Analytical Geometry is the branch of Mathematics which makes use of algebraic methods to evaluate measures and lengths to solve geometrical problems. Euclidean Geometry defines concept, approaches problems theoretically and proves properties, postulates and theorems using logical statements. Empowered with the algebraic techniques of simplification and evaluation, the geometrical problems are solved with relative ease in analytical geometry.
Analytical geometry is often referred to as Coordinate Geometry, as it primarily uses coordinates.
Starting from Rene Descarte’s classical Cartesian system of coordinates, other systems like Polar, Cylindrical and spherical systems of coordinates are widely used in solving many mathematical problems.

Formulas derived for different systems of coordinates bring in the notable ease in solving problems in analytical geometry. All problems are solved making use of at least one of these formulas. We have listed here some of the commonly used formulas, which are indispensible in understanding and mastering the subject.

Distance Formula:
The distance between two points P (x1, y1) and Q (x2, y2) on a coordinate plane is given by
Distance PQ = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Distance between two points P ($x_1, y_1, z_1$) and Q ($x_2, y_2, z_2$) in space is given by
Distance PQ = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}$

Midpoint Formula:
The midpoint M of the line segment joining the two points A (x1, y1 ) and B (x2, y2)
M = $ (\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$

Centroid Formula:
The centroid G of a triangle with vertices (x1, y1) , (x2, y2) and (x3 , y3) is,
G = $ (\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3})$

Area of a Triangle Formula:
Area of a triangle with vertices (x1, y1) , (x2, y2) and (x3 , y3) is,
Area of $\triangle$ = $\frac{1}{2}$ |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

This can be written using determinant as, Area of triangle = $\frac{1}{2}$ $\begin{vmatrix}
x_{1} & y_{1} &1\\
x_{2} & y_{2} & 1\\
x_{3} & y_{3} & 1
\end{vmatrix}$
The conic sections are important two dimensional curves discussed in analytical geometry. A conic is defined to be a curve traced by a point which moves in such a way, that its distance from a fixed point bears a constant ratio to its distance from a fixed line. The fixed point is called the Focus and the fixed line is called the Directrix. The constant ratio is known as the eccentricity of the conic and is denoted by 'e'.

A parabola is traced when e = 1. This means any point on a parabola is equidistant from the Directirx and the Focus.

You have already learned in algebra, the graph of a quadratic function is a parabola with a vertical axis of symmetry. In analytical geometry, the general form of equations for parabolas with vertical and horizontal orientations are derived, using the coordinates of the vertex and the distance between the vertex and focus as parameters. The general equations and the characteristics of both types of parabola are given below along with a general sketch.

Horizontal Parabola
General Equation: (y - k)2 = 4p(x - h)
Vertex: ( h, k)
Focus: ( h + p, k)
Axis of symmetry: y = k
Equation to Directrix: x = h - p
The parabola opens right, when p is positive
and opens left, if p is negative

          
Analytical Geometry Parabola
Vertical Parabola
General Equation: (x - h)2 = 4p(y - k)
Vertex: (h, k)
Focus: (h, k + p)
Axis of symmetry: x = h
Equation to Directrix: y = k - p
The parabola opens up, when p is positive
and opens down, when p is negative.
  Analytical Geometry Parabolas

Given below are some of the problems on analytical geometry.

Solved Example

Question: The coordinates of triangle ABC are given as A (-4, 1), B (4, 4) and C (0,6).
  1. Find the distance AB and the length of the altitude on AB. Also, calculate the area of the triangle ABC.
  2. Verify the answer got in (1), by finding the area of the triangle using area formula.
 
     Analytical Geometry Problems
Solution:
The slope of line AB = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ = $\frac{4 - 1}{4 - (-4)}$   = $\frac{3}{8}$
Equation to line AB,
y - 1 = $\frac{3}{8}$ (x + 4)

y - 1 = $\frac{3}{8}$ x + $\frac{3}{2}$   

3x - 8y + 20 = 0
'h' is the length of the perpendicular from C(0, 6) to the line AB, 3x - 8y + 20 = 0

h = | $\frac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}}$ |

   = | $\frac{-28\sqrt{73}}{73}$ | = $\frac{28\sqrt{73}}{73}$

Length of AB = Distance AB = $\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$

                                             = $\sqrt{(4+4)^{2}+(4-1)^{2}}$ = $\sqrt{73}$

Area of triangle ABC = $\frac{1}{2}$ x base x height = $\frac{1}{2}$  x $\sqrt{73}$ x $\frac{28\sqrt{73}}{73}$
                                                                     = 14 Sq.units