Find the equation of the line which satisfy the given conditions passing through ,$$(2,2\sqrt{3})$$) and inclined with the x-axis at an angle of 75°.

Asked by Pragya Singh | 1 year ago |  110

##### Solution :-

Given: point (2, $$2\sqrt{3}$$) and θ = 75°

Equation of line: (y – y1) = m (x – x1)

where, m = slope of line = tan θ and (x1, y1) are the points through which line passes

m = tan 75°

75° = 45° + 30°

Applying the formula:

$$\dfrac{tan 45°+tan 30°}{1-tan 45°.tan 30°}$$

Substitute the values,

$$1+\dfrac{1}{\sqrt{3}}.1-\dfrac{1}{\sqrt{3}}$$

tan 45° = $$\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$$

Let us rationalizing we get,

$$\dfrac{3+1+2\sqrt{3}}{3-1}$$

$$2+\sqrt{3}$$

We know that the point (x, y) lies on the line with slope m through the fixed point (x1, y1), if and only if, its coordinates satisfy the equation y – y1 = m (x – x1)

Then, y –$$2\sqrt{3}$$ = ( $$2+\sqrt{3}$$) (x – 2)

y – $$2\sqrt{3}$$ = 2 x – 4 + $$\sqrt{3}x-$$$$2\sqrt{3}$$

y = 2 x – 4 +$$\sqrt{3}x$$

$$2+\sqrt{3}$$) x – y – 4 = 0

The equation of the line is ($$2+\sqrt{3}$$) x – y – 4 = 0.

Answered by Abhisek | 1 year ago

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