The equation x^2-5xy+py^2+3x-8y+2=0 represent | vedclass

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(A) Comparing the given equation $x^2-5xy+py^2+3x-8y+2=0$ with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=-\frac{5

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$\frac{1}{\sqrt{50}}$

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(A) Comparing the given equation $x^2-5xy+py^2+3x-8y+2=0$ with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=-\frac{5}{2}, b=p, g=\frac{3}{2}, f=-4, c=2$. The equation represents a pair of straight lines if $abc+2fgh-af^2-bg^2-ch^2=0$. Substituting the values: $1(p)(2) + 2(-4)(\frac{3}{2})(-\frac{5}{2}) - 1(-4)^2 - p(\frac{3}{2})^2 - 2(-\frac{5}{2})^2 = 0$. $2p + 30 - 16 - \frac{9p}{4} - \frac{25}{2} = 0$. Multiplying by $4$: $8p + 120 - 64 - 9p - 50 = 0$ $\Rightarrow -p + 6 = 0$ $\Rightarrow p=6$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight|$. $	an 	heta = \left| \frac{2\sqrt{(-\frac{5}{2})^2 - 1(6)}}{1+6} ight| = \left| \frac{2\sqrt{\frac{25}{4}-6}}{7} ight| = \left| \frac{2\sqrt{\frac{1}{4}}}{7} ight| = \frac{2(\frac{1}{2})}{7} = \frac{1}{7}$. Since $	an 	heta = \frac{1}{7}$,we can form a right-angled triangle with opposite side $1$ and adjacent side $7$. The hypotenuse is $\sqrt{1^2+7^2} = \sqrt{50}$. Therefore,$\sin 	heta = \frac{1}{\sqrt{50}}$.

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

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