Question

The equation $x^2-5xy+p{y}^2+3x-8y+2=0$ represent a pair of straigth lines.

If $\theta$ is the angle between them, then $\sin \theta =?$

• A. $\frac{1}{\sqrt{50}}$
• B. $\frac{1}{7}$
• C. $\frac{1}{5}$
• D. $\frac{1}{\sqrt{10}}$

Answer 1

The correct option is A $\frac{1}{\sqrt{50}}$

$a{x}^2+2nxy+b{y}^2+2gx+2fy+c=0$

$x^2-5xy+p{y}^2+3x-8y+2=0$

$\begin{array}{cc}a=1& a=1\\ 2h=-5& h=-5/2\\ b=p& b=p\\ 2g=3& g=\frac{3}{2}\\ 2f=-8& f=-4\\ c=2& c=2\end{array}$

$\left|\begin{array}{ccc}1& \frac{-5}{2}& \frac{3}{2}\\ \frac{-5}{2}& p& -4\\ \frac{3}{2}& -4& 2\end{array}\right|=0$

$\left|\begin{array}{cc}p& -4\\ -4& 2\end{array}\right|+\frac{5}{2}\left|\begin{array}{cc}\frac{-5}{2}& -4\\ \frac{3}{2}& 2\end{array}\right|+\frac{3}{2}\left|\begin{array}{cc}-\frac{5}{2}& p\\ \frac{3}{2}& -4\end{array}\right|=0$

$(2p-16)+\frac{5}{2}(-5+4\times \frac{3}{2})+\frac{3}{2}(\frac{5}{2}\times -\frac{3p}{2})=0$

$(2p-16)+\frac{5}{2}(-5+6)+\frac{3}{2}[\frac{20}{2}-\frac{3p}{2}]=0$

$(2p-16)+\frac{5}{2}+\frac{3}{4}(20-3p)=0$

$\frac{8p-64+10+60-9p}{4}=0$

$\Rightarrow 8p-9p+6=0$

$-p=-6$

$p=6$

$tan\theta =\frac{2\sqrt{\frac{25}{4}-p}}{1+p}$

$=\frac{2\sqrt{\frac{25}{4}-6}}{1+6}$

$\frac{2\sqrt{\frac{25-24}{4}}}{7}$

$=\frac{2\times \frac{1}{2}}{7}=\frac{1}{7}$

$tan\theta =\frac{1}{7}$

$sin\theta =\frac{1}{\sqrt{50}}$