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

During an adiabatic process,the pressure of a gas is proportional to the cube of its temperature. The value of $C_p / C_V$ for that gas is

When a non-volatile solute is added to a solvent,the vapor pressure of the solvent decreases by $10 \ mm \ Hg$. The mole fraction of the solute in the solution is $0.2$. If the vapor pressure is to be decreased by $20 \ mm \ Hg$,what will be the mole fraction of the solvent?

$x A_{(s)} \rightleftharpoons y B_{(g)} + z C_{(g)}$. If $\frac{K_c}{K_p} = (RT)^{-2}$,then which is correct?

In the reaction,$C_2H_5OC_2H_5 + CO \xrightarrow[150^{\circ}C, 500 \text{ atm pressure}]{BF_3} X$. What is $X$?

If $e^{f(x)}=\frac{10+x}{10-x}, x \in(-10,10)$ and $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$,then $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module