COORDINATE GEOMETRY

Consider the two points $$A(x_1, y_1) \quad\mbox{and}\quad B(x_2, y_2)$$
The distance between the points $A$ and $B$ is $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\quad\quad\quad \mbox{DISTANCE}$$ 
The coordinates of the midpoint of the interval joining the points $A$ and $B$ are $$\left( {x_1+x_2 \over 2 }, {y_1+y_2 \over 2}\right)\quad\quad\quad \mbox{MIDPOINT}$$ 
The slope or gradient of the line through the points $A$ and $B$ is $$m={y_2-y_1 \over x_2-x_1 }={ \mbox{rise}\over \mbox{run}}\quad\quad\quad \mbox{GRADIENT}$$ 
The GRADIENT-INTERCEPT form of the equation of a line is $$y=mx+b$$ where $m$ is the gradient or slope of the line, and $b$ is the $y$-intercept. 

The POINT-GRADIENT form of a line is $$y-y_1=m(x-x_1)$$ where $m$ is the gradient of the line, and $(x_1,y_1)$ is ANY point on the line. 

The GENERAL FORM OF A LINE is $$ax+by+c=0$$ where $a,b,c$ are constants.The slopes or gradients of parallel lines are equal, that is, $m_1=m_2$.

The slopes of perpendicular lines are negative reciprocals of each other. This can be written as $$m_2=-{1\over m_1}\quad\quad\mbox{or}\quad\quad m_1\times m_2=-1$$ 
The $x$-intercept is found by putting $y=0$ and solving for $x$ in the equation of the line. The $y$-intercept is found by putting $x=0$ and solving for $y$ in the equation of the line.

VERTICAL LINES have the form, $$x=k$$ where $k$ is a constant (number).

HORIZONTAL LINES have the form, $$y=k$$ where $k$ is a constant (number). }