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Worked Examples · Example 7

Q.Find the equation of a line whose perpendicular distance from the origin is 2 units and the angle between the perpendicular segment and the positive direction of the x-axis is 240∘240^\circ.

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Use the normal form of a line, xcos⁡α+ysin⁡α=px\cos\alpha+y\sin\alpha=p, with the given perpendicular distance p=2p=2 and inclination α=240∘\alpha=240^\circ.

Normal form of a line:

xcos⁡α+ysin⁡α=px\cos\alpha + y\sin\alpha = p

where p>0p>0 is the perpendicular distance from the origin to the line, and α\alpha is the angle the perpendicular (normal) makes with the positive xx-axis.

  1. Given values.

p=2,α=240∘p = 2, \qquad \alpha = 240^\circ

  1. Find cos⁡240∘\cos 240^\circ and sin⁡240∘\sin 240^\circ. 240∘=180∘+60∘240^\circ = 180^\circ+60^\circ lies in the third quadrant, where both cosine and sine are negative:

cos⁡240∘=−cos⁡60∘=−12,sin⁡240∘=−sin⁡60∘=−32\cos 240^\circ = -\cos 60^\circ = -\frac{1}{2}, \qquad \sin 240^\circ = -\sin 60^\circ = -\frac{\sqrt3}{2}

  1. Substitute into the normal form.

x(−12)+y(−32)=2x\left(-\frac12\right) + y\left(-\frac{\sqrt3}{2}\right) = 2

  1. Multiply both sides by −2-2 to clear fractions and simplify signs. …

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