Math, asked by koral1697, 1 month ago

find the equation of straight line with the normal form p =2√2, a=5π/4

Answers

Answered by patidarmaya47
0

Answer:

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Answered by pulakmath007
1

SOLUTION

TO DETERMINE

The equation of straight line with the normal form

 \displaystyle \sf{p = 2 \sqrt{2}  \:  \:  \:,  \:  \alpha  =  \frac{5\pi}{4} }

EVALUATION

We know that the equation of straight line with the normal form is

 \sf{x \cos\alpha  + y \sin \alpha  = p  }

Now it is given that

 \displaystyle \sf{p = 2 \sqrt{2}  \:  \:  \:,  \:  \alpha  =  \frac{5\pi}{4} }

Thus we get

 \displaystyle \sf{x \cos \bigg(\frac{5\pi}{4} \bigg) +   y  \sin \bigg(\frac{5\pi}{4} \bigg) = 2 \sqrt{2} }

 \displaystyle \sf{ \implies \:  -  \frac{x}{ \sqrt{2} }  -  \frac{y}{ \sqrt{2} }  = 2 \sqrt{2} }

 \displaystyle \sf{ \implies \:   - x - y = 2 \sqrt{2} \times  \sqrt{2}  }

 \displaystyle \sf{ \implies \:   - x - y = 4}

 \displaystyle \sf{ \implies \:  x  + y  + 4 = 0}

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