Question

Find the largest value of $k$ for which the points $(k+1,1),(2k+1,3),(2k+2,2k)$ are collinear

• A. -2
• B. 2
• C. 1
• D. 3

Answer 1

The correct option is B 2

Three points given are $A(k+1,1),B(2k+1,3),C(2k+2,2k)$

These points will be collinear if slope of line AB= slope of line BC

Slope $m$ of a line joining two points $(x_1,y_1)$ and $(x_2,y_2)$ is,

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

Now, slope of line $AB$= slope of line $BC$

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

$\frac{2}{k}=\frac{2k-3}{1}$

$2=2k^2-3k$

$2k^2-3k-2=0$

$(2k+1)(k-2)=0$

$k=2$ or $k=-0.5$

So largest values of $k$ is $=2$