### Given either a point and a slope or two points, how do you find the equation of a line in both point-slope and standard form?

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17 months ago by

Write the point-slope form of an equation of the line that passes
through the given point and has the given slope:

(1, 3), m = -2 and (-6, 3), m = -2/3

Write the standard form of an equation of the line that passes through
the given point and has the given slope:

(2, 13), m = 4 and (8, 2), m = -2/5

Write the point-slope form of an equation of the line that passes
through each pair of points:

(-5, 2), (4, -1) and (4, -2), (8, -2)

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17 months ago by

All of your questions revolve around how to get an equation of a line
from a point and a slope or two points, so let's start with a quick
review of the formulas for a line.

We have two basic forms of a line, the point-slope and the standard
(the standard is sometimes also called the slope-intercept form).
They are:

Standard: y = mx+b
m is the slope
b is the y-intercept (the value of y when x=0)

Point-Slope: (y-y1) = m(x-x1)
m is the slope
(x1,y1) is the given point (some point on the line)

When I was learning this, I found it very useful to look at the
relation between these two forms of the equation of a line. In
fact, the two equations are easily obtained from one another.
How, you ask? Well, let's look at the point slope form and see if we
can show how you get the standard form from it.

Point-Slope:

(y-y1) = m(x-x1)
y-y1 = mx - mx1 distributive property
y = mx - mx1 +y1 additive inverse (y1-y1=0) and adding the
same quantity to both sides (y1 in this case)
y = mx + (y1-mx1) commutative and associative properties of
y = mx + b let b = (y1 - mx1)

Why bother with this, you might ask (and it is a very good question
to ask). Well, it shows that the two forms of the equation of the
line are equivalent and it helps you to see the connection. For
instance, we now have b = y1 - mx1, which is the formula for
calculating the y-intercept of a line from any point on the line and
the slope of the line. We can also note that by taking b to the other
side of the equation we have y-b = mx, which is a special case of the
point-slope form (with the point (0,b)). A final point for why the
relation between the two forms is useful is that it helps you to
remember them (you can get one from the other if you know the
relation).

Enough theory. Let's try some problems:

1) You gave me the following point and slope: (1, 3), m = -2. Let's
find the line in both forms. The point-slope form is easiest to get,
so let's do that first:

We have (x1,y1) = (1,3) and m=-2 and the form is (y-y1) = m(x-x1).

Thus by direct substitution we have y-3 = -2(x-1) as the solution for
the point slope form.

Now we want to get the standard form, we can do the algebra again
(if we forgot the form of b we found above) or we can use the form we
found, namely b = y1-mx1. Let's do both for clarity

With algebra we have:

y-3 = -2(x-1)
y-3 = -2x+2
y = -2x+2+3
y = -2x+5 This is our answer

By formula we have:

b = (y1-mx1)
b = (3-(-2)(1))
b = (3+2)
b = 5

By direct substitution we have y = -2x+5, which is the same.

The point slope form is obviously very easy to get when you have a
point and the slope of the line (which is the whole reason for this
form). The standard form takes a little more work but is very useful
for drawing and analyzing your line. Hence the need for both. You
can use either way you like to go between the two forms. It's all a
matter of personal preference. We will do one more quickly and you
should be able to answer the rest.

Let's get the two forms of the line with point and slope given by:
(8, 2), m = -2/5.

By direct substitution, we can get the point-slope form:

y-2 = (-2/5)(x-8)

By algebra we can get the standard form:

y-2 = (-2/5)x+16/5
y = (-2/5)x+26/5

By substitution we can also get the standard form:

b = y1-mx1
b = 2-(-2/5)8
b = 2+16/5
b = 26/5

So we have y = (-2/5)x+26/5

Now you also had some problems where you were given two points and
asked to give the equation of the line. This requires you to
calculate the slope of the line from these points and then use the
slope and one of the points as we did above to get the equation of a
line. So we need a formula for the slope given two points. There are
several ways to get this. You can remember that the slope is the rise
(change in y) divided by the run (change in x) or you could calculate
this form from the point-slope form of a line.

If we have points (x1, y1) and (x2, y2), by memory we have:

m = (y2-y1)/(x2-x1)

By algebra on point-slope form:

(y-y1) = m(x-x1) which holds for all (x,y) on the line.
In particular, it holds for
(x,y) = (x2,y2)
(y2-y1) = m(x2-x1)

Dividing by x2-x1 (which cannot be zero for distinct points because a
vertical line is not a function and a line must be a function) we get:

m = (y2-y1)/(x2-x1)

Thus we can use our equation for calculating the slope given two
points and then pick either point (it does not matter which) and we
can get the equation of the line as we did in the first part (we now
have a point and the slope). Lets try an example:

You gave the following two points: (-5, 2), (4, -1) So what is the
slope?

m = (y2-y1)/(x2-x1)
m = (-1-2)/(4--5)
m = (-3)/(9)
m = -1/3

Or by algebra on the point-slope equation of a line:

(y2-y1) = m(x2-x1)
(-1-2) = m(4--5)
(-3) = m(9)
-1/3 = m

Then we can use either {(-5,2), m = -1/3} or {(4,-1), m = -1/3} and
solve for the equation of the line in either form as we did before.

I have given you a couple of ways of solving the problems, so let's
put the options in a little step by step review:

Given points (x1,y1) and (x2,y2) find the equation of the line.
Option 1
1) get the slope by m = (y2-y1)/(x2-x1).
2) pick either (x1,y1) or (x2,y2) as the point to use with your
slope.
3) follow the given point and slope steps below for the form you
want

Option 2
1) get the slope by algebraic solution of point-slope equation
(y2-y1) = m(x2-x1). (Note this already has the substitution
(x,y)=(x2,y2).)
2) pick either (x1,y1) or (x2,y2) as the point to use with your
slope.
3) follow the given point and slope steps below for the form you
want.

Given point (x1,y1) and slope m and you want point-slope form:
1) substitute values into the equation (y-y1) = m(x-x1)
2) you're done

Given point (x1,y1) and slope m and you want standard form:
Option 1
1) get point-slope equation through above steps
2) use algebra to manipulate into the form y = mx+b
3) you're done

Option 2
1) calculate the y intercept (b) by b = y1-mx1
2) directly substitute into the formula y = mx+b
3) you're done

As you can see, the most important part of working with lines is
understanding what the relations between the forms are. Each of the
options listed calculate the same thing, some by formula and some by
algebraic manipulation, so it really comes down to what you like and
feel comfortable with. I would suggest trying both and seeing what you
think.