# Straight Lines Notes

## What is Coordinate System

• We require two perpendicular axes to locate a point in the plane. One of them is horizontal and other is Vertical.The plane is called Cartesian plane and axis are called the coordinates axis
• The horizontal axis is called x-axis and Vertical axis is called Y-axis
• The point of intersection of axis is called origin.
• The distance of a point from y axis is called x –coordinate or abscissa and the distance of the point from x –axis is called y – coordinate or Ordinate
• The x-coordinate and y –coordinate of the point in the plane is written as (x, y) for point and is called the coordinates of the point

## Distance Formula

For two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ in the Cartesian plane, the distance between them is given by:
$D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

## Section Formula

For two points $A(x_1, y_1)$ and $B(x_2, y_2)$, the coordinates $(x, y)$ of a point $P$ dividing the line segment $AB$ internally in the ratio $m:n$ are:
$x = \frac{m x_2 + n x_1}{m + n}$
$y = \frac{m y_2 + n y_1}{m + n}$

## Mid Point Formula

For two points $A(x_1, y_1)$ and $B(x_2, y_2)$, the coordinates $(x, y)$ of a mid point $P$ is :
$x = \frac{x_2 + x_1}{2}$
$y = \frac{y_2 + y_1}{2}$

## Area of Triangle

Area of triangle ABC of coordinates $A(x_1, y_1)$ , $B(x_2, y_2)$ and $C(x_3, y_3)$
$A=\frac {1}{2}|x_1 (y_2-y_3 )+x_2 (y_3-y_1 )+x_3 (y_1-y_2 )|$

## Formula For Collinearity

For point A, B and C to be collinear, the value of A should be zero
$|x_1 (y_2-y_3 )+x_2 (y_3-y_1 )+x_3 (y_1-y_2 )|=0$

## Definition of Slope of the Straight line

• If $\theta$ is the inclination of a line l, then $\tan \theta$ is called the slope or gradient of the line l.
• The slope of a line whose inclination is 90° is not defined.
• The slope of a line is denoted by m.
• Thus, $m = \tan \theta$, $\theta \ne 90$

## Slope Formula

Case A : Line Having Acute angle

1. Consider the two given points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the Cartesian plane.
2. Draw perpendiculars from both points to the x-axis. Let them intersect the x-axis at points $P$ and $Q$ respectively.
3. The length of segment $AP$ is $y_1$, and the length of segment $BQ$ is $y_2$. Also, the horizontal distance between $A$ and $B$ is $x_2 - x_1$.
4. Draw a perpendicular AM on BQ
4. $m = \tan(\theta)$
From the triangle $ABM$,
$\tan(\theta) = \frac{\text{Opposite side}}{\text{Adjacent side}}$
5. The length of the opposite side (or vertical change) is the difference in the y-coordinates, $y_2 - y_1$. The length of the adjacent side (or horizontal change) is the difference in the x-coordinates, $x_2 - x_1$.
$\tan(\theta) = \frac{y_2 - y_1}{x_2 - x_1}$
6. Since $m = \tan(\theta)$,
$m = \frac{y_2 - y_1}{x_2 - x_1}$

Case B : Line Having Obtuse angle

1. Consider the two given points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the Cartesian plane.
2. Draw perpendiculars from both points to the x-axis. Let them intersect the x-axis at points $P$ and $Q$ respectively.
3. The length of segment $AP$ is $y_1$, and the length of segment $BQ$ is $y_2$. Also, the horizontal distance between $A$ and $B$ is $x_2 - x_1$.
4. Draw a perpendicular AM on BQ
4. $m = \tan(\theta)$
From the triangle $ABM$,
$\tan(180 -\theta) = \frac{\text{Opposite side}}{\text{Adjacent side}}$
5. The length of the opposite side (or vertical change) is the difference in the y-coordinates, $y_2 - y_1$. The length of the adjacent side (or horizontal change) is the difference in the x-coordinates, $x_1 - x_2$.
$\tan(180 -\theta) = \frac{y_2 - y_1}{x_1 - x_2}$
or
$-\tan(\theta) = \frac{y_2 - y_1}{x_1 - x_2}$
or
$\tan(\theta) = -\frac{y_2 - y_1}{x_1 - x_2}$
or
$-\tan(\theta) = \frac{y_2 - y_1}{x_2 - x_1}$

6. Since $m = \tan(\theta)$,
$m = \frac{y_2 - y_1}{x_2 - x_1}$
Therefore The slope or gradient of a line through the points $(x_1, y_1)$ and $(x_2, y_2)$ is:
$m = \frac{y_2 - y_1}{x_2 - x_1}$

## Parallel Lines

When the two lines are Parallel, the angle of incliniation will be same
$\alpha = \beta$
Taking Tan on both sides
$\tan \alpha = \tan \beta$
$m_1 = m_2$
So Two lines are parallel if their slopes are equal

## Perpendicular Lines

Suppose two lines are perpendicular as shown in figure
$\alpha + \beta=90$
$\alpha = 90 - \beta$
Taking Tan on both sides
$\tan \alpha = -cot \beta$
$\tan \alpha \tan \beta=-1$
( m_1 m_2 = -1 \)
Two lines are perpendicular if the product of their slopes is -1: $m_1 m_2 = -1$.

## Angle Between Two Lines

If $m_1$ and $m_2$ are the slopes of two lines, then the angle $\theta$ between them is:
$\tan \theta =| \frac{m_2 - m_1}{1 + m_1 m_2}|$

## Equation of a Line

### Point-slope form

For a line with slope $m$ passing through the point $(x_1, y_1)$:

Here slope will be
$m= \frac {y-y_1}{x-x_1}$
Hence line equation will be
$y - y_1 = m(x - x_1)$

### Two-point form

For a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$:

Here slope will be
$m= \frac {y_2-y_1}{x_1-x_1}$
Now from Point slope form, equation can be written as
$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)$

### Intercept form

what is intercept
The intercept of a straight line refers to the point(s) or distance where the line intersects or "cuts off" a coordinate axis.
X-intercept
The x-intercept is the point where the line intersects or touches the x-axis. At this point, the y-coordinate is 0.
If a line intersects the x-axis at the point $(a, 0)$, then $a$ is the x-intercept.
Y-intercept
The y-intercept is the point where the line intersects or touches the y-axis. At this point, the x-coordinate is 0.
If a line intersects the y-axis at the point $(0, b)$, then $b$ is the y-intercept.

Lets take three cases here
Case I : Y- Intercept Form
A line of slope m intercept at y axis at (0,c)

Here slope will be
$m= \frac {y-c}{x-0}$
Here equation can be written as
$y =mx + c$

Case II : X- Intercept Form
A line of slope m intercept at x axis at (x,0)

Here slope will be
$m= \frac {y-0}{x-d}$
Here equation can be written as
$y =m(x -d)$

Case III : Intercept Form
For a line intercepting the x-axis at $a$ and the y-axis at $b$:

Here slope will be
$m= \frac {b-0}{0-a}$
Now from Point slope form, equation can be written as
$\frac{x}{a} + \frac{y}{b} = 1$

### Normal form

For a line making an angle $\omega$ with the positive x-axis and having a perpendicular distance $p$ from the origin:

we have $ON= p cos \omeaga$ and $OM = p sin \omega$, so that the coordinates of the point A are $(p cos \omega, p sin \omega)$.
Now OA and line are perpendicular, Therefore slope of line will be
$Slope= \frac {-1}{tan \omega}$
Now from Point slope form, equation can be written as
$y - p sin \omega =\frac {-1}{tan \omega} (x - p cos \omega)$
$x \cos \omega + y \sin \omega = p$

## General Equation of Line

Any equation of the form $Ax + By + C = 0$, with A and B are not zero, simultaneously, is called the general linear equation or general equation of a line

Let's convert this general equation into various forms:
Slope-Intercept Form: $y = mx + c$
To express the general equation in slope-intercept form:
1. Solve for $y$ in terms of $x$.
2. Identify the slope $m$ and y-intercept $c$.
From $Ax + By + C = 0$:
$y = -\frac{A}{B}x - \frac{C}{B}$
Here, the slope $m$ is $-\frac{A}{B}$ and the y-intercept $c$ is $-\frac{C}{B}$.

Intercept Form: $\frac{x}{a} + \frac{y}{b} = 1$
To convert to intercept form:
1. Find the x-intercept (the value of $x$ when $y = 0$) which gives the value of $a$.
2. Find the y-intercept (the value of $y$ when $x = 0$) which gives the value of $b$.
3. Write the equation using the intercepts.
From $Ax + By + C = 0$:
The x-intercept (when $y = 0$) is $a = -\frac{C}{A}$
The y-intercept (when $x = 0$) is $b = -\frac{C}{B}$

Normal Form: $x \cos \theta + y \sin \theta = p$
To express in this form:
1. The perpendicular distance $p$ from the origin to the line is given by:
$p = \frac{C}{\sqrt{A^2 + B^2}}$
2. The angle $\theta$ that the normal to the line makes with the positive x-axis is given by the angle whose cosine is $\frac{A}{\sqrt{A^2 + B^2}}$ and whose sine is $\frac{B}{\sqrt{A^2 + B^2}}$.
The above conversions provide multiple ways of expressing the equation of a straight line, each providing unique insights or simplifications for different geometrical interpretations and problems.

## Distance of a Point From a Line

The distance of a point from a line is the length of the perpendicular drawn from the point to the line. Let L : Ax + By + C = 0 be a line, whose distance from the point
P (x1, y1) is d. Then d is given by
$d= \frac {|Ax_1 + By+1 + C|}{\sqrt {A^2 + B^2}}$

## Distance between two Parallel Lines

We know that Parallel lines have same slopes,So let assume we have two parallel given as
$y=mx + c_1$
$y=mx+ c_2$
If d id the distance between them,then d is given by
$d= \frac {|c_1 -c_2|}{\sqrt {1+ m^2}}$

• Notes