# Co-ordinate Geometry

Coordinate Geometry is the unification of algebra and geometry in which algebra is used in the study of geometrical relations and geometrical figures are represented by means of equations.
The most popular coordinate system is the rectangular cartesian system.

Coordinates of a point are the real variables associated in an order to describe its location in space.

Here we consider the space to be two-dimensional.

Through a point O, referred to as the origin, we take two mutually perpendicular lines XOX’ and YOY’ and call them x and y axes respectively.

The position of a point is completely determined with reference to these axes by means of an ordered pair of real numbers (x, y) called the coordinates of P where |x| and |y| are the distances of the point P from the y-axis and the x – axis respectively.

x is called the x-coordinate or the abscissa of P and y is called the y-coordinate or the ordinate of the point P.

### Distance between two points:

Let A and B be two given points, whose coordinates are given by A(x1, y1) and B(x2, y2) respectively.

Then $\displaystyle AB = \sqrt{(x_1 – x_2)^2 + (y_1-y_2)^2}$

### Section formula:

Coordinates of the point P dividing the join of two points A(x1, y1) and B(x2, y2) internally in the given ratio λ1 : λ2 are

$\displaystyle ( \frac{\lambda_1 x_2 + \lambda_2 x_1}{\lambda_1 + \lambda_2} , \frac{\lambda_1 y_2 + \lambda_2 y_1}{\lambda_1 + \lambda_2} )$

Coordinates of the point P dividing the join of two points A(x1, y1) and B(x2, y2) externally in the ratio of λ1 : λ2 are

$\displaystyle ( \frac{\lambda_1 x_2 – \lambda_2 x_1}{\lambda_1 – \lambda_2} , \frac{\lambda_1 y_2 – \lambda_2 y_1}{\lambda_1 – \lambda_2} )$

In both the cases,λ12 is positive.

Notes:
⋄ If the ratio, in which a given line segment is divided, is to be determined, then sometimes, for convenience (instead of taking the ratio λ1 : λ2), we take the ratio k : 1. If the value of k turns out to be positive, it is an internal division otherwise it is an external division.

⋄ The coordinates of the mid-point of the line-segment joining (x1, y1) and (x2, y2) are

$\displaystyle ( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} )$

Definitions Related to Co-Ordinate Geometry

Centroid: The point of concurrency of the medians of a triangle is called the centroid of the triangle. The centroid of a triangle divides each median in the ratio 2 :1.
The coordinates are given by

$\displaystyle ( \frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3} )$

Orthocentre: The point of concurrency of the altitudes of a triangle is called the orthocentre of the triangle.
The co-ordinates of the orthocentre of the triangle A(x1, y1), B(x2, y2), C(x3,y3) are

$\displaystyle ( \frac{x_1 tanA + x_2 tanB + x_3 tanC}{tanA + tanB + tanC} , \frac{y_1 tanA + y_2 tanB + y_3 tanC}{tanA + tanB + tanC} )$

Incentre : The point of concurrency of the internal bisectors of the angles of a triangle is called the incentre of the triangle. The coordinates of the incentre are given by

$\displaystyle ( \frac{a x_1 + b x_2 + c x_3}{a + b + c} , \frac{a y_1 + b y_2 + c y_3}{a + b + c} )$

where a, b and c are length of the sides BC, CA and AB respectively

Circumcentre : The point of concurrency of the perpendicular bisectors of the sides of a triangle is called circumcentre of the triangle.

The coordinates of the circumcentre of the triangle with vertices A(x1, y1), B(x2, y2), C(x3, y3) are

$\displaystyle ( \frac{x_1 sin2A + x_2 sin2B + x_3 sin2C}{sin2A + sin2B + sin2C} , \frac{y_1 sin2A + y_2 sin2B + y_3 sin2C}{sin2A + sin2B + sin2C} )$

⋄ If the triangle is right angle then circumcentre is the mid point of the hypotenuse

Note:

⋄ Centriod G, orthocentre H and Circumcentre P of a ΔABC are collinear and G Divides HP in the ratio 2 :1. i.e. HG:GP =2:1. Also, AH = 2PD.

Example : If midpoints of the sides of a triangle are (0, 4), (6, 4) and (6, 0), then find the vertices of triangle, centroid and circumcentre of triangle.

Solution: Let points A (x1, y1), B (x2, y2) and C (x3, y3) be vertices of ΔABC.

x1 + x3 = 0 , y1 + y3 = 8

x2 + x3 = 12 , y2 + y3 = 8

x1 + x2 = 12, y1 + y2 = 0

Solving we get A (0, 0), B (12, 0) and C (0, 8)

Hence ΔABC is right angled triangle.
∠A = π/2
Circumcentre is midpoint of hypotenuse which is (6, 4) itself and centroid

$\displaystyle (\frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3} )$

= (4 , 8/3)

Illustration : Prove that the incentre of the triangle whose vertices are given by A(x1,y1),B(x2,y2), C(x3, y3) is

$\displaystyle ( \frac{a x_1 + b x_2 + c x_3}{a + b + c} , \frac{a y_1 + b y_2 + c y_3}{a + b + c} )$

where a, b, and c are the sides opposite to the angles A, B and C respectively.

Solution:

By geometry, we know that

$\displaystyle \frac{BD}{DC} = \frac{AB}{AC}$ (since AD bisects ∠A)

If the lengths of the sides AB, BC and AC are c, a and b respectively, then

$\displaystyle \frac{BD}{DC} = \frac{AB}{AC} = \frac{c}{b}$

Coordinates of D are

$\displaystyle (\frac{b x_2 + c x_3}{b + c} , \frac{b y_2 + c y_3}{b + c} )$

Since

$\displaystyle \frac{BD}{DC} = \frac{c}{b}$

$\displaystyle BD = \frac{a c}{b+c}$

IB bisects ∠B. Hence

$\displaystyle \frac{ID}{IA} = \frac{BD}{BA} = \frac{a}{c+b}$

Let the coordinates of I be (x , y)

Then

$\displaystyle ( x = \frac{a x_1 + b x_2 + c x_3}{a + b + c} , y = \frac{a y_1 + b y_2 + c y_3}{a + b + c} )$

(using section formula)