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

Straight Lines

Solution:

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