# What is the length between PO, PQ, OQ have to be when the points form a equilateral triangle?

## There are 2 lines on a cartesian axis. One line is defined by Y=-2X+1 and point O and Q are an element of it. The other line is described as y=-2X+2 and point P is an element of it. I would appreciate it if you could explain it step by step with the use formula's.

Jul 18, 2018

There are 2 lines on a cartesian axis. One line is defined by Y=-2X+1 and point O and Q are an element of it. The other line is described as y=-2X+2 and point P is an element of it. I would appreciate it if you could explain it step by step with the use formula's.

Given that eqution of line-1on which the vertices O and Q of the equilateral triangle POQ lie, is
$y = - 2 x + 1 \mathmr{and} 2 x + y = 1$

It is also given that the vertex $P$ of the triangle lies on the line-2 having eqution $y = - 2 x + 2 \mathmr{and} 2 x + y = 2$ These two lines are parallel as their slopes are same.So distance between them must be height ($h$) of the equilateral $\Delta P O Q$

The normal forms of these lines are

${\text{Line}}_{1} \to \frac{2}{\sqrt{5}} x + \frac{1}{\sqrt{5}} y = \frac{1}{\sqrt{5}}$

And

${\text{Line}}_{2} \to \frac{2}{\sqrt{5}} x + \frac{1}{\sqrt{5}} y = \frac{2}{\sqrt{5}}$

Hence the perpendicular distance of these parallel straight linse or the height of the equilateral $\Delta P O Q$ will be $h = \frac{2}{\sqrt{5}} - \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}}$

If $P O = O Q = P Q = a$ then height of $\Delta P O Q$

$h = \frac{\sqrt{3}}{2} a$

So

$\frac{\sqrt{3}}{2} a = \frac{1}{\sqrt{5}}$

Hence
$P O = O Q = P Q = a = \frac{2}{\sqrt{15}}$