$\displaystyle y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})$

any ideas, just confused with the gradient

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- Sep 8th 2008, 07:50 AMjvignaciosolving this equation.
$\displaystyle y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})$

any ideas, just confused with the gradient - Sep 8th 2008, 07:56 AMKrizalid
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? - Sep 8th 2008, 07:56 AMSimplicity
An equation of a line is in the form $\displaystyle y = mx + c$

Where:

- $\displaystyle y, x$ = 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. - Sep 8th 2008, 08:04 AMjvignacio
- Sep 8th 2008, 08:06 AMjvignacio
- Sep 8th 2008, 08:13 AMSimplicity
$\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}$

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 $\displaystyle y=mx+c$. - Sep 9th 2008, 03:41 AMjvignacio