What is the maximum distance from point a (1, - 3) to the line xsin β + ycos β = 2?
From the distance formula, d = | 1 * sin β + (- √ 3) cos β | / √ (1 & sup2; + (- √ 3) & sup2;) = | sin β - √ 3cos β - 2 | / 2, and
Sin β - √ 3cos β = 2Sin (β - π / 3), so the maximum value of D is 0.5 * | - 2-2 | = 2
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