Find equation of the line through the point (0, 2) making an angle $$\dfrac{ 2π}{3}$$ 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.

Asked by Abhisek | 11 months ago |  81

##### Solution :-

Point (0, 2) and θ = $$\dfrac{2π}{3}$$

We know that m = tan θ

m = tan ($$\dfrac{2π}{3}$$) = $$- \sqrt{3}$$

We know that the point (x, y) lies on the line with slope m through the fixed point (x0, y0), if and only if, its coordinates satisfy the equation y – y0 = m (x – x0)

y – 2 = $$- \sqrt{3}$$(x – 0)

y – 2 = $$- \sqrt{3}x$$

$$-\sqrt{3}x$$ + y – 2 = 0

Given, equation of line parallel to above obtained equation crosses the y-axis at a distance of 2 units below the origin.

So, the point = (0, -2) and m = $$- \sqrt{3}$$

From point slope form equation,

y – (-2) = $$- \sqrt{3}$$ (x – 0)

y + 2 = $$- \sqrt{3}x$$

$$\sqrt{3}x$$ + y + 2 = 0

The equation of line is $$\sqrt{3}x$$+ y – 2 = 0 and

the line parallel to it is $$\sqrt{3}x$$ + y + 2 = 0.

Answered by Pragya Singh | 11 months ago

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