The sine of the angle between the pair of lin | vedclass

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(C) The given equation is $x^2-7xy+12y^2=0$. Comparing this with the general form $ax^2+2hxy+by^2=0$,we have $a=1$,$2h=-7$,and $b=12$. The angle $	he

Vedclass · Accepted Answer

$\frac{1}{\sqrt{170}}$

Vedclass · Answer

(C) The given equation is $x^2-7xy+12y^2=0$. Comparing this with the general form $ax^2+2hxy+by^2=0$,we have $a=1$,$2h=-7$,and $b=12$. The angle $	heta$ between the pair of lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight|$. Substituting the values,we get $	an 	heta = \left| \frac{2\sqrt{(-7/2)^2 - 1 \cdot 12}}{1+12} ight| = \left| \frac{2\sqrt{49/4 - 12}}{13} ight| = \left| \frac{2\sqrt{1/4}}{13} ight| = \frac{2 \cdot (1/2)}{13} = \frac{1}{13}$. Since $	an 	heta = \frac{1}{13}$,we can construct a right-angled triangle with opposite side $1$ and adjacent side $13$. The hypotenuse is $\sqrt{1^2+13^2} = \sqrt{1+169} = \sqrt{170}$. Therefore,$\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{1}{\sqrt{170}}$. Hence,option $(C)$ is correct.

The sine of the angle between the pair of lines represented by the equation $x^2-7xy+12y^2=0$ is

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