The equation of a straight line which cuts off an intercept of $5$ units on negative direction of y-axis and makes an angle $120^{\circ}$ with positive direction of x-axis is

y + \sqrt{3} x + 5 = 0

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y - \sqrt{3} x + 5 = 0

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y + \sqrt{3} x - 5 = 0

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y - \sqrt{3} x - 5 = 0

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Solution

The correct option is A $y + \sqrt{3} x + 5 = 0$
Equation of line
$y = m x + c$
$m = S l o p e$
$c = I n t e r c e p t$
$m = t a n 120 = - \sqrt{3}$
$c = - 5$ (given)
Putting value of $m$ and $c$
we get $y = - \sqrt{3} x - 5$
or $y + \sqrt{3} + 5 = 0$
So correct answer will be option A

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The equation of a straight line which cuts off an intercept of 5 units on negative direction of y-axis and makes an angle 120∘ with positive direction of x-axis is

The correct option is A y+√3x+5=0Equation of line y=mx+cm=Slopec=Interceptm=tan120=−√3c=−5 (given)Putting value of m and cwe get y=−√3x−5or y+√3+5=0So correct answer will be option A

The equation of a straight line which cuts off an intercept of $5$ units on negative direction of y-axis and makes an angle $120^{\circ}$ with positive direction of x-axis is

The correct option is A $y + \sqrt{3} x + 5 = 0$
Equation of line
$y = m x + c$
$m = S l o p e$
$c = I n t e r c e p t$
$m = t a n 120 = - \sqrt{3}$
$c = - 5$ (given)
Putting value of $m$ and $c$
we get $y = - \sqrt{3} x - 5$
or $y + \sqrt{3} + 5 = 0$
So correct answer will be option A