trig identities

[esmith]

i really need help with this problem--i'm sooo lost that i don't even know what it is asking
(and don't worry, i plan on contributing to society through other means than math)

Write the equation 2y-3x=6 in normal form. Then, find the length of the normal and the angle it makes with the positive x-axis. Explain how you determined the angle.

 

[masters]

i really need help with this problem--i'm sooo lost that i don't even know what it is asking
(and don't worry, i plan on contributing to society through other means than math)

Write the equation 2y-3x=6 in normal form. Then, find the length of the normal and the angle it makes with the positive x-axis. Explain how you determined the angle.

Hi esmith,

The standard form of a linear equation, Ax+By+C=0,
can be changed to normal form by dividing each term by \(\displaystyle \pm \sqrt{A^2+B^2}\).

The sign is chosen opposite the sign of C.

When in Normal form, you can find the length of the Normal directly from the equation.
You can find the angle by using the relation \(\displaystyle \tan \theta=\frac{\sin \theta}{\cos \theta}\)

When in standard form, the coefficient of x is equal to \(\displaystyle \cos \theta\)
and the coefficient of y is equal to \(\displaystyle \sin \theta\)
The quadrant can be found by checking the signs of the sine and cosine.

Normal Form: \(\displaystyle x \cos \theta + y \sin \theta - p=0\), where p is the length of the Normal and \(\displaystyle \theta\) is the angle between the Normal line and the x-axis.

\(\displaystyle -3x+2y-6=0\)

\(\displaystyle \pm \sqrt{A^2 + B^2}=\pm \sqrt{(-3)^2+(2)^2}=\pm \sqrt{13}=\sqrt{13}\)

\(\displaystyle \frac{-3}{\sqrt{13}}x+\frac{2}{\sqrt{13}}y-\frac{6}{\sqrt{13}}=0\)

Normal Form: \(\displaystyle -\frac{3\sqrt{13}}{13}x+\frac{2\sqrt{13}}{13}y-\frac{6\sqrt{13}}{13}=0\)

Length of the Normal: \(\displaystyle \boxed{p = \frac{6\sqrt{13}}{13}}\)

\(\displaystyle \cos \theta=-\frac{3\sqrt{13}}{13} \:\:\:\sin \theta=\frac{2\sqrt{13}}{13}\)

\(\displaystyle \tan \theta=-\frac{2}{3}\)

\(\displaystyle \tan^{-1}\left(-\frac{2}{3}\right)=-33.7^{\circ}\)

Cosine is negative, Sine is positive, angle is in quadrant II

\(\displaystyle \boxed{\theta=-33.7^{\circ}+180^{\circ}=146.3^{\circ}}\)

 

[esmith]

thank you so much!!! that was quite informative!