Question

A straight line has y -intercept -5. If it makes $120^{\circ}$ with the x-axis, then the equation of the line is

• A. $\sqrt{3}x+y+20=0$
• B. $\sqrt{3}x+y+10=0$
• C. $\sqrt{3}x-y+10=0$
• D. $\sqrt{3}x+y-10=0$
• E. $\sqrt{3}x+y+5=0$

Hint:

Math Tip: Remember your common trigonometric values for obtuse angles. Since an angle of $120^{\circ}$ points into the second quadrant, its slope (tangent) will always be negative.

Answer 1

Answer: (E)

Concept:
Coordinate Geometry - Equation of a Straight Line (Slope-Intercept Form).
Step 1: Identify the given parameters.
The y-intercept ($c$) is given as $-5$. So, $c = -5$.
The angle of inclination ($\theta$) with the positive x-axis is $120^{\circ}$.
Step 2: Calculate the slope (m) of the line.
The slope $m$ is the tangent of the angle of inclination: $$ m = \tan(\theta) = \tan(120^{\circ}) $$ Use the supplementary angle identity $\tan(180^{\circ} - \alpha) = -\tan(\alpha)$: $$ m = \tan(180^{\circ} - 60^{\circ}) $$ $$ m = -\tan(60^{\circ}) $$ $$ m = -\sqrt{3} $$
Step 3: Use the slope-intercept form.
The standard slope-intercept form for the equation of a line is: $$ y = mx + c $$
Step 4: Substitute the values and rearrange.
Substitute $m = -\sqrt{3}$ and $c = -5$ into the equation: $$ y = -\sqrt{3}x - 5 $$ Bring all terms to one side to match the standard general form $Ax + By + C = 0$: $$ \sqrt{3}x + y + 5 = 0 $$