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Question

(C) The equations of the lines passing through the origin with slopes $m_1 = \frac{2}{3}$ and $m_2 = -\frac{2}{3}$ are given by $y = m_1 x$ and $y = m

Vedclass · Accepted Answer

$4 x^2-9 y^2=0$

Vedclass · Answer

(C) The equations of the lines passing through the origin with slopes $m_1 = \frac{2}{3}$ and $m_2 = -\frac{2}{3}$ are given by $y = m_1 x$ and $y = m_2 x$.Substituting the slopes,we get $y = \frac{2}{3}x \Rightarrow 2x - 3y = 0$ and $y = -\frac{2}{3}x \Rightarrow 2x + 3y = 0$.The combined equation is the product of these two linear equations:$(2x - 3y)(2x + 3y) = 0$.Using the identity $(a-b)(a+b) = a^2 - b^2$,we get:$(2x)^2 - (3y)^2 = 0 \Rightarrow 4x^2 - 9y^2 = 0$.

The combined equation of the lines passing through the origin and having slopes $\frac{2}{3}$ and $-\frac{2}{3}$ is

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