The joint equation of two lines passing throu | vedclass

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

Question

(C) The lines pass through the origin and make an angle of $30^{\circ}$ with the $Y$-axis. The angles these lines make with the positive $X$-axis are

Vedclass · Accepted Answer

$3x^{2}-y^{2}=0$

Vedclass · Answer

(C) The lines pass through the origin and make an angle of $30^{\circ}$ with the $Y$-axis. The angles these lines make with the positive $X$-axis are $90^{\circ}-30^{\circ}=60^{\circ}$ and $90^{\circ}+30^{\circ}=120^{\circ}$. The slopes of the lines are $m_{1}=	an 60^{\circ}=\sqrt{3}$ and $m_{2}=	an 120^{\circ}=-\sqrt{3}$. The equations of the lines are $y=\sqrt{3}x$ and $y=-\sqrt{3}x$,which can be written as $\sqrt{3}x-y=0$ and $\sqrt{3}x+y=0$. The joint equation is $(\sqrt{3}x-y)(\sqrt{3}x+y)=0$. Expanding this,we get $(\sqrt{3}x)^{2}-y^{2}=0$,which simplifies to $3x^{2}-y^{2}=0$.

The joint equation of two lines passing through the origin,each making an angle of $30^{\circ}$ with the $Y$-axis,is

Similar Questions

$\sim (p \vee q)$ is equal to

Which of the following is a copper-releasing $IUD$?

An electron beam travels with a velocity of $1.6 \times 10^7 \ m/s$ perpendicularly to a magnetic field of intensity $0.1 \ T$. Calculate the radius of the path of the electron beam. (Given: $m_e = 9 \times 10^{-31} \ kg$,$e = 1.6 \times 10^{-19} \ C$)

The logically equivalent proposition of $p \Leftrightarrow q$ is

The refractive index of the material of a double convex lens is $1.5$ and its focal length is $5 ~cm$. If the radii of curvature are equal,the value of the radius of curvature (in $cm$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module