# Coordinate Geometry

Sub Topics
Coordinate geometry is part of the geometry in which the position of a point in a plane is shown in the form of an ordered pair of numbers. It is also called the Analytic geometry.

The Cartesian plane consists of 4 quadrants, namely I quadrant, II quadrant, III quadrant and IV quadrant. It is also called the coordinate plane or xy plane. The axes in the coordinate plane are called the coordinate axes.

Ordered pair: ordered pair consists of x-coordinate and the y-coordinate on the Cartesian plane, denoted as (x, y)

x-coordinate: The perpendicular distance from the y-axis when measured along the x-axis is called the x-coordinate (positive along the positive direction of the x-axis and negative along the negative direction of the x-axis). It is also called the abscissa.

y-coordinate: The perpendicular distance from the x-axis when measured along the y-axis is called the y-coordinate (positive along the positive direction of the y-axis and negative along the negative direction of the y-axis). It is also called the ordinate.

In denoting the coordinates of a point, the x-coordinate is written first and the y-coordinate is written second placed in a parenthesis. Coordinates of a point are denoted as (x, y)

Example:

The point P on the graph below denotes the ordered pair(5, 8), 5 being the x-coordinate and 8 being the y-coordinate
the point A denotes the ordered pair (-5, -5), -5 being the x-coordinate and -5 being the y-coordinate

The coordinates of origin is (0,0)
The signs of the coordinates in the different quadrants:

1.    In the 1st quadrant, a point is denoted as (+,+) as the x-axis and y-axis both are positive
2.    In the 2nd quadrant, a point is denoted as (-,+) as the x-axis is negative and the y-axis is positive
3.    In the 3rd quadrant, a point is denoted as (-,-) as the both the x-axis and the y-axis are negative
4.    In the 4th quadrant, a point is denoted as(+,-) as the x-axis is positive and the y-axis is positive

Example: Locate the following points on the Cartesian plane.

1.    A(3, -5)
2.    B(4, 6)
3.    C(-2, -4)
4.    D(-5, 1)
5.    E(0, 6)
6.    F(7, 0)

Solution: The graph below shows the plotting of the different points given

## Coordinate Geometry Formulas

Distance formula : The distance between two points P(x1, y1) and Q(x2, y2) is given by the formula,
$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
In a special case when one of the points is the origin, then the distance formula is.

P= (x, y) and O(0, 0)
$\sqrt{x^2+y^2}$

Section Formula : The coordinates of the point P(x, y) which divides the line segment joining the two points Q(x1, y1) and R(x2, y2) internally in the ratio m:n is,
$\left ( \frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2} \right )$

This is the section formula

If the given ratio in which P divides in k:1, then the coordinates of the point P is given by the formula,

$\left ( \frac{kx_2+x_1}{k+1},\frac{ky_2+y_1}{k+1} \right )$

The coordinates of the midpoint of the line segment joining the points P(x1, y1) and Q(x2, y2) is given by

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

Area of a triangle: Area of a triangle with vesrtices A (x1,y1), B(x2,y2) and C(x3,y3) is given by the formula

$\frac{1}{2}[x_1(y_2-y_3+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Equation of a Line:

y = mx + c   (m =slope of the line, c = y-intercept)
y = mx , when the line passes through the origin
y = k,  (when the line is parallel to the x-axis through (0, k))
y = 0   line is the x-axis
x = k , (when the line is parallel to the y-axis through (k, 0))
x = 0     line is the y-axis
ax + by + c = 0 is the general equation of a straight line
slope, m = (y- y1)/(x- x1)
Equation of a line passing through a point (x1, y1) with slope ‘m’ is given by
(y - y1) = m (x - x1)
Equation of a line when two points are given
(y - y1)/(y- y1)= (x - x1)/(x- x1)

Parallel and Perpendicular lines:

Slopes of Parallel lines:

When two lines are parallel then the product of their slopes is equal to 1

If the slopes of the parallel lines are m1 and m2

Then, m1 x m2 = 1

Slopes of Perpendicular lines:

When two lines are perpendicular to each other, then the product of their slopes is equal to -1

If the slopes of the perpendicular lines are m1 and m2

Then, m1 x m2 = -1

## Application of Coordinate Geometry

Coordinate geometry has many uses in the real life, in business, presentation graphs are used to show the performance of the company, coordinate geometry is used to draw the graphs.

Analytical geometry is used to find the position of a place on earth by using the pictures taken by the satellite.

## History of Coordinate Geometry

Coordinate geometry is also called the Analytical geometry. It was first attributed by Rene Decarthes, who made significatn progress with methods in one of his three essays by name ‘Discourse on Method’.This was written in French, this provided the foundation for Infinitesimal calculus in Europe. Peirra Fermat was also one of the mathematician who pioneered the development of coordinate geometry. The difference between these two mathematicians work was the viewpoint.

## Coordinate Geometry Problems

Below you could see coordinate geometry problems

## Coordinate Geometry Proofs

1. Prove that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a square

Proof: Given points,A (1, 7),B (4, 2),C (-1, -1) and D(-4, 4)

( property of a square, all the sides are of equal length and the two diagonals are of equal length in a square.)

Distance formula :  $\sqrt{(x_2-x_1)+(y_2-y_1)^2}$

AB = $\sqrt{(1-4)^2+(7-2)^2}$ = $\sqrt{(9+25)}$ = $\sqrt{34}$
BC = $\sqrt{(4+1)^2+(2+1)^2}$ = $\sqrt{(25+9)}$ = $\sqrt{34}$
CD = $\sqrt{(-4+1)^2+(-4-1)^2}$ = $\sqrt{(9+25)}$ = $\sqrt{34}$
DA = $\sqrt{(1+4)^2+(7-4)^2}$ = $\sqrt{(25+9)}$ = $\sqrt{34}$
AC = $\sqrt{(1+1)^2+(7+1)^2}$ = $\sqrt{(4+64)}$ = $\sqrt{68}$
BD = $\sqrt{(4+4)^2+(2-4)^2}$ = $\sqrt{(64+4)}$ = $\sqrt{68}$

From the above, we can say that AB = BC = CD = DA, the four sides of the given quadrilateral are equal.

Diagonals AC = BD, hence the given points are the vertices of a square as they satisfy the properties of a square. Hence, ABCD is a square.

2. Find the relation between x and y such that the point (x, y) is equidistant from the points P(1, 7) and Q(3, 5)

Solution: Given points P(1, 7) and Q(3, 5)

Let R(x,y) be the point equidistant from P and Q
We are given, PR = QR
(PR) ² =(QR) ²
(x-7) ²+(y-1) ² = (x-3) ² + (y-5) ²
On simplifying the above equation, we get
x – y = 2 is the required relation

3. Find the coordinates of the point which divides the line segments joining the points (4,-3) and (8,5) in the ratio 3:1 internally

Solution: Let A(x,y) be the point.Section formula is given by,

$\left ( \frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2} \right )$
Substituting the given values of m1 and m2 and the coordinates, we get
x =$\frac{[3(8)+1(4)]}{(3+1)}$ = 7
y =$\frac{[(3(5)+1(-3)]}{(3+1)}$ = 3

The required point is (7, 3)

4. If the points P(6, 1), Q(8, 2), R(9, 4) and S(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

Solution: In a parallelogram, the diagonals bisect each other.

So, the coordinates of the midpoint of PR= the coordinates of the midpoint of QS
$\frac{[(6+9)}{2},\frac{(1+4)}{2}$ = $\frac{[(8+p)}{2},\frac{(2+3)}{2}$
$[\frac{15}{2},\frac{5}{2}]$ = $[\frac{(8+p)}{2},\frac{5}{2}]$
$\frac{15}{2}$ = $\frac{(8+p)}{2}$
15 = 8 + p
p = 15 - 8
p = 7 is the required answer!

5.  Prove that the points A(3, 2), B(-2, -3) and (2, 3) form the vertices of a right angled triangle

Proof: Given points A(3, 2), B(-2, -3) and C(2, 3)

Distance formula : $\sqrt{(x_2-x_1)+(y_2-y_1)^2}$

Using the above formula, let us find the distances AB, BC and CA
AB²= (-2-3) ² +(-3-2) ² = 25 +25 = 50
BC²=(2+2) ² + (3+3) ² = 16 + 36 =52
CA²=(3-2) ² + (2-3) ²= 1 +1 = 2
AB² + CA² = 50 + 2 = 52=BC²
According to the converse of Pythagorean theorem
we get, ∟A=90°
Therefore, the given points are the vertices of a right angled triangle, hence proved

## Coordinate System

A coordinate system a point in a plane specifies the position of the point by  an ordered pair (x, y) each of which is a numerical coordinate. x, being the numerical x – coordinate and y, being the numerical y-coordinate.

The distance of a point from the y-axis is called its x-axis or abscissa. The distance of a point from the x-axis is called its y-axis or ordinate

There are two reference lines in a coordinate plane. Each line in the coordinate system is called the coordinate axis. The point at which the two axis meet is the origin (0, 0).

In denoting the coordinates of a point, the x-coordinate is written first and the y-coordinate is written second, placed in a bracket. Coordinates of a point are denoted as (x, y)

In a coordinate system, the coordinates of x are denoted as (x, 0) and the coordinates of y are denoted as (0, y).

A coordinate system specifies a point in a plane by a an ordered pair (x, y) each of which is a numerical coordinate. x, being the numerical x – coordinate and y, being the numerical y-coordinate.
There are two reference lines in a coordinate plane. Each line in the coordinate system is called the coordinate axis. The point at which the two axis meet is the origin (0, 0).
The distance of a point from the y-axis is called its x-axis or abscissa. The distance of a point from the x-axis is called its y-axis or ordinate
In denoting the coordinates of a point, the x-coordinate is written first and the y-coordinate is written second placed in a parenthesis. Coordinates of a point are denoted as (x, y)
In a coordinate system, the coordinates of x are denoted as (x, 0) and the coordinates of y are denoted as (0, y)

In the graph of the coordinate plane given, we have two points, P and A
The coordinates of P(x, y) = (5, 8)
The coordinates of A(x, y) = (-5, -5)