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The equation of a straight line which cuts off an intercept of 5 units on negative direction of yaxis and makes an angle 120° with positive direction of x-axis is​
  • a)
    y + √3x + 5 = 0
  • b)
    y − √3x + 5 = 0
  • c)
    y + √3x − 5 = 0
  • d)
    y − √3x − 5 = 0
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The equation of a straight line which cuts off an intercept of 5 units...

m = tan 120° = −√3
y+5 = −√3x⟹ y+√3x+5 = 0
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Most Upvoted Answer
The equation of a straight line which cuts off an intercept of 5 units...
To find the equation of a straight line, we need both the slope and the y-intercept.

Given that the line cuts off an intercept of 5 units on the negative y-axis, we can determine the y-intercept is (0, -5).

To find the slope, we can use the information that the line makes an angle of 120° with the positive x-axis.

The slope of the line is given by the tangent of the angle:

slope = tan(120°)

Using the tangent function, we have:

slope = tan(120°) = √3

Therefore, the slope of the line is √3.

Now, we can write the equation of the line using the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept:

y = √3x - 5

So, the equation of the straight line is y = √3x - 5.
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The equation of a straight line which cuts off an intercept of 5 units on negative direction of yaxis and makes an angle 120° with positive direction of x-axis is​a)y + √3x + 5 = 0b)y − √3x + 5 = 0c)y + √3x − 5 = 0d)y − √3x − 5 = 0Correct answer is option 'A'. Can you explain this answer?
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