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# Find equation of the line through the point (0, 2) making an angle 2π/3 with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin

**Solution:**

The slope of the line making an angle 2π/3 with the positive x-axis is m = tan (2π/3) = - √3

Now, the equation of the line passing through points (0, 2) and having a slope - √3 is

( y - 2) = - √3 ( x - 0)

√3x + y - 2 = 0

The slope of line parallel to line √3x + y - 2 = 0 is √3.

It is given that the line parallel to line √3x + y - 2 = 0 crosses the y-axis 2 units below the origin, i.e., it passes through point (0, 2).

Hence, the equation of the line passing through points (0, 2) and having a slope - √3 is

y - (-2) = - √3 (x - 0)

y + 2 = - √3

√3x + y + 2 = 0

Thus, the equation of the line is √3x + y + 2 = 0.

Now we will find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin

The equation of the line parallel to √3x + y + 2 = 0 is of the form √3x + y + k = 0 ... (1)

Since this line crosses the y-axis at a distance of 2 units below the origin, its y-intercept is (0, 2). Substituting x = 0 and y = 2 in the above equation,

√3(0) + 2 + k = 0

k = -2

Substituting this in (1),

√3x + y - 2 = 0.

Thus, the equations of the required lines are √3x + y + 2 = 0 and √3x + y - 2 = 0

NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.2 Question 14

## Find equation of the line through the point (0, 2) making an angle 2π/3 with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin

**Summary:**

The equation of the line through the point (0, 2) making an angle 2π/3 with the positive x-axis is √3x + y + 2 = 0. The equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin is √3x + y - 2 = 0

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