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
$d= \frac {|c_1 -c_2|}{\sqrt {1+ m^2}}$
