Eucledean Geometry uses Propositions, Theorems and Properties to establish geometrical facts in a logical manner. The French Mathematician and Philosopher Rene Descartes was successful in using Algebraic techniques in solving Geometrical Problems. This blending of Algebra and Geometry had branched out as a subject in Mathematics known as Coordinate Geometry.

It is also commonly known as Analytical Geometry implying its algebraic empowerment. The coordinate system consisting of intersecting perpendicular lines at a point called Origin. The intersecting mutually perpendicular lines which determined the plane or space are called the axes of the Coordinate System.

 Coordinate Plane         
Coordinate Geometry
In 2D Coordinate System the two axes X and Y intersect at the Origin O. A point is represented by an ordered pair (x, y). x represents the distance of the point measured horizontally while y the distance measured in the vertical direction. The origin O is represented by the ordered pair
(0, 0).
   3D Coordinate System consists of mutually perpendicular
axes X, Y and Z. A point is represented by an ordered
triplet (x, y, z) each representing the distance measured along the corresponding axes.
 

The power of Coordinate Geometry lies in the formulas used to determine geometrical lengths and measures. The algebraic aspect of coordinate Geometry can be seen clearly in applying these formulas. Let us look at some of the frequently used formulas in Coordinate Geometry.

Formulas - Cartesian Coordinate System

1. Distance Formula
Distance formula is perhaps the most frequently used formula in Coordinate Geometry. The distance PQ between two points P (x1, y1) and Q (x, y) is given by,
Distance PQ = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
The 3D counterpart of the above formula is
Distance PQ = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

2. Midpoint Formula
The coordinates of the mid point M of the line segment joining the points A (x1, y1) and B (x2, y2)
M = $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$
The corresponding midpoint formula in 3D Coordinate System is
M = $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2},\frac{z_{1}+z_{2}}{2})$

3. Slope Formula
While studying the linear equations, we have learned to compute the slope of a line, if two points on the line are known.
Slope of a line = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

3. Area of a triangle
The area of a triangle can be found by evaluating the determinant formed using the coordinates of the vertices of the triangle.
Suppose A (x1, y1) , B (x2 , y2) and C (x3 , y3) are the vertices of Δ ABC. Then,
Area of triangle ABC = $\frac{1}{2}$ $\begin{vmatrix}
x_{1} & y_{1} & 1\\
x_{2} & y_{2} & 1\\
x_{3}& y_{3} & 1
\end{vmatrix}$
The above formula can be extended to find the area of any polygon with known vertices, choosing the appropriate order of the determinant.

4. Centroid of a Triangle
The coordinates of the centroid of a triangle with vertices (x1 , y1) , (x2 , y2) and (x3 , y3) is given by
Centroid = $(\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3})$
The corresponding formula in 3D geometry is,
Centroid = $(\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3},\frac{z_{1}+z_{2}+z_{3}}{3})$

5. The distance of a point P (x, y, z) from a line ax + by +z = 0 is given by
$\frac{ax_{1}+by_{1}+cz_{1}}{\sqrt{a^{2}+b^{2}+c^{2}}}$.

Solved Examples

Question 1: A diameter of a circle has end points A (2, 5) and B(-1, 1). Find the circumference of the circle.
Solution:
 
The length of the diameter can be calculated using the distance formula.
    Diameter d = $\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$
                     = $\sqrt{(2-(-1))^{2}+(5- 1)^{2}}$
                     = $\sqrt{3^{2}+4^{2}}$
                     = $\sqrt{25}$ = 5 units
    Radius of the circle = 2.5 units
    Circumference of the circle = 2 x $\pi$ x 2.5 = 5 $\pi$ units.
 

Question 2: The vertices of a triangle are given as A (-4, 5) B (6, 1) and C (-8, -5).
Solution:
 
Verify using distance formula, the fact that ABC is a right triangle. Also identify the right angle.
    The lengths of the sides of the triangle can be calculated using distance formula.
    d = $\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$
    Length AB = $\sqrt{(6-(-4))^{2}+(1-5)^{2}}$
                    =$\sqrt{10^{2}+(-4)^{2}}$
                    = √116
    Length BC = $\sqrt{(-8-6)^{2}+(-5-1)^{2}}$
                    = $\sqrt{(-14)^{2}+(-6)^{2}}$
                    = √232
    Length CA = $\sqrt{(-4-(-8))^{2}+(5-(-5))^{2}}$
                    = $\sqrt{4^{2}+10^{2}}$
                    = √116
   Squaring the shorter lengths CA and AB and adding them we get,
   CA2 + AB2 = 116 + 116 = 232
                                      = BC2                Square of the longer length
   Hence by the converse of Pythagorean theorem Δ ABC is a right triangle.
   The shorter sides CA and AB has the common point A.
   Hence the triangle is right angled at the vertex A.