transform the following equations into normal form 1) √3x+y+10=0
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transform the following equations into normal form 1) √3x+y+10=0
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Answer:
12+36-32@$666655589+=321456
SOLUTION
TO DETERMINE
To transform the following equation √3x+y+10=0 into
normal form
CONCEPT TO BE IMPLEMENTED
General form
The general form of any line is ax + by = c
Normal form
[tex] \sf{x \cos \alpha + y \sin \alpha = p}[/tex]
Slope - Intercept form
y = mx + c
Intercept form
[tex] \displaystyle \sf{ \frac{x}{a} + \frac{y}{b} = 1}[/tex]
EVALUATION
Here the given equation of the line is
[tex] \sf{ \sqrt{3}x + y + 10 = 0 }[/tex]
Normal form
Here the given equation of the line is
[tex] \sf{ \sqrt{3}x + y + 10 = 0 }[/tex]
The given equation of the line can be rewritten as
[tex]\displaystyle \sf{ - \frac{ \sqrt{3} x}{ \sqrt{3 + 1} } - \frac{y}{ \sqrt{3 + 1} } = \frac{10}{ \sqrt{3 + 1} } }[/tex]
[tex] \displaystyle \sf{ \implies \: - \frac{ \sqrt{3} }{2 }x - \frac{y}{2} = 5 }[/tex]
[tex] \displaystyle \sf{ \implies \: x \cos \frac{7\pi}{6} + y \sin \frac{7\pi}{6} = 5 }[/tex]
Which is of the form
[tex] \sf{x \cos \alpha + y \sin \alpha = p}[/tex]
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