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: 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 (x1y1) and (x2y2) on the line and using the formula:

2. The point-slope form:
yy1 = m (xx1).
(x1, y1) is a point on the line. The slope m can be found from a second point, (x2y2), 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 = x1 + t
y = y1 + 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. (x1y1) 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 = x1
y = y1
or the ordered pair (x1y1), and when t = 1 then we get the point
x = x1 + 1
y = y1 + m
or the ordered pair (x1 + 1, y1 + 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 (x1, y1) = (0, 15) and (x2, y2) = (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 (015) as “the” point together with m = −5:
y15 = −5 (x0)       or simplifying:       y − 15 = −5 x.
Or we could instead use the other point, (30) and get:
y0 = −5 (x3)       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 (015) 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, (30) 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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