Finding the Equation of a Line Connecting Two Points in the Plane, Study notes of Algebra

How to find the equation of a line that connects two points in the plane using two different methods. The first method involves finding the slope of the line and then using it to derive the equation. The second method involves setting up a linear function that becomes an identity when the coordinates of the points are plugged in. Both methods result in the same equation for the line.

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Pre 2010

Uploaded on 03/10/2009

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Line Through Two Points In The Plane
Math 98
Lines
It is easy to write the equation of the unique line that connects two points
in the plane, but you have to do it right. Hereโ€™s two ways (even though they
are essentially the same).
If you have two points in the plane, say (a, b), and (c, d), it is easy, for
example from a picture, to find the slope of the line connecting them: indeed,
that has to be
m=dโˆ’b
cโˆ’a=bโˆ’d
aโˆ’c
Since a line has an equation of the form
y=mx +k
we only need to find k, and that we can do by โ€œforcingโ€ the line to go through
one of the point (the pother one will also follow, since we have picked the right
slope. For example, we can plug in x=a, y =b, and see that kmust be such
that
b=ma +k=bโˆ’d
aโˆ’ca+k
k=bโˆ’abโˆ’d
aโˆ’c=b(aโˆ’c)โˆ’a(bโˆ’d)
aโˆ’c=ab โˆ’bc โˆ’ab +ad
aโˆ’c=ad โˆ’bc
aโˆ’c
and the equation now becomes
y=mx +k=bโˆ’d
aโˆ’cx+ad โˆ’bc
aโˆ’c=(bโˆ’d)x+ (ad โˆ’bc)
aโˆ’c(1)
You really would not want to memorize (1). It is much faster to retrace the
steps we made to find k!
Another way is to write a linear function that will become an identity when
we plug in the two pairs describing the points. One standard way to do this
is to make sure we get 0 = 0 when, say, x=a, y =b. This means that the
function must look like
A(yโˆ’b) = B(xโˆ’a)
for some numbers Aand B. Now, we make sure that the two sides are both
equal to, say 1, when x=c, y =d. I.e., we need
A(dโˆ’b)=1, B (cโˆ’a) = 1
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Line Through Two Points In The Plane

Math 98

Lines

It is easy to write the equation of the unique line that connects two points in the plane, but you have to do it right. Hereโ€™s two ways (even though they are essentially the same). If you have two points in the plane, say (a, b), and (c, d), it is easy, for example from a picture, to find the slope of the line connecting them: indeed, that has to be

m =

d โˆ’ b c โˆ’ a

b โˆ’ d a โˆ’ c Since a line has an equation of the form

y = mx + k

we only need to find k, and that we can do by โ€œforcingโ€ the line to go through one of the point (the pother one will also follow, since we have picked the right slope. For example, we can plug in x = a, y = b, and see that k must be such that

b = ma + k =

b โˆ’ d a โˆ’ c a + k

k = b โˆ’ a

b โˆ’ d a โˆ’ c

b (a โˆ’ c) โˆ’ a (b โˆ’ d) a โˆ’ c

ab โˆ’ bc โˆ’ ab + ad a โˆ’ c

ad โˆ’ bc a โˆ’ c

and the equation now becomes

y = mx + k = b โˆ’ d a โˆ’ c

x + ad โˆ’ bc a โˆ’ c

(b โˆ’ d) x + (ad โˆ’ bc) a โˆ’ c

You really would not want to memorize (1). It is much faster to retrace the steps we made to find k! Another way is to write a linear function that will become an identity when we plug in the two pairs describing the points. One standard way to do this is to make sure we get 0 = 0 when, say, x = a, y = b. This means that the function must look like A (y โˆ’ b) = B (x โˆ’ a)

for some numbers A and B. Now, we make sure that the two sides are both equal to, say 1, when x = c, y = d. I.e., we need

A (d โˆ’ b) = 1, B (c โˆ’ a) = 1

that is

A =

d โˆ’ b

, B =

c โˆ’ a

so that the equation is y โˆ’ b d โˆ’ b

x โˆ’ a c โˆ’ a

Of course, our choice of 0 = 0, and 1 = 1 was just one of the many possible, and we could also have started working with the other point first. This just underlines that there are several ways in which to write an equation for the same line. Obviously, (1), and (2) define the same line. We can go from one to the other with a few algebraic steps. For instance, starting with (2), we can write

y โˆ’ b =

d โˆ’ b c โˆ’ a (x โˆ’ a)

y = b +

d โˆ’ b c โˆ’ a

(x โˆ’ a)

y =

b (c โˆ’ a) + (d โˆ’ b) (x โˆ’ a) c โˆ’ a

y =

bc โˆ’ ba โˆ’ da + ba + x (d โˆ’ b) c โˆ’ a

y = (bc โˆ’ da) + x (d โˆ’ b) c โˆ’ a

which is precisely (1), when you multiply both numerator and denominator by โˆ’1.