The equation $x^2-5 x y+p y^2+3 x-8 y+2=0$ represents a pair of straight lines. If $ heta$ is the angle between them, then $\sin heta$ is equal to

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The equation $x^2-5 x y+p y^2+3 x-8 y+2=0$ represents a pair of straight lines. If $	heta$ is the angle between them, then $\sin 	heta$ is equal to

Aicharya · Accepted Answer

Comparing the given equation $ x^2-5 x y+p y^2+3 x-8 y+2=0 $ with $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ we get $ a=1, h=\frac{-5}{2}, b=p, g=\frac{3}{2}, f=-4 \text { and } c=2 $ Eq. (i) represents a pair of straight lines, if $ \begin{array}{cc} & a b c+2 f g h-a f^2-b g^2-c h^2=0 \\ \Rightarrow & 1 \times p \times 2+2 \times(-4) \times \frac{3}{2} \times\left(\frac{-5}{2}\right)-1 \times(-4)^2 \\ & -p \times\left(\frac{3}{2}\right)^2-2 \times\left(\frac{-5}{2}\right)^2=0 \\ \Rightarrow & p=6 \quad \therefore p=6 \end{array} $ $\therefore$ Required angle, $ \tan \theta=\frac{2 \sqrt{h^2-a b}}{a+b}=\frac{2 \sqrt{\left(\frac{-5}{2}\right)^2-1 \times 6}}{1+6} $

$ \begin{array}{cc} \Rightarrow & \tan \theta=\frac{1}{7} \Rightarrow \theta=\tan ^{-1} \frac{1}{7} \\ \therefore & \sin \theta=\frac{1}{\sqrt{50}} \end{array} $

The equation $x^2-5 x y+p y^2+3 x-8 y+2=0$ represents a pair of straight lines. If $\theta$ is the angle…

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