Lines and Their Equations

Since the Cartesian plane uniquely identifies every point in the plane with a set of numerically ordered coordinates, we can create algebraic statements or equations to describe the geometric shapes and mathematical relations or functions which we use daily in our calculations. In other words, we can find equations to represent the relations between the x and y coordinates of the points which make up a given shape -- be it a line, curve or polygon.

For instance, y = 0 is the equation of the x-axis because every single point on the x-axis has a second coordinate of 0. Since y is the name we give to the 2nd coordinate, y = 0 describes every point on the x-axis.

Similarly x = 0 is the equation of the y-axis.

Later in this lesson, we'll learn to find the equation of the line AC seen above.

One of the greatest sources of confusion for students new to this material is all those x's and y's with and without subscripts (those little 0's and 1's below the letters like x₀ , y₀ and x₁ , y₁).

The notation P( x, y ) represents the general point or any point in the plane.

The notation A( x ₀ , y ₀ ) represents a unique and defined point in the plane.

If there are no subscripts with the coordinates, the point is non-specific.

It represents any point in the plane.

If there are subscripts with the coordinates, the point is unique and defined.

It represents one and only one point in the plane.

The subscripts define the order of the ordered pairs.

They pair the 1st value of x with the 1st value of y

and the 2nd value of x with the 2nd value of y.

Note: The zero subscript is called "nought" pronounced " not " so the initial point ( x₀ , y₀ ) is called (ex-not, why-not). Mathematicians when asked "why", have often answered " y₀? "

So, if we're given A(– 3, 6) and B(4, –7)

x ₀ = – 3 and y ₀ = 6.

x ₁ = 4 and y ₁ = – 7

Note that ( x ₀ , y ₀ ) is A and

( x ₁ , y ₁ ) is B.

The Slope of a Line

The y-axis measures movement up and down and the x-axis measures movement left and right, so we say that a change in y measures rise and a change in x measures run.

If the change in y is a drop instead of a rise, we call it a negative rise.

Similarly, if a change in x is from right to left, we call it a negative run.

Example

Find the slope of the line segment with endpoints A (5, 2) and B(– 3, – 4).

If we let A be P₀, and B be P₁ then: slope of AB =

If we reverse and let B be P₀, and A be P₁ then: slope of AB =

We can see that the slope is the same no matter which way we go.

However, we must be consistent -- we can't change direction in mid stream.

So, if we said y₀ is the y coordinate of A , then we must consider x₀ to be the x coordinate of A.

Note that when we started at A and went to B, both the rise and the run were negative numbers. That's because B is below and left of A.

When we go the opposite way, we get positive values for both the rise and the run, since A is above and right of B.

Example

Find the slope of the line through T(a, b) and R(c, d).

Solution

The slope of TR is

Example

Find the slope of the line through A( x, y) and B(– 3, – 7).

Solution

The slope of AB is

Note: A line parallel to the x-axis has a slope of 0 since there is no change in y.
This makes the numerator of the slope fraction zero.

A line parallel to the y-axis has an undefined slope since the denominator of the slope fraction is zero.

Finding the Equation of a Line

1/ The Point/Slope Method

point D

In the previous example, we found the slope of the line joining A(5, 2) to B(–3,– 4) is .

This means that to get from any point on the line to any other point on the line, the ratio of rise to run must be 3 : 4. For every 3 units of rise, there must be 4 units of run. So, 1.5 units of rise corresponds to 2 units of run; 9 units of rise corresponds to 12 units of run.

In the diagram, CD is 3 units and BC is 4, so D is a point on AB.

Similarly, PA is 6 units and BP is 8 units.

This illustrates the fact that the slope of a line is constant.

Since the slope between any two points on a straight line equals the slope between any other two points on the same line,

to write the equation of a line given a point and the slope of the line:

make an algebraic statement (an equation) which states that

the slope between any point (x, y) and the given point (x ₀, y ₀)

equals the slope of the line.

Example

Let's find the equation of the the line AB in the previous example.

(5, 2) is a point on AB and we found that its slope = ¾,

so we're looking for the equation of a line through the point (5, 2) with slope = 3/4.

Solution

Let (x, y) be the coordinates of any point on the line.

The slope between (x, y) and (5, 2) = ¾,

so we write that as an algebraic statement like this:

because represents the slope between the points (x, y) and (5, 2).

Now, we have the point/slope or "generic" form of the equation of the line.

What we do now depends on what we have to do next.

If we want the general form of the equation, we cross multiply, collect terms to get

4(y – 2) = 3(x – 5) becomes 3x – 4y – 7 = 0.

If we want the standard or function form (or slope/y-intercept form)

we multiply through by (x – 5) and transpose the –2 to get

Both of these are the equation of the line AB.

This means that every point on this line will have coordinates (x, y) such that if you substitute the x and y values into this equation, it will be a true statement.

Let's see if (–3, –4) satisfies the first equation, since (–3, –4) is also a point on AB.

Set x = –3 and y = –4. The equation becomes 3(–3) – 4(–4) – 7 = – 9 + 16 – 7 = 0.

This proves that (–3, –4) is a point on the line.

In the last diagram, point D with coordinates (1, –1) is on the line AB..

Let's test D's coordinates in the 2nd equation we found.

so D (1, –1) is also a point on AB.

This is known as the point/slope method of finding the equation of a line,

since we are given P(x₀, y₀), a point on the line and m, the slope of the line.

This is the fundamental method to use unless that line is parallel to one of the axes. This is the most efficient way to find the equation of a line. Once we have the equation, we can put it into whatever form is needed be it function, general or symmetric form.

2/ The Two Point Method

This method, almost identical to the point/slope method, is used when we are given the coordinates of two points on the line whose equation we wish to find.

Once we find the slope between the two points, we're right back to the point/slope situation. So, really, this method is identical to the first one except, here, we have to find the slope first.

Example

Let's find the equation of the line AB again from the previous example.

We know two points on the line: A(5, 2) and B(–3, –4), so we define any other point P(x, y) and write an equation which says: slope PA = slope AB. The equation is:

Once we cross multiply and collect terms, we end up with exactly the same equation we had before: 3x – 4y – 7 = 0.

Our results would have been exactly the same had we used the coordinates of B instead of those of A on the left side of the equation.

Notice that we're doing exactly the same as we did in the point/slope method, only here, we have to calculate the value of the slope by finding .

Example

(a) Find the equation of the line through P(–2, 7) with slope –3/5.

(b) Find the equation of the line through P(4, 13) and Q(–4, 1).

(c) Determine whether the point R(3, 4) is on the lines in (a) and (b).

Solution

(a) 5(y – 7) = –3(x + 2) which is 3x + 5y – 29 = 0

(b) 3(x – 4) = 2(y – 13) becomes 3x – 2y + 14 = 0.

(c) Test R in 3x + 5y – 29 = 0 we get 3(3) + 5(4) – 29 = 0 ü

Test R in 3x – 2y + 14 = 0 we get 3(3) – 2(4) – 29 = –28 0

So R(3, 4) is a point on the line in (a) but not in (b).

By "satisfy the equation" we mean the equation becomes a true statement when the values for x and y are substituted into it.

For instance, 3x – 2y = 11 is true at A(5, 2) since 3(5) – 2(2) is equal to 11.

However it is not true at B(5, 7) since 3(5) – 2(7) is not equal to 11.

Geometrically, this means that A is a point on the line 3x – 2y = 11 but B is not.

Practice

1) Find the slopes of AB, BC, CD and AD.

A(–3, 2)

B(4, 7)

C(–2, –5)

D(3, –6)

2) Write the equations (in Ax + By + C = 0 form wherever possible) of the line:

a) through (–1, 6) with slope –3/7	b) through (2, –3) and (–1, –5)
c) through ( 2, 5) and (2, –17)	d) through (–3, 4) and (4, 4)
e) through (3, 0) and (0, –6)

Solutions

1)Find the slopes of AB, BC, CD and AD.

slope of AB =

slope of BC =

slope of CD =

slope of AD =

2) Write the equations (in Ax + By + C = 0 form wherever possible) of the line:

a) 3x + 7y – 39 = 0

b)2x – 3y – 13 = 0

c) since both abscissas = 2, the equation is x = 2

d) since both ordinates = 4, the equation is y = 4

e) 2x – y – 6 = 0

( Analytic Geometry MathRoom Index )

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