The combined equation of the lines passing th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:$(2x)^2 - (3y)^2 = 0$.Therefore,$4x^2 - 9y^2 = 0$.

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

Similar Questions

The equation $x^2 + kxy + y^2 - 5x - 7y + 6 = 0$ represents a pair of straight lines,then $k$ is

The joint equation of the pair of lines passing through the origin and making an angle of $\frac{\pi}{6}$ with the line $3x+y-6=0$ is

If the equation $kxy + 5x + 3y + 2 = 0$ represents a pair of lines,then $k=$

The lines represented by the equation $a{x^2}(b - c) - xy(a(b - c) + c(a - b)) + c{y^2}(a - b) = 0$ are

If one of the lines represented by $ax^2+2hxy+by^2=0$ is perpendicular to $mx+ny=18$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module