The joint equation of the lines passing throu | vedclass

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

Question

(C) The first quadrant is the region between the positive $x$-axis $(	heta = 0)$ and the positive $y$-axis $(	heta = \frac{\pi}{2})$.Since the lines

Vedclass · Accepted Answer

$\sqrt{3}x^2-4xy+\sqrt{3}y^2=0$

Vedclass · Answer

(C) The first quadrant is the region between the positive $x$-axis $(	heta = 0)$ and the positive $y$-axis $(	heta = \frac{\pi}{2})$.Since the lines trisect the first quadrant,they divide the $90^\circ$ angle into three equal parts of $30^\circ$ each.Thus,the angles made by the lines with the positive $x$-axis are $	heta_1 = 30^\circ = \frac{\pi}{6}$ and $	heta_2 = 60^\circ = \frac{\pi}{3}$.The equation of the first line is $y = 	an(\frac{\pi}{6})x$ $\Rightarrow y = \frac{1}{\sqrt{3}}x$ $\Rightarrow x - \sqrt{3}y = 0$.The equation of the second line is $y = 	an(\frac{\pi}{3})x$ $\Rightarrow y = \sqrt{3}x$ $\Rightarrow \sqrt{3}x - y = 0$.The joint equation is $(x - \sqrt{3}y)(\sqrt{3}x - y) = 0$.Expanding this,we get $\sqrt{3}x^2 - xy - 3xy + \sqrt{3}y^2 = 0$.$\Rightarrow \sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$.

The joint equation of the lines passing through the origin and trisecting the first quadrant is . . . . . .

Similar Questions

$A$ tension of $20 ~N$ is applied to a copper wire of cross-sectional area $0.01 ~cm^2$. The Young's modulus of copper is $1.1 \times 10^{11} ~N/m^2$ and the Poisson's ratio is $0.32$. The decrease in the cross-sectional area of the wire is:

If the pair of lines $2x^2 + 3xy + y^2 = 0$ makes angles $\theta_1$ and $\theta_2$ with the positive direction of the $X$-axis,then $|\tan(\theta_1 - \theta_2)| = $

The area (in sq units) of the triangle formed by the tangent at $(\sqrt{3}, 0)$ to the hyperbola $x^2-3y^2=3$ with the pair of asymptotes of the hyperbola is

If $\frac{1}{2} \leq x \leq 1$,then $\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right)$ is equal to

$\int_0^2 \frac{2x-2}{2x-x^2} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module