Question

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

Answer 1

Answer:

x - y + 6 = 0

Step-by-step 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 $p = 3\sqrt{2}$ and $\theta = 135$ Degree,

Therefore, the normal form of the straight line equation is  

$x \cos 135 + y \sin 135 = 3\sqrt{2}$

⇒ $x \cos (180 - 45) + y \sin (180 - 45) = 3\sqrt{2}$

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

⇒$- \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 3\sqrt{2}$

⇒ - x + y = 3√2 × √2 = 6

⇒ x - y + 6 = 0  

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