What are the equations of the two lines passi | vedclass

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

Question

(A) The slope of the given line $x - \sqrt{3}y - 2\sqrt{3} = 0$ is $m = \frac{1}{\sqrt{3}}$.Let the slope of the required lines be $m'$. The angle bet

Vedclass · Accepted Answer

$x = 7, x + \sqrt{3}y = 7 + 9\sqrt{3}$

Vedclass · Answer

(A) The slope of the given line $x - \sqrt{3}y - 2\sqrt{3} = 0$ is $m = \frac{1}{\sqrt{3}}$.Let the slope of the required lines be $m'$. The angle between the lines is $	heta = 60^{\circ}$.Using the formula $	an 	heta = \left| \frac{m' - m}{1 + m'm} ight|$,we have $	an 60^{\circ} = \left| \frac{m' - 1/\sqrt{3}}{1 + m'/\sqrt{3}} ight|$.$\sqrt{3} = \left| \frac{\sqrt{3}m' - 1}{\sqrt{3} + m'} ight|$.Case $1$: $\sqrt{3} = \frac{\sqrt{3}m' - 1}{\sqrt{3} + m'}$ $\Rightarrow 3 + \sqrt{3}m' = \sqrt{3}m' - 1$ $\Rightarrow 3 = -1$ (Impossible,implies vertical line).$A$ vertical line passing through $(7, 9)$ is $x = 7$.Case $2$: $-\sqrt{3} = \frac{\sqrt{3}m' - 1}{\sqrt{3} + m'}$ $\Rightarrow -3 - \sqrt{3}m' = \sqrt{3}m' - 1$ $\Rightarrow 2\sqrt{3}m' = -2$ $\Rightarrow m' = -\frac{1}{\sqrt{3}}$.The equation of the line is $y - 9 = -\frac{1}{\sqrt{3}}(x - 7)$ $\Rightarrow \sqrt{3}y - 9\sqrt{3} = -x + 7$ $\Rightarrow x + \sqrt{3}y = 7 + 9\sqrt{3}$.Thus,the equations are $x = 7$ and $x + \sqrt{3}y = 7 + 9\sqrt{3}$.

What are the equations of the two lines passing through the point $(7, 9)$ and making an angle of $60^{\circ}$ with the line $x - \sqrt{3}y - 2\sqrt{3} = 0$?

Similar Questions

The angle between the lines whose intercepts on the axes are $a, -b$ and $b, -a$ respectively,is

The equation of a straight line passing through the point $(1, 2)$ and inclined at $45^{\circ}$ to the line $y = 2x + 1$ is

If the equation of a line having a slope $m$ $(m \in \mathbb{Z})$,passing through $(1, 1)$ and making an angle of $\tan^{-1}\left(\frac{5}{7}\right)$ with the line $x + y - 3 = 0$ is $ax + y + c = 0$,then $ac =$

Find the equation of the line passing through the point $(4, 5)$ and making equal angles with the lines $3x = 4y + 7$ and $5y = 12x + 6$.

$A$ straight line $(\sqrt{3} - 1)x = (\sqrt{3} + 1)y$ makes an angle of $75^{\circ}$ with another straight line which passes through the origin. Then the equation of the line is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module