- 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

\[ D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \]

\[ x = \frac{m x_2 + n x_1}{m + n} \]

\[ y = \frac{m y_2 + n y_1}{m + n} \]

\[ x = \frac{x_2 + x_1}{2} \]

\[ y = \frac{y_2 + y_1}{2} \]

$A=\frac {1}{2}|x_1 (y_2-y_3 )+x_2 (y_3-y_1 )+x_3 (y_1-y_2 )|$

$|x_1 (y_2-y_3 )+x_2 (y_3-y_1 )+x_3 (y_1-y_2 )|=0$

- 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, $m = \tan \theta$, $\theta \ne 90$

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 \( m = \tan(\theta) \),

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

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:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

$\alpha = \beta$

Taking Tan on both sides

$\tan \alpha = \tan \beta$

\( m_1 = m_2 \)

So Two lines are parallel if their slopes are equal

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: \( m_1 m_2 = -1 \).

\[ \tan \theta =| \frac{m_2 - m_1}{1 + m_1 m_2}| \]

Here slope will be

$m= \frac {y-y_1}{x-x_1}$

Hence line equation will be

\[ y - y_1 = m(x - x_1) \]

Here slope will be

$m= \frac {y_2-y_1}{x_1-x_1}$

Now from Point slope form, equation can be written as

\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \]

The intercept of a straight line refers to the point(s) or distance where the line intersects or "cuts off" a coordinate axis.

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.

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

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

\[ y =mx + c \]

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

\[ y =m(x -d) \]

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

\[ \frac{x}{a} + \frac{y}{b} = 1 \]

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

\[ y - p sin \omega =\frac {-1}{tan \omega} (x - p cos \omega)$ \]

\[ x \cos \omega + y \sin \omega = p \]

Let's convert this general equation into various forms:

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 \( -\frac{C}{B} \).

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 \( b = -\frac{C}{B} \)

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.

P (x1, y1) is d. Then d is given by

$d= \frac {|Ax_1 + By+1 + C|}{\sqrt {A^2 + B^2}}$

$y=mx + c_1$

$y=mx+ c_2$

If d id the distance between them,then d is given by

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

**Notes**