Question

) Find the angle which the line √3x + y + 5 = 0 makes with the positive directi
x-axis.​

Answer 1

Given Equation,

$\sf \: \sqrt{3} x + y + 5 = 0 \\ \\ \dashrightarrow \sf \: y = - \sqrt{3} x - 5 - - - - - (1)$

Slope Intercept form of a line,

$\sf y = mx + c - - - - - - - - - (2)$

On comparing (1) and (2),

$\sf \: m = - \sqrt{3} \\ \\ \dashrightarrow \sf \: tan \theta = - \sqrt{3} \\ \\ \dashrightarrow \sf \: tan \theta = - tan \dfrac{\pi}{3} \\ \\ \: \dashrightarrow \sf \: tan \theta = - tan \dfrac{\pi}{3} \\ \\ \dashrightarrow \sf \: tan \theta = tan(\pi - \dfrac{\pi}{3} ) \\ \\ \dashrightarrow \boxed{ \boxed{\sf \: \theta = \dfrac{2\pi}{3} }}$

Answer 2

Given:

• Line $\rm{ \sqrt{3} x + y + 5 = 0}$ makes with the positive directive X-Axis.

To Find:

• Value of angle $\rm{ \theta}$ = ?

Solution:

We have the given equation,

$\dashrightarrow\bf{ \sqrt{3} x + y + 5 = 0} \\ \\ \dashrightarrow\bf{y = - \sqrt{3x} - 5 \: .... \: (1)}$

We know that the Slope Intercept form of a line is given by,

$\dashrightarrow\bf{y = mx + c \: .... \: (2)}$

From (1) & (2),

$\dashrightarrow\bf{m = \sqrt{3} } \\ \\ \dashrightarrow\bf{tan \theta = - \sqrt{3} } \\ \\ \dashrightarrow\bf{tan \theta = - \: tan \: \frac{\pi}{3} } \\ \\ \dashrightarrow\bf{tan \theta = tan \: (\pi - \frac{\pi}{3} )} \\ \\ \dashrightarrow{\boxed{\sf{\red{ \theta = \frac{2\pi}{3} }}}} \: \bigstar$