Question

The coordinates of two consecutive vertices A and B of a regular hexagon ABCDEF are (1,0) and (2,0)respectively.The equation of the diagonal CE is

Answer 1

using trigonometry get coordinates of C i.e (2.5,(squrt(3)/2))
and coordinate of E is (1,square root 3)
by 2 point form of straight line get its equation as y-(sqrt 3/2)=(x-1)*(-(1/sqrt 3))

Answer 2

The equation of the diagonal CE is $x+\sqrt{3} y=4$

Given:

Hexagon ABCDEF

It’s diagonals are, A(1,0) and B(2,0)

To find:

The equation of the diagonal CE  

Solution:

From the above figure,

$C=\left(2+1 . \cos 60^{\circ}, 1 . \sin 60^{\circ}\right)=\left(\frac{5}{2}, \frac{\sqrt{3}}{2}\right)$

$E=\left(1.1 . \sin 60^{\circ}, 1 . \sin 60^{\circ}\right)=(1, \sqrt{3})$

The equation of the diagonal CE is,

$y-\sqrt{3}=\frac{\left(\sqrt{3}-\frac{\sqrt{3}}{2}\right)}{\left(1-\frac{5}{2}\right)} \cdot(x-1)$

$y-\sqrt{3}=\frac{\left(\frac{2 \sqrt{3}-\sqrt{3}}{2}\right)}{-\frac{3}{2}} \cdot(x-1)$

$y-\sqrt{3}=\frac{\sqrt{3}}{-3} \cdot(x-1)$

$y-\sqrt{3}=\frac{1}{\sqrt{3}}(1-x)$

$\sqrt{3} y-3=(1-x)$

$x+\sqrt{3} y=4$

The equation of the diagonal CE is,

$x+\sqrt{3} y=4$