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Write the standard form of an equation with p=3 sqrt2, theta= 135°​

User Spasm
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1 Answer

4 votes

Answer:

x - y + 6 = 0

Explanation:

In normal form of a straight line, the equation is given by


x\cos \theta + y\sin \theta = p

where p is the perpendicular distance of the line from the origin and
\theta is the angle between the perpendicular line and the positive direction of the x-axis.

Here, in our case
p = 3√(2) and
\theta = 135 Degree,

Therefore, the normal form of the straight line equation is


x \cos 135 + y \sin 135 = 3√(2)


x \cos (180 - 45) + y \sin (180 - 45) = 3√(2)


- x \cos 45 + y \sin 45 = 3√(2) {Since, Cos (180 - Ф) = - Cos Ф and Sin (180 - Ф) = Sin Ф}


- (x)/(√(2)) + (y)/(√(2)) = 3√(2)

⇒ - x + y = 3√2 × √2 = 6

x - y + 6 = 0

So, the standard form of the equation is x - y + 6 = 0. (Answer)

User Mayhewsw
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8.8k points