Graphing Linear Equations, PointSlope Form, and
Parallel/Perpendicular lines
REVIEW
Algebra Honors
Mr Smith
Objective
• By the end of this lesson you should be
able to take given points or slope and
graph and write it in point-slope, slopeintercept, and standard forms.
• By the end of this lesson you should be
able to write equations for parallel and
perpendicular lines.
Slope- Intercept form
• To _______, it is best to have the linear
equation in slope-intercept form.
• Slope intercept form is y = ___ + ___
• The 4 variables and definitions of slope int
form are:
___ = _______
____ = ________
Pause!
___ = _______
____ = ________
Slope- Intercept form
• To _Graph_, it is best to have the linear
equation in slope-intercept form.
• Slope intercept form is y = mx + b
• The 4 variables and definitions of slope int
form are:
_y_ = total
_m_ = slope_
x = units_
b = y - intercept
Writing in slope- intercept
• Slope = 5, y-int = -7
• 4x + 3y = 18
(-4, 6) (3, -8)
Pause!
Writing and Graphing in slopeintercept form
1. Slope = 5, y-int = -7
2. 4x + 3y = 18
-4x
-4x
3y = -4x + 18
3
3
y=
−4
x
3
+6
y = 5x - 7
3. (-4, 6) (3, -8)=
m=
−14
=
7
−8 −6
3 −(−4)
-2
y=mx+b
6=-2(-4)+b
6=8+b
b = -2
y=-2x-2
Graphing
For the following, the first slide is the problem,
the second is the solution
The first step, which we just did, is to put the
info into slope-int form, and then graph.
Now, Graph y = 5x - 7
Graph y = 5x - 7
Graph y =
−4
x
3
+6
Graph y =
−4
x
3
+6
Graph y=-2x-2
Now, Graph y=-2x-2
Point-Slope Form
• You can write the equation of a line using
point slope form, even if you do not know the
second point on a line.
Write in point-slope
y – y1 = m (x – x1)
A line passing through point (4, -5) with a slope
of -2
Pause!
Write in point-slope
y – y1 = m (x – x1)
A line passing through point (4, -5) with a slope
of -2
y + 5 = -2 (x – 4)
Next, write this in slope- int, and then standard
form
Pause!
To slope – int
To standard
y + 5 = -2 (x – 4)
y + 5 = -2x + 8
-5
-5
y = -2x + 3
y = -2x + 3
+2x +2x
2x + y = 3
Write in point-slope
y – y1 = m (x – x1)
A line passing through (3, -4) and (-6, -1)
Pause!
Write in point-slope
y – y1 = m (x – x1)
A line passing through (3, -4) and (-6, -1)
1. Find the slope
−1 −−4 3
=
−6 −3 −9
=
−1
3
next, put into slope-int
and standard
1. Pick a point and the slope
y - -4
y+4
−1
= (x – 3)
3
−1
= (x – 3)
3
Pause!
To slope – int
To standard
−1
(x
3
−1
x
3
y+4 =
– 3)
y=
+
y+4=
-4
y=
−1
x
3
−1
x
3
+1
-4
+3
1
x
3
(3)
+
1
x
3
+3
1
x
3
+ y = 3(3)
x + 3y = 9
Parallel and Perpendicular
Parallel lines Have the same slope, but different
y-intercepts
Perpendicular Lines have Opposite Reciprocal
slopes, but could have the same intercept.
Example: a slope of 3 4 has opposite reciprocal
slope of -4 3
Write the equation of a line that is…
Parallel to y = 5x + 6, and goes through (-5, 9)
Slope is the same, so m=5, solve for b:
y = mx+b > 9 = 5(-5) + b
9 = -25 + b
+25 +25
34 = b, so …
y = 5x + 34 is the answer
Write the equation of a line that is…
Perpendicular to y=5x+6, and goes through(-3,6)
Slope is opposite reciprocal, so m = 1
5
y = mx + b > 6 = - (-3) + b > 6
2
5
5
3
5
3
= +
5
3
5
1
5
b
1
5
= b, so… y = - +
Yes, you can have intercepts that are not integers
2
5
5
Things to Remember
• To Graph, you should use slope-intercept
form. Start at the y-intercept, and then plot
the slope from there.
• Moving between forms really comes down to
moving terms around, and watching your
signs.
• Parallel lines have the same slope,
perpendicular lines are opposite reciprocals.