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

x - y + 6 = 0

Step-by-step explanation:

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

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

Here, in our case and Degree,

Therefore, the normal form of the straight line equation is  

⇒

⇒ {Since, Cos (180 - Ф) = - Cos Ф and Sin (180 - Ф) = Sin Ф}

⇒

⇒ - x + y = 3√2 × √2 = 6

⇒ x - y + 6 = 0  

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