Question

Find the values of k, if the points A(k+1, 2k), B(3k, 2k+3) and C(5k-1, 5k) are collinear.

• A. k = 5
• B. k = $\frac{1}{2}$
• C. k = 3
• D. k = 2

Answer 1

The correct option is D

k = 2

Given, points A, B and C are collinear.

Then, slope of the line segment AB = slope of the line segment BC

We know that, Slope(m) of the line formed by joining the points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$m=\frac{y_2-y_1}{x_2-x_1}$

So, slope of the line segment AB = slope of the line segment BC implies,

$\frac{(2k+3)-\left(2k\right)}{3k-(k+1)}$ = $\frac{\left(5k\right)-(2k+3)}{(5k-1)-3k}$

$⟹\frac{3}{2k-1}=\frac{3k-3}{2k-1}$

$⟹\frac{1}{2k-1}=\frac{k-1}{2k-1}$

$⟹1=k-1$

$⟹k=2$

Hence, the required value of k = 2.