Question

Find the value of $k$ for which the points with coordinates $(2,5)$ and $\left(k,\frac{11}{2}\right)$ and $(4,6)$ are colinear.

Answer 1

Finding the value of $k$:

Given points $(2,5),$$\left(k,\frac{11}{2}\right)$& $(4,6)$are colinear.

Then area from these points is equal to zero.

We know the area of the triangle with vertices $(x_1,y_1),(x_2,y_2)and(x_3,y_3)$.

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

$∆=\frac{1}{2}\left|\left\{2\left(\frac{11}{2}-6\right)+k(6-5)+4\left(5-\frac{11}{2}\right)\right\}\right|$

$\Rightarrow 0=\frac{1}{2}\left|-1+k-2\right|$

$\Rightarrow 0=k-3$

$\therefore k=3$

Hence, the value of $k$ is $3$.