(y_2-y_3+x_2(y_3-y_1)+x_3(y_1-y_2)\right\}\right|$

Step 2: Finding the area of a triangle:

$∆=\frac{1}{2}\left|\left\{3\left(6-\frac{15}{2}\right)+m(\frac{15}{2}-5)+\frac{1}{2}(5-6)\right\}\right|$

$\Rightarrow \frac{1}{2}\left|\frac{-9}{2}+\frac{5m}{2}-\frac{1}{2}\right|=0$

$\Rightarrow \frac{5m}{2}-\frac{10}{2}=0$

$\Rightarrow 5m-10=0$

$\therefore m=2$

Hence, the value of $m$ is $2$.