The angle between the lines x - sqrt{3}y + 5 | vedclass

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

Question

(B) The given line is $x - \sqrt{3}y + 5 = 0$.Rearranging it into slope-intercept form $y = mx + c$,we get $\sqrt{3}y = x + 5$,which implies $y = \fra

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) The given line is $x - \sqrt{3}y + 5 = 0$.Rearranging it into slope-intercept form $y = mx + c$,we get $\sqrt{3}y = x + 5$,which implies $y = \frac{1}{\sqrt{3}}x + \frac{5}{\sqrt{3}}$.The slope of this line is $m_1 = \frac{1}{\sqrt{3}}$.The $y$-axis is a vertical line,so its slope $m_2$ is undefined (or we can consider the angle it makes with the $x$-axis as $90^o$).The angle $	heta$ between a line with slope $m$ and the $y$-axis is given by $	an 	heta = |\frac{1}{m}|$.Here,$	an 	heta = |\frac{1}{1/\sqrt{3}}| = \sqrt{3}$.Therefore,$	heta = 	an^{-1}(\sqrt{3}) = 60^o$.

The angle between the lines $x - \sqrt{3}y + 5 = 0$ and the $y$-axis is ....$^o$

Similar Questions

The angle between the lines $2x + 11y - 7 = 0$ and $x + 3y + 5 = 0$ is

If the lines $y = (2 + \sqrt{3})x + 4$ and $y = kx + 6$ are inclined at an angle of $60^{\circ}$ to each other,then the value of $k$ will be

The angle between the lines $2x - y - 15 = 0$ and $3x + y + 4 = 0$ is .....$^o$.

The equations of the lines passing through the origin and making an angle of $60^{\circ}$ with the line $x + y\sqrt{3} + 3\sqrt{3} = 0$ are

Find the angles between the lines $\sqrt{3}x + y = 1$ and $x + \sqrt{3}y = 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module