The joint equation of a pair of lines passing | vedclass

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(A) The slope of the line $3x + 2y - 8 = 0$ is $m_1 = -\frac{3}{2}$.Let $m$ be the slope of the lines passing through the origin that make an angle of

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$5x^2 + 24xy - 5y^2 = 0$

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(A) The slope of the line $3x + 2y - 8 = 0$ is $m_1 = -\frac{3}{2}$.Let $m$ be the slope of the lines passing through the origin that make an angle of $\frac{\pi}{4}$ with the given line.Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + mm_1} ight|$,we have:$	an \frac{\pi}{4} = \left| \frac{m - (-3/2)}{1 + m(-3/2)} ight|$$1 = \left| \frac{2m + 3}{2 - 3m} ight|$Squaring both sides:$(2 - 3m)^2 = (2m + 3)^2$$4 - 12m + 9m^2 = 4m^2 + 12m + 9$$5m^2 - 24m - 5 = 0$Since the lines pass through the origin,their equations are $y = mx$,so $m = \frac{y}{x}$.Substituting $m = \frac{y}{x}$ into the auxiliary equation:$5(\frac{y}{x})^2 - 24(\frac{y}{x}) - 5 = 0$Multiplying by $x^2$,we get:$5y^2 - 24xy - 5x^2 = 0$Rearranging,we get $5x^2 + 24xy - 5y^2 = 0$.

The joint equation of a pair of lines passing through the origin and making an angle of $\frac{\pi}{4}$ with the line $3x + 2y - 8 = 0$ is

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