maths test questions

**poniekid** · April 26th 2008, 09:02 PM

I'm really stuck practicing for my test and i cant get these questions.
Please show working out.

1. Show that the triangle whose vertices are (0,1), (2,-1) and (-sqrt(3)+1,-sqrt(3)) is equilateral.

2. ImageShack - Hosting :: mathxw9.jpg
I'm sorry it is upside down

Thanks

**Moo** · April 26th 2008, 11:57 PM

Hello,

For 1), calculate the norm of the vectors joining two points... do it 3 times

If $A(x_A;y_A)$ and $B(x_B;y_B)$ , then :

$||\vec{AB}||=\underbrace{AB}_{\text{distance AB}}=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}$

This formula can be retrieved by using the Pythagorean theorem while dealing with a graph.

So let's take an example :

$A(0;1)$ and $C(-\sqrt{3}+1,-\sqrt{3})$

$\begin{array}{ccc} AC & = & \sqrt{(0-(-\sqrt{3}+1))^2+(1-(-\sqrt{3}))^2} \\ & = & \sqrt{(\sqrt{3}-1)^2+(1+\sqrt{3})^2} \\ & = & \sqrt{3-2\sqrt{3}+1+1+2\sqrt{3}+3} \\ & = & \sqrt{8} \\ & = & 2 \sqrt{2} \end{array}$

Do the same for the others

**Gusbob** · April 27th 2008, 12:05 AM

For number 2:

Use this formula:

Let acute angle between two lines = $x$
$tan x = |\frac{m_1-m_2}{1+m_1m_2}|$

where $m_1$ the gradient of one of the lines, and $m_2$ is the gradient of the other line. (1 and $\frac{1}{\sqrt3}$ in your case)

**earboth** · April 27th 2008, 12:15 AM

Originally Posted by poniekid

...

2. ImageShack - Hosting :: mathxw9.jpg
I'm sorry it is upside down

to #2:

Let a denote the angle which is included by a straight line and the x-axis. Then the slope m (or gradient) of the line is:

$\tan(a) = m$

If $\tan(a) = 1~\implies~a=45^\circ$

If $\tan(a) = \frac1{\sqrt{3}}~\implies~a=30^\circ$

I've attached your drawing (not upside down!) with the results given above.

I'll leave the rest for you.

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