Question

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

Answer 1

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)