The equation of a straight line which makes an angle of 60° with x-axis and passes through the point (√3, 2) is given bya)y = √3x + 1b)√3x- y =1c)x - √y = √3d)x-√2y=√2Correct answer is option 'B'. Can you explain this answer?

Question

The equation of a straight line which makes an angle of 60&deg; with x-axis and passes through the point (&radic;3, 2) is given bya)y = &radic;3x + 1b)&radic;3x- y =1c)x - &radic;y = &radic;3d)x-&radic;2y=&radic;2Correct answer is option 'B'. Can you explain this answer?

Answer

The equation of a straight line that makes an angle of 60 degrees with the positive x-axis can be written in the form y = mx + b, where m is the slope of the line.

To find the slope, we use the fact that the tangent of the angle between the line and the x-axis is equal to the slope of the line. Since the angle is 60 degrees, the tangent of 60 degrees is √3.

Therefore, the slope of the line is √3.

The equation of the line can now be written as y = √3x + b, where b is the y-intercept of the line.

To find the value of b, we need a point on the line. Let's assume that the line passes through the point (0, c), where c is the y-coordinate of that point.

Substituting the coordinates of the point into the equation, we get c = √3(0) + b, which simplifies to c = b.

So, the y-intercept of the line is c.

Therefore, the equation of the straight line that makes an angle of 60 degrees with the positive x-axis is y = √3x + c.

Answer

The required equation
(i) makes an angle of 60° with x-axis
(ii) passes through the point (√3 , 2)
Now
(i) ⇒ equation of the straight line will be
y = tan 60° x + c
or y = √3 x + c ...(i)
(ii) ⇒ the coordinates of the point will satisfy equation (i)

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