Download 2.4 More on Slope

Section 2.4 Notes Page 1
2.4 More on Slope
Parallel lines have the same slope. These lines do not cross.
Perpendicular lines have opposite reciprocal slopes (opposite sign and one fraction is flipped over).
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Ex:
and  are opposite reciprocals. Also  and 2 are opposite reciprocals. Perpendicular lines will
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cross at a right angle (90 degrees).
EXAMPLE: Find the slope of a line that is perpendicular to y = -4x – 3.
The slope of this line is -4, and because it says perpendicular, we need to find the opposite reciprocal. The
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number -4 can be rewritten as the fraction  . Because it is a negative, the opposite sign will be positive. If
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we flip over the fraction we get , which is our answer.
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EXAMPLE: Use the given conditions to write an equation for each line in point-slope form and slope-intercept
form.
a.) Passing through (-2, -7) and parallel to the line whose equation is y  5 x  4
Since we want a line parallel, then this will have the same slope as our given equation, so we know m = -5. We
are also given a point. From here, we will plug this information into the point-slope formula. You should get:
y  (7)  5( x  (2)) . Simplifying gives you y  7  5( x  2) . This is our first answer. To get this into the
slope-intercept form, we need to solve for y. First distribute: y  7  5 x  10 . Now subtract 7 from both
sides: y  5 x  17 . This is our answer in slope-intercept form.
b.) Passing through (-4, 2) and perpendicular to the line whose equation is y 
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x7
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Since we want a line perpendicular, then this will have the opposite reciprocal slope as our given equation, so
we know m = -3. We are also given a point. From here, we will plug this information into the point-slope
formula. You should get: y  2  3( x  (4)) . Simplifying gives you y  2  3( x  4) . This is our first
answer. To get this into the slope-intercept form, we need to solve for y. First distribute: y  2  3x  12 .
Now add 2 to both sides: y  3 x  10 . This is our answer in slope-intercept form.
c.) Passing through (5, -9) and perpendicular to the line whose equation is x  7 y  12  0
Since we want a line perpendicular, then this will have the opposite reciprocal slope as our given equation. We
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first need to solve our given equation for y: y   x  . The slope of this line is  . A line perpendicular
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to this one will have a slope of m = 7. From here, we will our slope and given point into the point-slope
formula. You should get: y  (9)  7( x  5) . Simplifying gives you: y  9  7( x  5) . This is our first
answer. To get this into the slope-intercept form, we need to solve for y. First distribute: y  9  7 x  35 .
Now subtract 9 from both sides: y  7 x  44 . This is our answer in slope-intercept form.
Section 2.4 Notes Page 2
General Form: Ax + By + C = 0. In the general form, everything is set equal to zero. The constants, A, B,
and C should be written as integers if possible, and A must be written as a positive number.
EXAMPLE: Given y  9  7 x  35 , write this in general form.
In order to solve this, we must bring everything over to one side of the equation and set it equal to zero. I will
move everything from the right side to the left: y  9  7 x  35  0 . Now simplify and write the x and y terms
first:  7 x  y  44  0 . The A constant must be written as a positive number, so we will multiply the whole
equation by -1 to get: 7 x  y  44  0 .
Average Rate of Change (A.R.C.)
The A.R.C. is an estimate of the slope between x and c. Basically how much does something change between x
and c. The formula is as follows and is derived from the slope formula, much like the difference quotient.
y f ( x 2 )  f ( x1 )

x
x 2  x1
EXAMPLE: Find the A.R.C. for the f ( x)  3 x 2  3 from x1  0 to x 2  2 .
We can substitute these numbers into our general A.R.C. formula:
f (2)  f (0)
Now we will work this out. Find f (2) and f (0)
20
93
 6 So the slope is -6 between the x values of 0 and 2.
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EXAMPLE: Find the A.R.C. for the f ( x)  x 3  x  2 from x1  1 to x 2  3 .
We can substitute these numbers into our general A.R.C. formula:
f (3)  f (1)
3 1
26  2
 12
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Now we will work this out. Find f (3) and f (1)
So the slope is 12 between the x values of 1 and 3.
EXAMPLE: Find the A.R.C. for the f ( x)  x from x1  9 to x 2  16 .
We can substitute these numbers into our general A.R.C. formula:
f (16)  f (9)
Now we will work this out. Find f (16) and f (9)
16  9
43 1
1

So the slope is between the x values of 9 and 16.
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