Question

Slope of line joining points $(5,3)$ and $(k^2,k+1)$ is $\frac{1}{2}$ then $k$ is

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

Answer 1

Answer: (B)

$1\pm \sqrt{2}$

$y=mx+c=>3=\frac{1}{2}\times 5+c=>2c=1=>c=\frac{1}{2}y=mx+c=>k+1=\frac{1}{2}{k}^2+\frac{1}{2}=>2k+2=k^2+1=>k^2-2k-1=0=>k=\frac{2\pm \sqrt{4+4}}{2}=\frac{2\pm 2\sqrt{2}}{2}=1\pm \sqrt{2}$