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Equation of line 3x+y8=0sqrt{3} x+y-8=0 can be represented in normal form as

a

3xy8=0sqrt{3} x-y-8=0

b

3x2y282=0frac{sqrt{3} x}{2}-frac{y}{2}-frac{8}{2}=0

c

3x2+y282=0frac{sqrt{3} x}{2}+frac{y}{2}-frac{8}{2}=0

d

3x+y4=0sqrt{3} x+y-4=0

Answer : Option C
Explanation :

Equation of line in normal form can be written as

axa2+b2+bya2+b2+ca2+b2=0frac{a x}{sqrt{a^{2}+b^{2}}}+frac{b y}{sqrt{a^{2}+b^{2}}}+frac{c}{sqrt{a^{2}+b^{2}}}=0

3x32+12+y32+11832+12=0Rightarrow frac{sqrt{3} x}{sqrt{3}^{2}+1^{2}}+frac{y}{sqrt{sqrt{3}^{2}+1^{1}}}-frac{8}{sqrt{sqrt{3}^{2}+1^{2}}}=0

3x4+y484=0frac{sqrt{3} x}{sqrt{4}}+frac{y}{sqrt{4}}-frac{8}{sqrt{4}}=0

3x2+y282=0Rightarrow frac{sqrt{3} x}{2}+frac{y}{2}-frac{8}{2}=0

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