Question

Slope of line joining points $(5,3)$ and $\left(k^2,k+1\right)$ is $1/2$, then $k$ is

• A. 1
• B. $1+\sqrt{2}$
• C. $\sqrt{2}-1$
• D. $-1-\sqrt{2}$

Answer 1

Answer: (B)

Slope $=\frac{k+1-3}{k^2-5}=\frac{1}{2}$
$\Rightarrow k=1\pm \sqrt{2}$