# Lines in the Plane and Slope

## Introduction

The simplest mathematical model for relating two variables is the linear equation $y=mx+b$ (Slope-intercept form). This equation is called Linear because its graph is a line. $m$ is the slope and $(0,b)$ is the y-intercept.

## Finding the slope $m$ For instance, suppose you want to find the slope of the line passing through the distinct points $(x_{1},x_{2})$ and $(y_{1},y_{2})$ .

 $Slope={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}={\frac {y_{1}-y_{2}}{x_{1}-x_{2}}}$ Exercises Find the slope of the line passing through the distinct points below

1) $(-6,2)$ and $(3,20)$ Solution:
$2$ 2)$(3,-7)$ and $(-3,-7)$ Solution:
$0$ 3)$(3,-2)$ and $(-3,1)$ Solution:
${\frac {-1}{2}}$ ## Writing the linear equation given a slope and a point on the line

 Point-Slope Form of the Equation of a Line
The equation of the line with slope  passing through the point $(x_{1},y_{1})$ is
$y-y_{1}=m(x-x_{1})$ Notice: In order to write this equation, we need a point and a slope given

Exercises Find the equation of the line line given the information below

1) slope $m=3$ and goes through $(1,2)$ Solution:
Apply the formula with $m=3$ , $x_{1}=1$ and $y_{1}=2$ to get the result
$y-2=3(x-1)$ ## Writing the linear equation given two points on the line

Given two point $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are on the line. To find the equation of this line:

First, use the formula to find the slope

Then, apply the point-slope formula with the slope we just found and one of the given points.

Exercises Find the equation of the line passing through the distinct points below

1) $(4,3)$ and $(0,-5)$ Solution:
$m=slope={\frac {3-(-5)}{4-0}}={\frac {8}{4}}=2$ Apply the point-slope formula with slope $m=2$ and the given point $(4,3)$ ( I choose $(4,3)$ in this case, but $(0,-5)$ will give the same result) to get
$y-3=2(x-4)$ ## Notes:

A vertical line goes through has equation of the form $x=a$ where $a$ is any constant.