$\displaystyle y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})$
any ideas, just confused with the gradient
Solve for $\displaystyle y$ or $\displaystyle x?$ This is a recent question that belongs to other thread, so I'll assume that you want to isolate $\displaystyle y:$
$\displaystyle y-2=2\sqrt{3}\left( x-\frac{\pi }{3} \right)\implies y=2\sqrt{3}x-\frac{2\sqrt{3}}{3}\pi +2=2\sqrt{3}x-\frac{2}{\sqrt{3}}\pi +2.$
Does that make sense?
An equation of a line is in the form $\displaystyle y = mx + c$
Where:
$\displaystyle y - 2 = 2\sqrt{3}\left(x-\frac{\pi}{3}\right)$
- $\displaystyle y, x$ = Variables
- $\displaystyle m$ = Gradient
- $\displaystyle c$ = $\displaystyle y$-intercept
$\displaystyle y-2 = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3}$
$\displaystyle y= 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + 2$
$\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}$
^ Compare this form with $\displaystyle y = mx + c$ to work out what your gradiant $\displaystyle \left(m\right)$ is.
$\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}$
Is...
$\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + \frac{6}{3}$
Which is...
$\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + 2$
By keeping it as $\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}$, I though it would be easier for you to compare the form with $\displaystyle y=mx+c$.