Question

If the slope of a line is $K$ and it make an angle of $\frac{\pi }{4}$ with another line which passess through $(K,0)$ and $(8,5)$ then the value(s) of $K$ is/are

• A. $6+\sqrt{39}$
• B. $6-\sqrt{39}$
• C. $12-\sqrt{39}$
• D. $12+\sqrt{39}$

Answer 1

Answer: (A), (B)

$6-\sqrt{39}$

Slope of first line $=K$
Slope of second line $=\frac{5}{8-K}$
Angle between them $=\frac{\pi }{4}$

$\therefore \tan \frac{\pi }{4}=\left|\frac{K-\frac{5}{8-K}}{1+K\times \frac{5}{8-K}}\right|\Rightarrow 1=\left|\frac{8K-K^2-5}{8-K+5K}\right|\Rightarrow 1=\left|\frac{-K^2+8K-5}{8+4K}\right|\Rightarrow 8+4K=\pm (-K^2+8K-5)$

Case$\left(1\right)$ When $8+4K=+(-K^2+8K-5)$
$\Rightarrow K^2-4K+13=0D=16-52=-36<0$
Hence, no real value.

Case$\left(2\right)$ When $8+4K=-(-K^2+8K-5)$
$\Rightarrow K^2-12K-3=0\Rightarrow K=6\pm \sqrt{39}$