### Forms for the equation of a straight line

Suppose that we have the graph of a straight line and that we wish to find its equation.
(We will assume that the graph has *x* and *y* axes and a linear scale.)
The equation can be expressed in several possible forms.
To find the equation of the straight line *in any form* we must be given either:
- two points,
(
*x*_{1}, *y*_{1}) and
(*x*_{2}, *y*_{2}), on the line; or
- one point,
(
*x*_{1}, *y*_{1}), on the line and the slope, *m*; or
- the
*y* intercept, *b*, and the slope, *m*.

In the first case where we are given two points, we can find *m* by using the formula:
Once we have one form we can easily get any of the other forms from it
using simple algebraic manipulations. Here are the forms:

**1. The slope-intercept form:**
*y* = *m x* + *b*.

The constant *b* is simply the *y* intercept of the line, found by inspection.
The constant *m* is the slope, found by picking any
two points (*x*_{1}, *y*_{1}) and
(*x*_{2}, *y*_{2}) on the line and using the formula:

**2. The point-slope form:**
*y* − *y*_{1} = *m* (*x* − *x*_{1}).

(*x*_{1}, *y*_{1}) is a point on the line.
The slope *m* can be found from a second point,
(*x*_{2}, *y*_{2}), and using the formula:

**3. The general form:**
*a x* + *b y* + *c* = 0.

*a*, *b* and *c* are constants.
This form is usually gotten by manipulating one of the previous two forms.
Note that any one of the constants can be made equal to 1 by dividing the equation
through by that constant.

**4. The parametric form:**
*x* = *x*_{1} + *t*

*y* = *y*_{1} + *m t*

This form consists of a pair of equations; the first equation gives the *x* coordinate
and the second equation gives the *y* coordinate of a point on the line
as functions of a parameter *t*.
(*x*_{1}, *y*_{1}) is a known point on the line
and *m* is the slope of the line.
Each value of *t* gives a different point on the line.

For example when *t* = 0 then we get the point
*x* = *x*_{1}

*y* = *y*_{1}

or the ordered pair (*x*_{1}, *y*_{1}), and when *t* = 1 then we get the point
*x* = *x*_{1} + 1

*y* = *y*_{1} + *m*

or the ordered pair (*x*_{1} + 1, *y*_{1} + *m*), and so on.

**Example:** Show all of these forms for the straight line shown to the right.

**Solution:** Two points on this line are (*x*_{1}, *y*_{1}) = (0, 15) and
(*x*_{2}, *y*_{2}) = (3, 0). Thus the *y* intercept is
*b* = 15

and the slope is
= −5

**1.** To get the **slope-intercept form**, we simply substitute in the two values
*m* = −5 and *b* = 15:
*y* = −5 *x* + 15.

**2.** To get the **point-slope form**, we could use the point
(0, 15)
as “the” point together with *m* = −5:
*y* − 15 = −5 (*x* − 0)
or simplifying: *y* − 15 = −5 *x*.

Or we could instead use the other point,
(3, 0) and get:
*y* − 0 = −5 (*x* − 3)
or simplifying: *y* = −5 (*x* − 3).

**3. ** To get the **general form**, take any of these three forms found so far and distribute
and collect all terms on the left-hand-side. The result is the same for all:
5 *x* + *y* −15 = 0.

Note that dividing both sides by, say 5, results in the equation
*x* + 0.2 *y* − 3 = 0,

which is also in general form and is equivalent in every way to the previous one.

**4.** To get the **parametric form**, we could use the point
(0, 15)
as “the” point together with *m* = −5:
*x* = 0 + *t*

*y* = 15 −5 *t*

With this choice, when *t* = 0 we are at the point (0, 15) and when *t* = 3 we are
at the point (3, 0). We could instead use the other point,
(3, 0) and get
another parametric form:
*x* = 3 + *t*

*y* = 0 −5 *t*

With this choice, when *t* = 0 we are at the point (3, 0) and when
*t* = −3 we are at the point (0, 15).
With either choice we will get all the points on the line as we let *t*
range through all values.

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