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:$(2x)^2 - (3y)^2 = 0 \Rightarrow 4x^2 - 9y^2 = 0$.

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

Similar Questions

If $|a|=1, |b|=2$ and the angle between $a$ and $b$ is $120^{\circ}$,then ${(a+3b) \times (3a-b)}^2$ is equal to

The equation of the locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$,where $z=x+iy$ is a complex number,is

The parabola with directrix $x+2y-1=0$ and focus $(1,0)$ is

Catadromous fish migrates from

Compound $A$ $(C_3H_6O)$ undergoes the following reactions to form $B$ and $C$. Identify $A$,$B$,and $C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module