Writing equations given slope and point
5.2
12/6/13
Solving Equations
• In any equation, we can solve for any variable
that is not known
1) Rewrite in terms of that variable
2) Find a value if all the other values are known
3) Can use this to find equations
Point-Slope Form of the
Equation of a Line
• The point-slope equation of a non-vertical
line of slope m that passes through the point
(x1, y1) is
y – y1 = m(x – x1).
Solving With Slope and a Point
1. Substitute known values into the
equation
2. Rewrite the equation y = mx + b
with slope and y-intercept values
Example:
Writing the Point-Slope Equation of a Line
Write the point-slope form of the equation of the line passing through (-1,3) with a slope
of 4. Then solve the equation for y.
Solution We use the point-slope equation of a line with m = 4, x1= -1, and
y1 = 3.
y – y1 = m(x – x1)
This is the point-slope form of the equation.
y – 3 = 4[x – (-1)]
Substitute the given values. Simply.
y – 3 = 4(x + 1)
We now have the point-slope form of the equation for the
given line.
We can solve the equation for y by applying the distributive property.
y – 3 = 4x + 4
Add 3 to both sides.
y = 4x + 7
Solving with Slope and Point
• Find the equation of the line passing through
the point (-3, 0 ) and has slope of 1/3
Solving with Slope and a Point
• Find the equation of the line
passing through point (-2, -1) with
slope of -3
Solving with Slope and Point
• Find the equation of the line
passing through the point (-3, 0 )
and has slope of 1/3
Solving with Slope and a Point
• Find the equation of the line
passing through (3 , -4) and is
parallel to the line y = -3x – 2
Slope and Perpendicular Lines
Two lines that intersect at a right angle (90°)
are said to be perpendicular. There is a
relationship between the slopes of
perpendicular lines.
90°
Slope and Perpendicular Lines
• If two non-vertical lines are perpendicular, then
the product of their slopes is –1.
• If the product of the slopes of two lines is –1,
then the lines are perpendicular.
• A horizontal line having zero slope is
perpendicular to a vertical line having undefined
slope.
Perpendicular Lines
3
m    3
1
1
m
3
Perpendicular lines have negative
reciprocal slopes so if you need the
slope of a line perpendicular to a given
line, simply find the slope of the given
line, take its reciprocal (flip it over) and
make it negative.
Example: Finding the Slope of a Line
Perpendicular to a Given Line
Find the slope of any line that is perpendicular to the line whose
equation is x + 4y – 8 = 0.
Solution
We begin by writing the equation of the given line in
slope-intercept form. Solve for y.
x + 4y – 8 = 0
This is the given equation.
4y = -x + 8
y = -1/4x +
2
To isolate the y-term, subtract x and add 8
on both sides.
Divide both sides by 4.
Slope is –1/4.
The given line has slope –1/4. Any line
perpendicular to this line has a slope that is
the negative reciprocal, 4.
Example: Writing the Equation of a Line
Perpendicular to a Given Line
Write the equation of the line perpendicular to x + 4y – 8 = 0 that passes
thru the point (2,8) in standard form.
Solution: The given line has slope –1/4. Any line perpendicular to this
line has a slope that is the negative reciprocal, 4.
So now we need know the perpendicular slope and are given a point
(2,8). Plug this into the point-slope form and rearrange into the
standard form.
y – y1 = m(x – x1)
y – 8 = 4[x – (2)]
y - 8 = 4x - 8
y1 = 8
m =4
x1 = 2
-4x + y = 0
4x – y = 0 Standard form
Let's look at a line and a point not on the line
Let's find the equation of a line parallel to
y = - x that passes through the point (2, 4)
y=-x
(2, 4)
What is the slope of the
first line, y = - 1x ?
This is in slope intercept
form so y = mx + b which
means the slope is –1.
y  4y1  m
-1 x  x21 
Distribute and then solve for y to
leave in slope-intercept form.
y  x  6
So we know the slope is –1
and it passes through (2, 4).
Having the point and the
slope, we can use the pointslope formula to find the
equation of the line
What if we wanted perpendicular instead of parallel?
Let's find the equation of a line
perpendicular to y = - x that passes
through the point (2, 4)
y=-x
(2, 4)
The slope of the first line is still –
1.
The slope of a line perpendicular
is the negative reciporical so
take –1 and "flip" it over and
make it negative.
y  y4 1  m
1 x  x21 
Distribute and then solve for y to
leave in slope-intercept form.
y  x2
1
1
 1 
 
 1
So the slope of a
perpendicular line is 1 and it
passes through (2, 4).
Homework:
• 5.2 worksheet #’s 1-15
• Due Monday, 12/9