#################################
Coordinate Geometry: The Line
#################################

Revision
=====================

In your previous course you will have coved some of the basics of the line such as

* Plot a point such as :math:`p(3,2)`
* Find the distance between two points
* Find the midpoint of a line
* Find the distance between two points

Distance between 2 points
^^^^^^^^^^^^^^^^^^^^^^^^^^
The formula for the distance between two points :math:`p(x1, y1)` and :math:`p(x2, y2)` is:

Length of :math:`\left [ ab \right ]` is :math:`\sqrt{(x_{2} -x_{1})^2 +(y_{2}-y_{1})^2}`


This is because the horizontal distance of the line is given as :math:`x2-x1` while the vertical distance of the line is :math:`(y2-y1)`.
Pythagorus tells us that the length of the hypotenuse of a right angled triangle is:

.. math::
    (hypotenuse)^2 = (base)^2 + (height)^2
    
    hypotenuse = \sqrt{(base)^2 + (height)^2}
    

Midpoint of a Line Segment
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

The midpoint, :math:`m`, of a line segment joining :math:`a\left(x_{1},y_{1}\right)` and :math:`b\left(x_{2},y_{2}\right)` is

.. math::

    \left(\frac{x_{1} +x_{2}}{2},\frac{y_{1} +y_{2}}{2}\right)




Example
^^^^^^^^^^^^^^^^^^^
If :math:`a\left(-1,3\right)` and :math:`b\left(5,7\right)` are 2 points in the plane, find (i) :math:`\mid ab \mid` and (ii) the midpoint of :math:`\left[ab\right]`.

Solution
^^^^^^^^^^^^^^^^
:math:`(i)` The points are :math:`a\left(-1,3\right)` which corresponds to :math:`a\left(x_{1},y_{1}\right)` and :math:`b\left(5,7\right)` which corresponds to :math:`b\left(x_{2},y_{2}\right)`

.. math::
   :nowrap:

	x_{1} = -1 \\y_{1} = 3\\x_{2} = 5\\y_{2} = 7 

.. math::
   :nowrap:
   
   \begin{eqnarray}
	\mid ab \mid  & = & \sqrt{(x_{2} -x_{1})^2 +(y_{2}-y_{1})^2}  \\
	              & = & \sqrt{(5-(-1))^2 +(7-3)^2} \\
	              & = & \sqrt{36 +16} \\
	              & = & \sqrt{52} 
	\end{eqnarray}
	
:math:`(ii)` Midpoint of :math:`\left[ab\right]` :

.. math::
	 & = & \left(\frac{x_{1} +x_{2}}{2},\frac{y_{1} +y_{2}}{2}\right) \\
	 & = & \left(\frac{-1  + 5 }{2},\frac{3 + 7}{2}\right)\\
	 & = & (2,5)
	 
	 
	
	
	


Extras
^^^^^^^^^^^
.. math::

   (a + b)^2 = a^2 + 2ab + b^2
   
   (a - b)^2 = a^2 - 2ab + b^2
   
   
   
   x^2=y^2
   

.. math::

   \cos (2\theta) = \cos^2 \theta - \sin^2 \theta
   
   
.. math::

   (a + b)^2  &=  (a + b)(a + b) \\
              &=  a^2 + 2ab + b^2

.. math::
   :nowrap:

   \begin{eqnarray}
      y    & = & ax^2 + bx + c \\
      f(x) & = & x^2 + 2xy + y^2
   \end{eqnarray}


.. math::
   :nowrap:

   \begin{eqnarray}
      \sqrt{   (k-0)^2  +( -2k + 2)^2} &=& \sqrt{(k +2)^2 +(-2k +6)^2} \\
      \Rightarrow k^2 + 4k^2 -8k +4 &=& k^2 +4k +4 +4k^2 -24k +36 \\
      \Rightarrow -8k +4 &=& 40 -20k \\
      \Rightarrow 12k &=& 36 \\
      \Rightarrow k &=& 3 \\
   \end{eqnarray}



.. math:: e^{i\pi} + 1 = 0
   :label: euler