# Introduction

Rene Descartes (1596-1650) ले 1600 मा बिकास गरेको Geometry लाई Analytic Geometry भनिन्छ। यसलाई Coordinate Geometry वा Cartesian Geometry पनि भनिन्छ ।

Geometrical Object हरुलाई Position र Value (स्थान र मान) दिनु नै यस Geometry को मुख्य बिशेषता थियो। यस Geometry को बिकासले गर्दा geometrical object हरुलाई जस्तै point, line, circle, …. आदिलाई algebraic expression बाट number को प्रयोग गरि ब्याख्या बिश्लेषण गर्न सकिने भयो ।

यसको लागी Descartes ले Geometry को अध्ययनमा Algebra को प्रयोग गरेका थिए। Descartes को यो योगदानलाई “the turning point of modern mathematics” पनि भनिन्छ। यस सताब्दी लाई गणितको बिकासमा सबैभन्दा धनी सताब्दी पनि मानिन्छ । Descartes ले बिकास गरको Analytic Geometry को मुख्य आधार “there is one-to-one correspondence between the points of a line and real numbers” भन्ने तथ्य हो ।

जस्तै, Two dimensional plane मा कुनै एउटा point A भएमा point A लाई X-axis बाट एउटा number x र Y-axis बाट एउटा number y correspond गर्न सकिन्छ। यसै number (x,y) लाई A को coordinate भनिन्छ।

In the figure above, there are two scales – One is the X-axis which is running across the plane and the other one is the Y-axis which is at the right angles to the X-axis. This is similar to the concept of the rows and columns.

Another mathematician Legendre is considered one of the fathers of modern analytic geometry, a geometry that incorporates all the inherent power of both algebra and calculus.

By the use of this coordinate geometry, number of things are possible, some of them are given below.

1. Determine the distance between these points.
2. Find the equation, midpoint, and slope of the line segment.
3. Determine if the given lines are perpendicular or parallel.
4. Find the perimeter and the area of the polygon formed by the points on the plane.
5. Transform the shape by reflecting, moving and rotating it.
6. Define the equations of ellipses, curves, and circles.

gradient =$\frac{\text{change in y}}{\text{change in x}}$

We know that the gradient of a straight line is a measure of how steep it is. To calculate the gradient of a straight line, we choose any two points on it and find the run and the rise from the first point to the second point. The run is the change in the x-coordinates, and the rise is the change in the y-coordinates, as illustrated in Figure below [Figure 1].

Then,
gradient =$\frac{rise}{run}$ (1)
The run and the rise from one point to another on a straight line can be positive, negative or zero, depending on whether the relevant coordinates increase, decrease or stay the same from the first point to the second point.
A straight-line graph that slopes down from left to right has a negative gradient, one that’s horizontal has a gradient of zero, and one that slopes up from left to right has a positive gradient.
If the run is zero, which happens when the line is vertical, then the gradient of the line is undefined, since division by zero isn’t possible. Vertical lines are the only straight lines that don’t have gradients.
If the two points that we choose to calculate the gradient of a line are
$(x_1, y_1)$ and $(x_2, y_2)$ as illustrated in Figure below (Figure 2).

Then
run = x2 − x1 and rise = y2 − y1.
Now,
gradient =$\frac{y_2-y_1}{x_2-x_1}$ (2)
Remember that when we use this formula to calculate the gradient of a straight line, it doesn’t matter which point we take to be the first point, $(x_1, y_1)$, and which you take to be the second point, $(x_2, y_2)$, as we get the same result either way.
For example, using the formula to calculate the gradient of the line through the points (1, 8) and (5, 2), which is shown in Figure below [Figure 3]

It gives either
gradient =$\frac{y_2-y_1}{x_2-x_1}=\frac{2-8}{5-1}=-\frac{3}{2}$
gradient =$\frac{y_2-y_1}{x_2-x_1}=\frac{8-2}{1-5}=-\frac{3}{2}$
The gradient of a straight line

The gradient of a line is a measure of its slope or steepness. It is defined as a ratio of its vertical displacement or Rise, to its horizontal displacement or Runs.

To determine the slope (or gradient), we can use:

1. direct measurement OR
2. count the number of units on the vertical and
horizontal lines when the line is drawn on the Cartesian plane or a grid OR
3. use slope formula as
$m=\frac{y_2-y_1}{x_2-x_1}$
if two points $(x_1,y_1)$ and $(x_2,y_2)$ is given.

#### Some Examples

The slope of a line can positive, negative, zero, or undefined.
1. Positive slope
If a line admits, $y$ increases as $x$ increases, then the slope upwards to the right. This slope will be a positive number. The line in Figure below has a slope of about $+0.5$ , it goes up about 0.5 for every step of 1 along the x-axis.
2. Negative slope
If a line admits, $y$ decreases as $x$ increases, then the slope downwards to the right. This slope will be a negative number. The line in Figure below has a slope of about $-0.5$, it goes down about $0.5$ for every step of 1 along the x-axis.
3. Zero slope
If a line admits, $y$ does not change as $x$ increases /decreases, then the line is exactly horizontal. The slope of any horizontal line is always zero. The line in Figure below goes neither up nor down as $x$ increases, so its slope is zero.
4. Undefined slope
If a line is exactly vertical, it does not have a defined slope. In such a line any two $x$ coordinates are the same, so the difference is zero. The line in the Figure below is exactly vertical, so it has no defined slope. We say "the slope is undefined".

#### Example 1

Find the slope of a line through the points (3, 4) and (5, 1).

Solution
We know that,slope of a line through the points $(x_1,y_1)$ and $(x_2,y_2)$ is
$slope=\frac{y_2-y_1}{x_2-x_1}$
So, using $(3, 4)$ as $(x_1,y_1)$ , the slope is
$slope=\frac{y_2-y_1}{x_2-x_1}=\frac{1-4}{5-3}=-\frac{3}{2}$
NOTE,
using $(5, 1)$ as $(x_1,y_1)$ , the slope is
$slope=\frac{y_2-y_1}{x_2-x_1}=\frac{4-1}{3-5}=-\frac{3}{2}$

# Straight line

A straight line, drawn on a Cartesian Plane can be described by an equation. Such equations have a general form and vary depending on where the line cuts the axes and its degree of slope.

For detail study of a striaght line, following key points are essential.

# Horizental Lines

Some linear equations have only one variable. They may have just y .

Let’s consider the equation y=2. This equation has only one variable, y. The equation says that y is always equal to 2, so its value does not depend on x. No matter what is the value of x, the value of y is always 2.

So to make a graph, draw a horizental line from y=2.

Horizontal lines are all parallel to the x-axis.
We know that the equation of the $x$ -axis is
$y = 0$ .
This is because all points on this axis have a $y$ -coordinate zero, regardless of their different x-coordinates. So, the horizontal line which cuts the y-axis at $2$ has equation
$y = 2$ .
Also, the horizontal line that cuts the y-axis at $-1$ has equation
$y = -1$
and so on.

In general, the equation of a horizontal line is
$y=a$
where $a$ is a constant.
This line cut the vertical axis at $a$ and all points on the line have a $y$ -coordinate of $a$

# Vertical Line

Some linear equations have only one variable. They may have just x .

Let’s consider the equation x=−3. This equation has only one variable, x. The equation says that x is always equal to−3, so its value does not depend on y. No matter what is the value of y, the value of x is always −3.

So to make a graph, draw a verticasl line from x=-3.

#### Vertical Lines

Vertical lines are all parallel to the $y$ -axis. We know that the equation of the $y$ -axis is
$x = 0$ .

This is because all points on this axis have an $x$ -coordinate of zero, regardless of their different y-coordinates.
So, the vertical line which cuts the x-axis at $2$ has equation
$x = 2$ .
Also, the vertical line that cuts the $x$ -axis at $-1$ has equation
$x =-1$
and so on.

In general, the equation of a vertical line is
$x=b$
where $b$ is a constant.
This line cuts the horizontal axis at $b$ and all points on the line have an $x$ -coordinate of $b$ .

# Point Slope Form

#### Point Slope Form

Let $l$ is a straight line passing through a point $A=(x_1,y_1)$ with slope $m$ , then the equation of straight line $l$ is
$y-y_1=m(x-x_1)$

Proof
Given that $l$ is a straight line passing through a point $A(x_1,y_1)$ with slope $m$ . Suppose that, $P(x,y)$ is arbitrary point on the straight line $l$ , then
slope of line = slope of PA
or $m=\frac{y_2-y_1}{x_2-x_1}$
We take the point $P(x,y)$ as $(x_2,y_2)$ , thus
$m=\frac{y-y_1}{x-x_1}$
or $y-y_1=m(x-x_1)$

# Two Point For

Let $l$ is a straight line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ , then the equation of straight line $l$ is
$y-y_1= \frac{y_2-y_1}{x_2-x_1} (x-x_1)$

Proof

Given that $l$ is a straight line passing through two points $A(x_1,y_1)$ and $B(x_2,y_2)$
Suppose that, $P(x,y)$ is arbitrary point on the straight line $l$ , then
slope of PA = slope of AB
Here, slope of PA is
slope of PA$=\frac{y_2-y_1}{x_2-x_1}$
We take the point $P(x,y)$ as $(x_2,y_2)$ , and $A(x_1,y_1)$ as $(x_1,y_1)$ , thus
slope of PA$=\frac{y-y_1}{x-x_1}$ (1)
Next,
slope of $AB=\frac{y_2-y_1}{x_2-x_1}$ (2)
Now, Equating (1) and (2), we get
slope of PA = slope of AB
or $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$
or $y-y_1= \frac{y_2-y_1}{x_2-x_1} (x-x_1)$

# Intercept Form

#### Intercepts

In coordinate geometry of two dimension, there are two types of intercepts. they are x-intercept and y-intercept. The x-intercepts are where the straight line crosses the x-axis [In the Figure below, a is x-intercept], and the y-intercepts are where the line crosses the y-axis [In the Figure below, b is y-intercept].

To clarify it algebraically,

1. an x-intercept is a point on the graph where y is zero, [x-intercept is a point in the equation where the y-value is zero]
2. a y-intercept is a point on the graph where x is zero.[y-intercept is a point in the equation where the x-value is zero]

#### Slope Intercept Form

Let l is a straight line with slope m and yintercept c, then the equation of straight line l is
y=mx+c

Proof
Given that l is a straight line with slope m andy intercept c
Then the line intersects y-axis at a point A(0,c)
Now, equation of the lie l is
$y-y_1=m(x-x_1)$
Taking the point A(0,c) as $(x_1,y_1)$ , we get
y-c=m(x-0)
or y=mx+c

#### Slope Intercept Form

Let l is a straight line with slope m and y intercept c , then the equation of straight line l is
y=mx+c

Proof
Given that ll is a straight line with slope m and y intercept c
Then the line intersects y-axis at a point A(0,c)
Now, take an arbitrary point P(x,y) on l, then slope of the lie l is
slope of l=slope of l
Taking the point A(0,c) and P(x,y), we get
$m= \frac{y_2-y_1}{x_2-x_1}$
or $m= \frac{y-x}{x-0}$
or mx=y-c
or y=mx+c

# Double Intercept Form

Let $l$ is a straight line whose x intercept is a and y intercept is b , then the equation of straight line $l$ is
$\frac{x}{a}+\frac{y}{b}=1$

Solution
Given that $l$ is a straight line whose x intercept is a and y intercept is b
Thus, the line intersects $x$ -axis at a point $A(a,0)$ and $y$ -axis at a point $B(0,b)$ [See Figure below

Now, equation of the lie $l$ is
$(y-y_1)=\frac{y_2-y_1}{x_2-x_1} (x-x_1)$
Taking the point $A(a,0)$ as $(x_1,y_1)$ , and taking the point $B(0,b)$ as $(x_2,y_2)$ , the equation of straight line is
$(y-0)=\frac{b-0}{0-a} (x-a)$
or $y=-\frac{b}{a} (x-a)$
or $ay=-b(x-a)$
or $ay=-bx+ab$
or $bx+ay=ab$
or $\frac{x}{a}+\frac{y}{b}=1$

# Normal Form

#### Normal Form

The equation of the straight with length of the perpendicular from the origin $p$ and this perpendicular makes an angle $\alpha$ with x-axis is
$x \cos \alpha + y \sin \alpha = p$

Proof
Given that $l$ is a straight line with length of the perpendicular from the origin $p$ and this perpendicular makes an angle $\alpha$
Suppose the line $l$ intersects the x-axis at $A(a,0)$ and the y-axis at $B(0,b)$ . Now from the origin $O$ draw $OD$ perpendicular to $l$ , whose $x$ intercept is $a$ and $y$ intercept is $b$ [See Figure below]

Thus, the equation of striaght line is
$\frac{x}{a}+\frac{y}{b}=1$ (A)
Here, from the right-angled $\triangle ODA$ , we get,
$\frac{OD}{OA} = \cos \alpha$
or $\frac{p}{a} = \cos \alpha$
or $a=\frac{p}{\cos \alpha}$ (1)
Again, from the right-angled $\triangle ODB$ , we get,
$\frac{OD}{OB} = \cos \left(\frac{\pi}{2}-\alpha \right )$
or$\frac{p}{b} = \sin \alpha$
or $b=\frac{p}{\sin \alpha}$ (2)
Now, using (1) and (2) in (A), the equation of the straight line is
$\frac{x}{a}+\frac{y}{b}=1$
or $\frac{x}{\frac{p}{\cos \alpha}}+\frac{y}{\frac{p}{\sin \alpha}}=1$
or $x \cos \alpha + y \sin \alpha = p$