Find the equation of the line which satisfy the given conditions perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.

Asked by Pragya Singh | 11 months ago |  70

##### Solution :-

Given: p = 5 and ω = 30°

We know that the equation of the line having normal distance p from the origin and angle ω which the normal makes with the positive direction of x-axis is given by x cos ω + y sin ω = p.

Substituting the values in the equation, we get

x cos30° + y sin30° = 5

x($$\dfrac{\sqrt{3}}{2}$$) + y($$\dfrac{1}{2}$$) = 5

$$\sqrt{3x}$$ + y = 5(2) = 10

$$\sqrt{3x}$$ + y – 10 = 0

The equation of the line is $$\sqrt{3x}$$ + y – 10 = 0.

Answered by Abhisek | 11 months ago

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