MATH SOLVE

2 months ago

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

Accepted Solution

A:

Answer:x - y + 6 = 0 Step-by-step explanation:In normal form of a straight line, the equation is given by [tex]x\cos \theta + y\sin \theta = p[/tex]
where p is the perpendicular distance of the line from the origin and [tex]\theta[/tex] is the angle between the perpendicular line and the positive direction of the x-axis.
Here, in our case [tex]p = 3\sqrt{2}[/tex] and [tex]\theta = 135[/tex] Degree,
Therefore, the normal form of the straight line equation is [tex]x \cos 135 + y \sin 135 = 3\sqrt{2}[/tex]
⇒ [tex]x \cos (180 - 45) + y \sin (180 - 45) = 3\sqrt{2}[/tex]
⇒ [tex]- x \cos 45 + y \sin 45 = 3\sqrt{2}[/tex] {Since, Cos (180 - Ф) = - Cos Ф and Sin (180 - Ф) = Sin Ф}
⇒[tex]- \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 3\sqrt{2}[/tex]
⇒ - x + y = 3√2 × √2 = 6
⇒ x - y + 6 = 0 So, the standard form of the equation is x - y + 6 = 0. (Answer)