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Worked Examples · Example 8

Q.Reduce the equation 3x+y+2=0\sqrt{3}x + y + 2 = 0 to the normal form xcos⁡α+ysin⁡α=px\cos\alpha + y\sin\alpha = p and hence find the value of α\alpha and pp.

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Divide the line's equation by a2+b2\sqrt{a^2+b^2}, adjusting the sign so the constant on the right is positive, to read off cos⁡α\cos\alpha, sin⁡α\sin\alpha, and pp.

For a line ax+by+c=0ax+by+c=0, the normal form xcos⁡α+ysin⁡α=px\cos\alpha+y\sin\alpha=p (with p≥0p\ge0) is obtained by dividing by ±a2+b2\pm\sqrt{a^2+b^2}, choosing the sign opposite to that of cc so the right-hand side is positive.

  1. Given line. 3 x+y+2=0\sqrt3\,x + y + 2 = 0, so a=3a=\sqrt3, b=1b=1, c=2c=2.

  2. Compute the normalizing factor a2+b2\sqrt{a^2+b^2}:

(3)2+12=3+1=4=2\sqrt{(\sqrt3)^2+1^2} = \sqrt{3+1} = \sqrt4 = 2

  1. Choose the sign. Since c=+2>0c=+2>0, divide by −a2+b2=−2-\sqrt{a^2+b^2}=-2 (opposite sign to cc) so that the constant term moves to the right-hand side as a positive pp:

3 x−2+y−2+2−2=0   ⟹   −32x−12y−1=0\frac{\sqrt3\,x}{-2} + \frac{y}{-2} + \frac{2}{-2} = 0 \ \implies\ -\frac{\sqrt3}{2}x - \frac12 y - 1 = 0

  1. Rearrange to isolate the constant on the right.

−32x−12y=1-\frac{\sqrt3}{2}x - \frac12 y = 1

This is now in the form xcos⁡α+ysin⁡α=px\cos\alpha + y\sin\alpha = p with

cos⁡α=−32,sin⁡α=−12,p=1\cos\alpha = -\frac{\sqrt3}{2}, \qquad \sin\alpha = -\frac12, \qquad p = 1 …

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