Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

(i) x – $$\sqrt{3}y$$ + 8 = 0

(ii) y – 2 = 0

(iii) x – y = 4

Asked by Pragya Singh | 1 year ago |  88

##### Solution :-

(i) x – $$\sqrt{3}y$$ + 8 = 0

Given:

The equation is x – $$\sqrt{3}y$$ + 8 = 0

Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.

So now, x – $$\sqrt{3}y$$ + 8 = 0

x – $$\sqrt{3}y$$ = -8

Divide both the sides by

$$\sqrt{(-1)^2 + (\sqrt{3})^2)}$$

= $$\sqrt{(1 + 3)} = \sqrt{4} = 2$$

$$\dfrac{x}{\sqrt{2}}-\dfrac{y}{\sqrt{2}}=\dfrac{4}{\sqrt{2}}$$

($$- \dfrac{1}{2}$$)x + $$\dfrac{\sqrt{3}}{2}y$$ = 4

This is in the form of: x cos 120° + y sin 120° = 4

The above equation is of the form x cos θ + y sin θ = p, where θ = 120° and p = 4.

Perpendicular distance of line from origin = 4

Angle between perpendicular and positive x – axis = 120°

(ii) y – 2 = 0

Given:

The equation is y – 2 = 0

Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.

So now, 0 × x + 1 × y = 2

Divide both sides by $$\sqrt{(0^2 + 1^2)}$$ =

$$\sqrt{1}=1$$

0 (x) + 1 (y) = 2

This is in the form of: x cos 90°+ y sin 90° = 2

The above equation is of the form x cos θ + y sin θ = p, where θ = 90° and p = 2.

Perpendicular distance of line from origin = 2

Angle between perpendicular and positive x – axis = 90°

(iii) x – y = 4

Given:

The equation is x – y + 4 = 0

Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.

So now, x – y = 4

Divide both the sides by $$\sqrt{(1^2 + 1^2)}$$

= $$\sqrt{(1+ 1)}$$$$2 \sqrt{2}$$

$$\dfrac{x}{\sqrt{2}}-\dfrac{y}{\sqrt{2}}=\dfrac{4}{\sqrt{2}}$$

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

This is in the form: x cos 315° + y sin 315° = $$2 \sqrt{2}$$

The above equation is of the form x cos θ + y sin θ = p, where θ = 315° and p = $$2 \sqrt{2}$$.

Perpendicular distance of line from origin = $$2 \sqrt{2}$$

Angle between perpendicular and positive x – axis = 315°

Answered by Abhisek | 1 year ago

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