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(A) Let the slopes of the two lines passing through the origin be $m_1$ and $m_2$. Since the lines are perpendicular,$m_1 m_2 = -1$,or $m_2 = -\frac{1

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

Vedclass · Answer

(A) Let the slopes of the two lines passing through the origin be $m_1$ and $m_2$. Since the lines are perpendicular,$m_1 m_2 = -1$,or $m_2 = -\frac{1}{m_1}$.Given the lines form an isosceles right-angled triangle with the line $2x + 3y = 6$,the angles the lines make with the base line are $45^{\circ}$.The slope of the line $2x + 3y = 6$ is $m = -\frac{2}{3}$.Using the formula $	an 	heta = \left| \frac{m_1 - m}{1 + m_1 m} ight|$,where $	heta = 45^{\circ}$:$1 = \left| \frac{m_1 - (-2/3)}{1 + m_1(-2/3)} ight| = \left| \frac{3m_1 + 2}{3 - 2m_1} ight|$.Squaring both sides: $(3 - 2m_1)^2 = (3m_1 + 2)^2$.$9 - 12m_1 + 4m_1^2 = 9m_1^2 + 12m_1 + 4$.$5m_1^2 + 24m_1 - 5 = 0$.This quadratic equation gives the slopes $m_1$ and $m_2$ of the two lines.The combined equation of the pair of lines $y = m_1x$ and $y = m_2x$ is $(y - m_1x)(y - m_2x) = 0$,which is $y^2 - (m_1 + m_2)xy + m_1m_2x^2 = 0$.From the quadratic equation $5m^2 + 24m - 5 = 0$,we have $m_1 + m_2 = -\frac{24}{5}$ and $m_1m_2 = -1$.Substituting these values: $y^2 - (-\frac{24}{5})xy + (-1)x^2 = 0$.$5y^2 + 24xy - 5x^2 = 0$,which simplifies to $5x^2 - 24xy - 5y^2 = 0$.

The equation of the pair of perpendicular lines passing through the origin and forming an isosceles right-angled triangle with the line $2x + 3y = 6$ is

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