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Q.Express the equation 3x+4y=123x + 4y = 12 in

(i) slope form
(ii) normal form and
(iii) intercept form.
Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2018Subjective· 4mImportance★★★★★
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Manipulate 3x+4y=123x+4y=12 algebraically into each of the three standard forms.

Given line: 3x+4y=123x + 4y = 12

  1. Slope form y=mx+cy = mx+c: solve for yy. 4y=−3x+12⇒y=−34x+34y = -3x + 12 \Rightarrow y = -\dfrac{3}{4}x + 3 So slope m=−34m = -\dfrac34 and y-intercept c=3c=3.
  2. Intercept form xa+yb=1\dfrac{x}{a}+\dfrac{y}{b}=1: divide both sides by 12. 3x12+4y12=1⇒x4+y3=1\dfrac{3x}{12} + \dfrac{4y}{12} = 1 \Rightarrow \dfrac{x}{4} + \dfrac{y}{3} = 1 So x-intercept a=4a=4, y-intercept b=3b=3.
  3. Normal form xcos⁡ω+ysin⁡ω=px\cos\omega + y\sin\omega = p: divide by 32+42=25=5\sqrt{3^2+4^2}=\sqrt{25}=5 (choosing the sign so p>0p>0, already satisfied here since the right side is positive). …

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