Section 6.5: Slope-Point form of equation of linear function
Slope-intercept form
Slope-point form
Y=mx + b
y – y1 = m(x-x1)
Slope-point form is useful in some situations. [like test marks]
BUT! When you don’t have a visible y-intercept … you have to know a
different method to use. 
By rearranging the slope equation, we find a way to write out the equation
of the above graph … without knowing the y-intercept.
𝑆𝑙𝑜𝑝𝑒 =
𝑚=
𝑟𝑖𝑠𝑒
𝑦2 − 𝑦1
=
𝑟𝑢𝑛
𝑥2 − 𝑥1
𝑦2 − 𝑦1
𝑥2 − 𝑥1
𝑚 x (𝑥2 − 𝑥1) =
Multiply both sides by (x2-x1)
… just to get rid of the fraction
𝑦2−𝑦1
𝑥2−𝑥1
x (𝑥2 − 𝑥1)
becomes
m(x2 – x1) = y2 – y1
then flip 180o to have the y in front
y2 –
y1 = m(x2 – x1)
 compare to equation at top of these notes: point-slope form is
where you use ANY 2 points and the slope to write the equa
tion.
 Textbook says equation is: y-
y1 = m(x-x1)
Perfect! When you put numbers into this equation,
just LEAVE IN the y and the x.
For graph above, ONE point is (3, 9) and slope is -5. (I calculated it)
So equation is:
y
-9 = m(x
-3)
y – y1 = m(x – x1)
Example 1 page 367: Graphing from slope-point form
Equation given: y – 2 =
(a)
(b)
1
3
(x + 4)
Describe the graph of this equation.
Graph this equation
Solution: (a) WELL! It’s not in slope-intercept form.
Is it in point-slope form?
y2 – y1 = m(x2 – x1)
y–2=
1
3
(x + 4)
Almost.
BUT … it’s a + 4 not a –4 This is REALLY important! The point-slope equation
says – X1.
So rewrite to make it – X1 .
1
y – 2 =3 (x – (- 4))
This still keeps it a +4, too, overall!
y2 – y1 = m(x2 – x1)
So to ‘describe’: this graph has a slope of
1
3
& a point at (-4, +2)
Yes! +2 … look!!
You only write the 2;
Not the –ve.
Solution (b) To graph, just put the point on a grid … and use the slope to
put another point up 1 and right 3, then draw the line.
Example 2 page 368: Writing an equation when given only 1 point on the line and
the slope of the line.
(a)
Write an equation in slope-point
form for this line.
(b)
Write the equation in slopeintercept form and sate the yintercept.
Soln (a) Pick coordinates of 1 point, then calc slope and plug #’s into the equation.
Pick point (-1, -2)
𝑟𝑖𝑠𝑒
Slope is
𝑟𝑢𝑛
3
=4
Formula:
y
–y
Sub #’s:
y – (-2) =
Simplify:
y+2=
3
4
3
4
1
= m(x – x1)
(x – (-
(x +
1))
1)
Soln (b): Either remake equation by looking at graph … or just rewrite
slope-point form into slope intercept form. (Unless you can tell the
EXACT y-intercept and slope from the graph, you must choose the
rewrite me
thod)
y+2=
3
(x + 1)
3(𝑥 + 1)
𝑦+2=
4
4
3𝑥 + 3
𝑦+2=
4
3𝑥 + 3
4(𝑦 + 2) =
x4
4
 aiming for y = mx + b
spread the 3 through
Multiply both sides by 4
4𝑦 + 8 = 3𝑥 + 3
4𝑦 + 8 − 8 = 3𝑥 + 3 − 8
Minus 8 both sides
4𝑦 = 3𝑥 − 5
4𝑦 3𝑥 − 5
=
4
4
3
5
𝑦= 𝑥−
4
4
Divide by 4, both sides
Thus:
3
𝑦 = 4 𝑥 − 1.2
5.
And the y-intercept is -1.25.
Example 3 page 369: Writing the equation if you only have 2 points; no
line.
Statement: The sum of the angles, s degrees, in a polygon is a linear function of the number of
sides, n, of the polygon. The sum of the angles in a triangle is 180 o. The sum of the angles in a
quadrilateral is 360o.
Question: (a) write a linear equation to represent this function.
(b) use the equation to determine the sum of the angles in a dodecagon.
Soln: First, translate each sentence into math.
See how it says that
“s is a linear function of n”
Follow the words & write
s
Rewrite in regular way
f(n)
A triangle
A quadrilateral
has 3 sides and sum is 180o .
has 4 sides & sum is 360o.
= f(n)
=s
f
(3)
f(4
= 180
) = 360
So we can see that they told us 2 points: (________) & (________)
Textbook graphs them, but not necessary. We were TOLD that it is linear.
Find slope and use point-slope to write equation.
𝑠𝑙𝑜𝑝𝑒=
𝑦2 − 𝑦1
360 − 180
180
=
= (
)
𝑥2 − 𝑥1
4−3
1
So slope = 180.
= 180
So equation is now: y – y1 = 180(x – x1)
Now pick a point to use in point-slope equation. Then simplify.
Formula:
y – y1 = 180(x – x1)
I picked pt (3, 180) to use.
y – (+180) = 180 (X – (+3))
y – 180 = 180 (x – 3)
y – 180 = 180 (x-3)
y – 180 = 180x – 540
+180
+180
Get: y = 180x - 360
Use ‘s’ and ‘n’ in place of x and y: since SUM depends on SIDE # … y is s
and x is n. then: S = 180n – 360.
Solution (b):
Formula:
Sub #’s:
Solve:
“dodecagon” so see that it has 12 sides.
Then sub number in to formula.
s=1
s = 180(1
s = 18
80n – 360
2) – 360
00
The sum of the angles in a dodecahedron is 1800o.
Example 4 page 370: writing an equation of a parallel or perpendicular to
a line.
a)
Write an equation of a line that passes through point
2
R(1, -1) and is parallel to 𝑦 = 3 𝑥 − 5
b)
Write an equation of a line that passes through point
2
R(1, -1) and is perpendicular to 𝑦 = 3 𝑥 − 5
Solution: (a) Parallel means SAME SL
OPE. Use point-slope form
equation then put in slope and new point. Can later rearrange into slopeintercept form if you want.
Formula:
y – y1 = m(x – x1)
sub:
y – y1 = 3(x – x1)
and:
y – (-1) = 3 (x – (1))
then:
y + 1 = 3(x – 1)
2
2
2
Solution: (b) Perpendicular means INVERSE, NEGATIVE SLOPE . Use pointslope form equation then put in NEW slope and new point. Can later
rearrange into slope-intercept form if you want.
Formula:
sub:
y – y1 = m(x – x1)
3
y – y1 = − 2(x – x1)
3
and:
y – (-1) = − 2 (x – (1))
then:
y + 1 = − 2(x – 1)
3
Assignment:
Pages 367 – 370
Pages 372 – 374
[CYU 1-4]
[# 4ace, 5, 6, 7ac, 8, 9(for b, rewrite into slopeintercept form … don’t just guess y-intercept from
graph), 10, 11ab, 12, 14, 16, 19, 20, 21.]
** Except for #9b, stick to point-slope form as much as possible, for the
practice.