The equation of the pair of straight lines,ea | vedclass

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(D) Any line passing through the origin is $y = mx$. If it makes an angle $\alpha$ with the line $y = x$ (which has slope $m_1 = 1$),then the angle be

Vedclass · Accepted Answer

$x^2 - 2xy \sec 2\alpha + y^2 = 0$

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(D) Any line passing through the origin is $y = mx$. If it makes an angle $\alpha$ with the line $y = x$ (which has slope $m_1 = 1$),then the angle between them is given by $	an \alpha = \left| \frac{m - 1}{1 + m(1)} ight|$.Squaring both sides,we get $	an^2 \alpha = \frac{(m - 1)^2}{(m + 1)^2}$.This simplifies to $(m + 1)^2 	an^2 \alpha = (m - 1)^2$.Expanding,we get $(m^2 + 2m + 1) 	an^2 \alpha = m^2 - 2m + 1$.Rearranging the terms to form a quadratic in $m$: $m^2(1 - 	an^2 \alpha) - 2m(1 + 	an^2 \alpha) + (1 - 	an^2 \alpha) = 0$.Dividing by $(1 - 	an^2 \alpha)$,we get $m^2 - 2m \left( \frac{1 + 	an^2 \alpha}{1 - 	an^2 \alpha} ight) + 1 = 0$.Since $\frac{1 + 	an^2 \alpha}{1 - 	an^2 \alpha} = \sec 2\alpha$,the equation becomes $m^2 - 2m \sec 2\alpha + 1 = 0$.Substituting $m = \frac{y}{x}$,we get $\left( \frac{y}{x} ight)^2 - 2 \left( \frac{y}{x} ight) \sec 2\alpha + 1 = 0$.Multiplying by $x^2$,we obtain the required equation $y^2 - 2xy \sec 2\alpha + x^2 = 0$.

The equation of the pair of straight lines,each of which makes an angle $\alpha$ with the line $y = x$,is

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