Answer: [tex]p = -10+\sqrt{3}[/tex]
When writing this on a keyboard, we could say p = -10+sqrt(3)
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Explanation:
We're told the angle between the line and x axis is [tex]\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)[/tex]
This is one of the infinitely many exact solutions to the equation [tex]\tan(\theta) = \frac{-1}{\sqrt{3}}[/tex] where [tex]\theta[/tex] (Greek letter theta) is the angle in question.
It turns out that the tangent ratio is exactly the slope of a line.
- slope = rise/run
- tan(angle) = opposite/adjacent
Both describe a ratio of how far something moves up or down, over how far it moves to the right.
So saying [tex]\tan(\theta) = \frac{-1}{\sqrt{3}}[/tex] means that the slope of our mystery line is exactly [tex]\frac{-1}{\sqrt{3}}[/tex]
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What we'll do is compute the slope on the points [tex](\ -p, -2\sqrt{3} \ ) \text{ and } (4, -1)[/tex]
That produces some algebraic expression in terms of p. That expression is set equal to the previously mentioned slope of [tex]\frac{-1}{\sqrt{3}}[/tex] which will allow us to solve for p.
Part 1
[tex]m = \text{slope}\\\\m = \frac{y_2-y_1}{x_2-x_1}\\\\m = \frac{-1-(-2\sqrt{3})}{4-(-p)}\\\\m = \frac{-1+2\sqrt{3}}{4+p}\\\\-\frac{1}{\sqrt{3}} = \frac{-1+2\sqrt{3}}{4+p} \ \ \text{ ... plug in the previously mentioned slope}\\\\-(4+p) = (\sqrt{3})(-1+2\sqrt{3}) \ \ \text{ ... cross multiply}\\\\[/tex]
Part 2
[tex]-4-p = -\sqrt{3}+2(\sqrt{3})^2\\\\-4-p = -\sqrt{3}+2(3)\\\\-4-p = -\sqrt{3}+6\\\\-p = -\sqrt{3}+6+4\\\\-p = 10-\sqrt{3}\\\\p = -10+\sqrt{3}\\\\[/tex]
