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Q.Transform the equation 3x+y+10=0\sqrt{3}x + y + 10 = 0 into

(i) slope-intercept form
(ii) normal form.
Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2023Subjective· 2mImportance★★★★★
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Isolate yy for slope-intercept form; divide by a2+b2\sqrt{a^2+b^2} (with sign chosen so the constant is positive) for normal form.

  1. Slope-intercept form: 3x+y+10=0⇒y=−3x−10\sqrt3x+y+10=0 \Rightarrow y=-\sqrt3x-10. Here slope m=−3m=-\sqrt3, yy-intercept =−10=-10.
  2. Normal form: we need xcos⁡α+ysin⁡α=px\cos\alpha+y\sin\alpha=p with p≥0p\ge0. Multiply the given line by −1-1: −3x−y−10=0⇒−3x−y=10-\sqrt3x-y-10=0 \Rightarrow -\sqrt3x-y=10. Divide by (3)2+12=4=2\sqrt{(\sqrt3)^2+1^2}=\sqrt{4}=2: −32x−12y=5-\dfrac{\sqrt3}{2}x-\dfrac12y=5 …

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