1. Equations of Lines Equations of Lines
Linear Equations
Equations of Lines
This module delves into the various descriptions of lines and their features.
Linear Equations
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2. Lines and Equations   Point-Slope Form
Equations of Lines
Point-Slope Form
Given line L and point (x1, y1) on L
Let (x, y) be any other point on L
Find slope y x L
(x, y) 
(x1, y1)
∆y = y – y1 m = ∆y ∆x 
Point-Slope Form
The point-slope form of a linear equation allows us to write the equation with minimal information: the slope and the coordinates of one point on the line.
In the illustration, we are given one point, namely (x1, y1), and we want a general equation for the line. We use the definition of the slope and any other arbitrary point (x, y) to form y and x. As soon as we have written the slope as the ratio of these two quantities, we have an equation of the line, in the sense that any point (x, y) on the line represents an ordered pair (x, y) that is a solution of the equation, and conversely.
The equation is called the point-slope form because it clearly represents the slope of the line and the coordinates of one point on the line. This equation represents the whole line because we chose (x, y) to be any other point on the line, a way of saying that it represents all other points on the line.
∆x = x – x1
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Linear Equations
Linear Equations
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3. Lines and Equations   Point-Slope Form m = = y – y1 = m(x –x1)
Equations of Lines
Point-Slope Form m = ∆y ∆x y x =
y – y1
x – x1 L
(x, y) 
y – y1 = m(x –x1)
(x1, y1)
∆y = y – y1 
Point-Slope Form
The point-slope form of a linear equation allows us to write the equation with minimal information: the slope and the coordinates of one point on the line.
In the illustration, we are given one point, namely (x1, y1), and we want a general equation for the line. We use the definition of the slope and any other arbitrary point (x, y) to form y and x. As soon as we have written the slope as the ratio of these two quantities, we have an equation of the line, in the sense that any point (x, y) on the line represents an ordered pair (x, y) that is a solution of the equation, and conversely.
The equation is called the point-slope form because it clearly represents the slope of the line and the coordinates of one point on the line. This equation represents the whole line because we chose (x, y) to be any other point on the line, a way of saying that it represents all other points on the line.
The final form shown is sometimes referred to as the modified point-slope form. The original form is simply
∆x = x – x1
Point-Slope Form
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Linear Equations
Linear Equations
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4. Lines and Equations   Slope-Intercept Form
Equations of Lines
Slope-Intercept Form
Consider a non-vertical line L
Locate y-intercept
Let (x, y) be any other point on the line
Find slope m y x L
(x, y) 
Lines and Equations: Slope-Intercept Form
Here, given a non-vertical line L we write the equation of the line in general, using the vertical intercept (y-intercept) as the specified point on the line. We don’t yet know both intercept coordinates, so we write the equation in point-slope form using b and 0. Rewriting this as
mx = y – b
we see that moving b to the other side of the equation produces the standard slope-intercept form
mx + b = y .
∆y = y – b
(0, b) 
∆x = x – 0
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Linear Equations
Linear Equations
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5. Lines and Equations  Slope-Intercept Form m = = m(x – 0) = y – b
Equations of Lines
Slope-Intercept Form m = ∆y ∆x y x L 
(0, b)
(x, y)
∆x = x – 0
∆y = y – b =
y – b
x – 0
m(x – 0) = y – b
Solving for y,
Lines and Equations: Slope-Intercept Form
Here, given a non-vertical line L we write the equation of the line in general, using the vertical intercept (y-intercept) as the specified point on the line. We don’t yet know both intercept coordinates, so we write the equation in point-slope form using b and 0. Rewriting this as
mx = y – b
we see that moving b to the other side of the equation produces the standard slope-intercept form
mx + b = y .
y = mx + b
Slope-Intercept Form
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Linear Equations
Linear Equations
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6. Examples  Point-Slope Form A line through (5, 10) with slope
Equations of Lines
Point-Slope Form
A line through (5, 10) with slope
Find the equation: 4 5 y x m = 4 5 =
y – 10
x – 5
m = 4/5 
Point-Slope Form Example
In this example we are given a specific point and a slope number and asked to find the equation of the line through the given point with the given slope. We can write down the equation immediately in point-slope form, one of the advantages of using this form.
This requires no intermediate calculations. This form is not unique, since the slope 4/5 can also be written as 8/10, or 12/15, or …. Any slope that reduces to 4/5 can be used, meaning there are infinitely many ways to write the equation.
(5,10)
Question:
Is this form unique ?
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Linear Equations
Linear Equations
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7. Examples  Slope-Intercept Form
Equations of Lines
Slope-Intercept Form
A line with intercept (0, 4) and slope -3
Find the equation:
y = mx + b = -3x + 4 y x 
Slope-Intercept Form Example
In this example, we are again given a specific point and a slope value. In this case we are asked to find the equation of the line, in slope-intercept form, through the point with the given slope. In this case, since the given point is the y-intercept, we can immediately write down the equation as
y = mx + b
the typical slope-intercept form. This form is unique, since there is only one value that will fit the slope number and the coefficient of y is simply 1.
Note that if, in the second example, the given point were not the y-intercept, we would have to do some calculations to arrive at the slope-intercept form.
(0, 4)
Question:
Is this form unique ?
m = -3
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Linear Equations
Linear Equations
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8. Lines and Equations Standard Form
Equations of Lines
Standard Form
Algebraic form not directly graph related
Useful for systems of linear equations
For constants A, B and C, with B ≠ 0
Ax + By = C
Can we still find slope and intercepts ?
Standard Form
Any linear equation can written in standard form, so equations in either point-slope form or slope-intercept form can be converted to standard form. The big advantage for standard form becomes apparent when dealing with systems of linear equations, especially those with more than two variables. This form is then adaptable to the matrix techniques of linear algebra.
Question:
Is this form unique ?
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Linear Equations
Linear Equations
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9. Lines and Equations a ≠ A Standard Form – Slope m
Equations of Lines
Standard Form – Slope
Rewriting the equation:
This has slope-intercept form with y = x A B – + C
Note:
a ≠ A
= mx + b
b ≠ B ,
Standard Form
Any linear equation can written in standard form, so equations in either point-slope form or slope-intercept form can be converted to standard form. The big advantage for standard form becomes apparent when dealing with systems of linear equations, especially those with more than two variables. This form is then adaptable to the matrix techniques of linear algebra.
Note that the slope and the coordinates of both intercepts can be read directly from the coefficients in the equation. The form is not unique, since all the coefficients can be multiplied by any constant k to produce an equivalent equation.
If B = 0 then the equation becomes
Ax = C or x = C/B
which means x is constant and the graph is a horizontal line.
Also note the constants B and b are do not represent the same number. As in all of mathematics, upper and lower case characters are generally used to represent different things.
Slope = A B – = m
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Linear Equations
Linear Equations
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10. ( ) Lines and Equations Standard Form – Intercepts m
Equations of Lines
Standard Form
Rewriting the equation:
– Intercepts y = x A B – + C
= mx + b
So … = m A B –
Note:
b ≠ B C B b =
Standard Form
Any linear equation can written in standard form, so equations in either point-slope form or slope-intercept form can be converted to standard form. The big advantage for standard form becomes apparent when dealing with systems of linear equations, especially those with more than two variables. This form is then adaptable to the matrix techniques of linear algebra.
Note that the slope and the coordinates of both intercepts can be read directly from the coefficients in the equation. The form is not unique, since all the coefficients can be multiplied by any constant k to produce an equivalent equation.
If B = 0 then the equation becomes
Ax = C or x = C/B
which means x is constant and the graph is a horizontal line.
Also note the constants B and b are do not represent the same number. As in all of mathematics, upper and lower case characters are generally used to represent different things. = b m – C B A
( ) = C A
Fractions
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Linear Equations
Linear Equations
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11. ( ) ( ) ( ) Lines and Equations Standard Form – Intercepts a ≠ A
Equations of Lines
Standard Form
Rewriting the equation:
Intercepts
– Intercepts y = x A B – + C
Note:
a ≠ A
( ) 0, C B =
y-intercept
= (0, b)
b ≠ B
Standard Form
Any linear equation can written in standard form, so equations in either point-slope form or slope-intercept form can be converted to standard form. The big advantage for standard form becomes apparent when dealing with systems of linear equations, especially those with more than two variables. This form is then adaptable to the matrix techniques of linear algebra.
Note that the slope and the coordinates of both intercepts can be read directly from the coefficients in the equation. The form is not unique, since all the coefficients can be multiplied by any constant k to produce an equivalent equation.
If B = 0 then the equation becomes
Ax = C or x = C/B
which means x is constant and the graph is a horizontal line.
Also note the constants B and b are do not represent the same number. As in all of mathematics, upper and lower case characters are generally used to represent different things.
( ) = b m – , =
( ) C A
, 0
(a, 0) =
x-intercept
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Linear Equations
Linear Equations
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12. Standard Form Example  Intercepts with Standard Form
Equations of Lines
Intercepts with Standard Form
Consider equation
6x + 5y = 30
When x = 0 ,
5y = 30
y = 6
Vertical intercept is (0, 6) x y 
(0, 6)
Finding Intercepts from Standard Form
Given the equation in standard form, it is a simple matter to find the intercepts by letting first one variable, and then the other, be zero. As shown, this easily establishes both intercepts.
By simply remembering that intercepts always occur on one axis or the other, we know that one of the variables is zero. By setting each variable to 0 and solving the remaining equation, we immediately see the coordinates of each of the intercepts. This reinforces the concept of the intercept as a point, not a number.
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Linear Equations
Linear Equations
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13. Standard Form Example   Intercepts with Intercept Form
Equations of Lines
Intercepts with Intercept Form
Consider equation
6x + 5y = 30
When y = 0,
6x = 30
x = 5
Horizontal intercept is (5, 0) x y 
(0, 6)
(5, 0) 
Finding Intercepts from Standard Form
Given the equation in standard form, it is a simple matter to find the intercepts by letting first one variable, and then the other, be zero. As shown, this easily establishes both intercepts.
By simply remembering that intercepts always occur on one axis or the other, we know that one of the variables is zero. By setting each variable to 0 and solving the remaining equation, we immediately see the coordinates of each of the intercepts. This reinforces the concept of the intercept as a point, not a number.
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Linear Equations
Linear Equations
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14. ( ) Lines and Equations Intercept Form +
Equations of Lines
Intercept Form
Consider standard form equation
Ax + By = C
For C ≠ 0 , this becomes A C x + B y = 1
Finding Intercepts from the Intercept Form
Given the equation in standard form, it is a simple matter to find convert it to intercept form by dividing by C, assuming of course the C ≠ 0. This yields
and by multiplying and dividing each term by its coefficient we can rewrite as
which is known as the intercept form. It is easy to see that the fraction dividing x is the x-coordinate of the x-intercept by just letting y = 0 and solving for x. Similarly, letting x = 0 and solving for y yields the y-coordinate of the y-intercept. Thus, the intercepts are
where a and b are the usual x and y coordinates for the intercepts.
The above calculations assume that C ≠ 0. But what if C = 0? Then either the line crosses both axes at the origin, that is, the intercepts are the same point, or the line lies on one of the axes (x-axis if A = 0, or y-axis if B = 0). OR
intercept form = 1 ( ) C A x + y B
Fractions
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Linear Equations
Linear Equations
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15. ( ) ) ( ( ) Lines and Equations   Intercept Form + + a
Equations of Lines
Intercept Form
Consider intercept form equation ( ) C A x y B + = 1 a x + b y = 1 x y
Note:
A ≠ a 0, C B ) ( 
( )
0, b =
Finding Intercepts from the Intercept Form
Given the equation in standard form, it is a simple matter to find convert it to intercept form by dividing by C, assuming of course the C ≠ 0. This yields
and by multiplying and dividing each term by its coefficient we can rewrite as
which is known as the intercept form. It is easy to see that the fraction dividing x is the x-coordinate of the x-intercept by just letting y = 0 and solving for x. Similarly, letting x = 0 and solving for y yields the y-coordinate of the y-intercept. Thus, the intercepts are
where a and b are the usual x and y coordinates for the intercepts.
The above calculations assume that C ≠ 0. But what if C = 0? Then either the line crosses both axes at the origin, that is, the intercepts are the same point, or the line lies on one of the axes (x-axis if A = 0, or y-axis if B = 0).
B ≠ b
, 0 C A ( )
Question:
(a, ) = 
What if C = 0 ?
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Linear Equations
Linear Equations
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16. Standard Form Example   Finding Intercepts with Intercept Form ( )
Equations of Lines
Finding Intercepts with Intercept Form
Consider equation
6x + 5y = 30
Vertical intercept is (0, 6)
Horizontal intercept is (5, 0) x y 
(0, 6) 30 1
6x + 5y
( )
( ) = x 5 y 6 + = 1
(5, 0) 
Finding Intercepts from Standard Form
Given the equation in standard form, it is a simple matter to find the intercepts by letting first one variable, and then the other, be zero. As shown, this easily establishes both intercepts.
By simply remembering that intercepts always occur on one axis or the other, we know that one of the variables is zero. By setting each variable to 0 and solving the remaining equation, we immediately see the coordinates of each of the intercepts. This reinforces the concept of the intercept as a point, not a number.
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Linear Equations
Linear Equations
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17. Horizontal and Vertical Lines
Equations of Lines
Horizontal Lines
Form: y = k
for some constant k
From standard form Ax + By = C when A = 0, B ≠ 0 C B y =
Horizontal Lines
Here we show why the slope of a horizontal line is 0. It should be emphasized that this does NOT mean there is no slope. The slope of 0 is a perfectly good slope number.
The lines y = 0 and x = 0 are in fact the x-axis and y-axis, respectively.
Question:
What is the line y = 0 called ?
The x-axis !
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Linear Equations
Linear Equations
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18. Horizontal and Vertical Lines
Equations of Lines
Horizontal Lines
Example: y = 3
Pick any points (x1, 3) , (x2, 3)
Slope m is then
Note:
Zero slope is NOT the same as no slope x y
(x1, 3)
(x2, 3)  
y = 0
y = 3   m = y x
3 – 3
x2 – x1 = x1 x2
Horizontal Lines
Here we show why the slope of a horizontal line is 0. It should be emphasized that this does NOT mean there is no slope. The slope of 0 is a perfectly good slope number.
The lines y = 0 and x = 0 are in fact the x-axis and y-axis, respectively.
x = x2 – x1
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Linear Equations
Linear Equations
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19. Horizontal and Vertical Lines
Equations of Lines
Vertical Lines
Form: x = k for some constant k
From standard form Ax + By = C when A ≠ 0, B = 0
Then Ax = C
Vertical Lines
Here again we attempt to compute the slope of the vertical line and find we must divide by 0 to do it. Since division by 0 is not allowed in our number system, we cannot compute a slope for a vertical line. This line truly has no slope.
One might ask why we are not allowed to divide by 0. The answer is not that we aren’t intelligent enough, but that we must assign some number to the result. It is easily shown that any number we might assign, including 0 itself, leads to one or more contradictions. That is, we can then prove something to be true that is already known (and proved) to be false. Hence, as a matter of mathematical philosophy, we choose not to divide by 0. A C x =
Question:
What is the line x = 0 ?
The y-axis !
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Linear Equations
Linear Equations
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20. Horizontal and Vertical Lines
Equations of Lines
Vertical Line
Example: x = 4
Pick any points (4, y1) , (4, y2)
Slope m is then x y
x = 4   y2
(4, y2) y   y1
(4, y1) m = y x =
4 – 4
y2 – y1
x = 0
Vertical Lines
Here again we attempt to compute the slope of the vertical line and find we must divide by 0 to do it. Since division by 0 is not allowed in our number system, we cannot compute a slope for a vertical line. This line truly has no slope.
One might ask why we are not allowed to divide by 0. The answer is not that we aren’t intelligent enough, but that we must assign some number to the result. It is easily shown that any number we might assign, including 0 itself, leads to one or more contradictions. That is, we can then prove something to be true that is already known (and proved) to be false. Hence, as a matter of mathematical philosophy, we choose not to divide by 0. =
y2 – y1
Undefined !
Note:
No slope is not the same as zero slope !
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Linear Equations
Linear Equations
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21. Parallel Lines Parallel Lines Horizontal Lines
Equations of Lines
Parallel Lines
Horizontal Lines
Zero slope, always parallel
Vertical Lines
No slope, always parallel
Other lines
Lines with same slope, always parallel
Parallel Lines
We note that the key feature of parallel lines is the same slope, or in the case of vertical lines, no slope.
In the illustration we find the equation of a line, given a point on the line and the equation of a parallel line. The parallel line gives us the slope of the line we must find. With the given point, we find the equation of the line just as we did in finding the equation of a line given its slope and a point on the line.
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Linear Equations
Linear Equations
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22. Parallel Lines  Parallel Lines Example
Equations of Lines
Parallel Lines Example
Find the equation of the line through
(4, 10) parallel to the line
Slope of new line is
Point-slope form is y 1 2 – =
x + 6 1 2 – x y
(4, 10)  =
y – 10
x – 4 1 2 –
Parallel Lines
We note that the key feature of parallel lines is the same slope, or in the case of vertical lines, no slope.
In the illustration we find the equation of a line, given a point on the line and the equation of a parallel line. The parallel line gives us the slope of the line we must find. With the given point, we find the equation of the line just as we did in finding the equation of a line given its slope and a point on the line.
y = – (½)x + 12
y = – (½)x + 6 OR =
y – 10
(x – 4) 1 2 –
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Linear Equations
Linear Equations
7/2/2013

23. Perpendicular Lines in General
Equations of Lines
Step 1
Geometry gives us x y L1 m1 = b1 a ,
b1 = am1 m2 c1 b1
am1 =
c12 = a2 + b12 a b2
– am2 = c2 m2 =
– b2 a
Perpendicular Lines in General
The relationship between perpendicular lines is a little more complex than that between parallel lines. Given perpendicular lines L1 and L2 with slopes m1 and m2, respectively, as shown in the illustration.
Pick a point on line L1 above the intersection point and drop a vertical line from this new point to the x-axis. Project a horizontal line through the intersection of L1 and L2 as shown. We can now form three distinct triangles.
The horizontal segment marked a forms a leg of two distinct triangles, as shown. We apply the Pythagorean Theorem to the upper triangle, with leg b1 lying on the vertical projection line and hypotenuse c1 lying on line L1, giving
c12 = a2 + b12
The lower triangle is similarly formed using the same leg a, leg b2 on the vertical projection line, and the hypotenuse c2 on line L2 . This gives us
c22 = a2 + b22
The third triangle, with legs c1 and c2 and hypotenuse b1 + b2 give us
c12 + c22 = (b1 + b2)2
Combining this with the first two equations we have
c12 + c22 = a2 + b12 + a2 + b22 = (b1 + b2)2
leading to
b12 + 2b1b2 + b22 = b12 + 2a2 + b22
b2 = – am2 , m1 L2
c22 = a2 + b22
c12 + c22 =
b12
+ 2a2 + b22
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Linear Equations
Linear Equations
7/2/2013

24. Perpendicular Lines in General
Equations of Lines
Step 2 x y
c12 + c22
= b12
+ b22
+ 2a2 L1
c12 + c22 = (b1 + b2)2 m2 c1
= b12 + 2b1b2 + b22 b1
b1 + b2 a
2a2 =
2b1b2 b2 c2
= 2(am1)(– am2)
Perpendicular Lines in General
The relationship between perpendicular lines is a little more complex than that between parallel lines. Given perpendicular lines L1 and L2 with slopes m1 and m2, respectively, as shown in the illustration.
Pick a point on line L1 above the intersection point and drop a vertical line from this new point to the x-axis. Project a horizontal line through the intersection of L1 and L2 as shown. We can now form three distinct triangles.
The horizontal segment marked a forms a leg of two distinct triangles, as shown. We apply the Pythagorean Theorem to the upper triangle, with leg b1 lying on the vertical projection line and hypotenuse c1 lying on line L1, giving
c12 = a2 + b12
The lower triangle is similarly formed using the same leg a, leg b2 on the vertical projection line, and the hypotenuse c2 on line L2 . This gives us
c22 = a2 + b22
The third triangle, with legs c1 and c2 and hypotenuse b1 + b2 give us
c12 + c22 = (b1 + b2)2
Combining this with the first two equations we have
c12 + c22 = a2 + b12 + a2 + b22 = (b1 + b2)2
leading to
b12 + 2b1b2 + b22 = b12 + 2a2 + b22 m1 L2
= – 2a2m1m2
1 = – m1m2 – 1 m2 m1 =
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Linear Equations
Linear Equations
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25. Perpendicular Lines Example
Equations of Lines
Example
Find the equation of the line through (2, 3) perpendicular to line
y = –(⅓)x + 3
Slope of the given line is m1 = –⅓
Slope of the new line is x y 
(2, 3)
y = –(⅓)x + 3
Perpendicular Lines – Example
We use the slope relationship we just derived for perpendicular lines, namely
m2 = –
We are given the line y = (–½)x + 6 we want to find the equation of the line through point (10,10) perpendicular to the given line. We know that the slopes have the negative reciprocal relationship shown above.
Since the given line has slope –½ then the perpendicular line has slope 2. This line goes through point (10,10) has a point-slope equation
y – 10 = 2(x – 10)
which can be rewritten in other forms, such as
y = 2x – 10
or in “standard form” this is
2x – y = –10 . m1 1 1 m1 – m2 = = – 1 ⅓ = 3
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Linear Equations
Linear Equations
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26. Perpendicular Lines Example
Equations of Lines
Example
Slope of the new line is x y
(2, 3) 
y = –(⅓)x + 3 = 3 m2 1 m1 – ⅓
y = 3x – 3
Alternate point-slope form
y – y1 = m(x – x1)
y – 3 = 3(x – 2)
Perpendicular Lines – Example
We use the slope relationship we just derived for perpendicular lines, namely
m2 = –
We are given the line y = (–⅓)x + 3 and want to find the equation of the line through point (2,3) perpendicular to the given line. We know that the slopes have the negative reciprocal relationship shown above.
Since the given line has slope –⅓ then the perpendicular line has slope 3. This line goes through point (2,3) has a point-slope equation
y – 3 = 3(x – 2)
which can be rewritten in other forms, such as
y = 3x – 3
or in “standard form” this is
3x – y = 3 .
Slope-intercept form m1 1
y = mx – mx1 + y1
y = 3x – 3
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Linear Equations
Linear Equations
7/2/2013

27. Variation Direct Variation
Equations of Lines
Direct Variation
A variable y varies directly as variable x if
y = kx
for some constant k
The constant k is called the constant of variation
K is also known as the constant of proportionality
Direct Variation
The way a dependent variable changes with changes in the independent variable is known as variation.
If y = kx for some constant k, then y varies directly with x. That is, if x increases, so does y. And if x decreases, so does y.
If y = k/x for some constant k, then y varies inversely with x. That, if x increases, then y decreases. And if y decreases, then x increases.
The constant k is known as the constant of variation or constant of proportionality.
The direct variation relationship, y = kx, is represented as a linear equation whose graph has a zero intercept (i.e. zero initial value) and slope k.
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Linear Equations
Linear Equations
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28. Variation Direct Variation Example k
Equations of Lines
Direct Variation Example
State sales tax t varies directly as the amount of sale s , i.e. t = ks
For tax of $200 on a $12.50 sale, what is the constant of variation ? k = t s s t
Direct Variation
The way a dependent variable changes with changes in the independent variable is known as variation.
If y = kx for some constant k, then y varies directly with x. That is, if x increases, so does y. And if x decreases, so does y.
If y = k/x for some constant k, then y varies inversely with x. That, if x increases, then y decreases. And if y decreases, then x increases.
The constant k is known as the constant of variation or constant of proportionality.
The direct variation relationship, y = kx, is represented as a linear equation whose graph has a zero intercept (i.e. zero initial value) and slope k. =
12.50
200.00
k = .0625
.0625 =
Question:
Does this look like y = mx + b ?
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Linear Equations
Linear Equations
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29. Variation Inverse Variation
Equations of Lines
Inverse Variation
Variable y varies inversely as variable x if
for constant of variation k
k is also known as the constant of inverse proportionality y = k x x y
Inverse Variation
As with direct variation, a dependent variable can change value as the independent variable changes, but in a different way.
If y = k/x for some constant k, then y varies inversely with x. That is, if x increases then y decreases. And if x decreases then y increases.
The inverse variation relationship, y = k/x, is a non-linear relationship. The graph for this relationship is of course a hyperbola.
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Linear Equations
Linear Equations
7/2/2013

30. Variation Inverse Variation Example
Equations of Lines
Inverse Variation Example
At constant temperature the pressure P of a gas in a balloon is inversely proportional to its volume V so that V P P = k V
Inverse Variation
As with direct variation, a dependent variable can change value as the independent variable changes, but in a different way.
Since both P and V are positive, then k must also be positive. Theoretically, as the volume approaches zero, the pressure increases without bound (i.e., “goes to infinity”). Conversely, as the volume V increases without bound, the pressure P approaches zero. In the real world, however, there are physical limits on volume and pressure, so that volume of gas cannot go to zero, and pressure cannot become infinite.
7/2/2013
Linear Equations
Linear Equations
7/2/2013

31. Think about it ! Equations of Lines Linear Equations 7/2/2013

32. Perpendicular Lines in General
Equations of Lines
Step 1
Geometry gives us x y L1 m1 = b1 a ,
b1 = am1
c12 = a2 + b12 m2 c1 b1
am1 =
= a2 + (am1)2 a b2
– am2 = c2 m2 =
– b2 a
b2 = – am2 ,
Perpendicular Lines in General
The relationship between perpendicular lines is a little more complex than that between parallel lines. Given perpendicular lines L1 and L2 with slopes m1 and m2, respectively, as shown in the illustration.
Pick a point on line L1 above the intersection point and drop a vertical line from this new point to the x-axis. Project a horizontal line through the intersection of L1 and L2 as shown. We can now form three distinct triangles.
The horizontal segment marked a forms a leg of two distinct triangles, as shown. We apply the Pythagorean Theorem to the upper triangle, with leg b1 lying on the vertical projection line and hypotenuse c1 lying on line L1, giving
c12 = a2 + b12
The lower triangle is similarly formed using the same leg a, leg b2 on the vertical projection line, and the hypotenuse c2 on line L2 . This gives us
c22 = a2 + b22
The third triangle, with legs c1 and c2 and hypotenuse b1 + b2 give us
c12 + c22 = (b1 + b2)2
Combining this with the first two equations we have
c12 + c22 = a2 + b12 + a2 + b22 = (b1 + b2)2
leading to
b12 + 2b1b2 + b22 = b12 + 2a2 + b22 m1 L2
c22 = a2 + b22
= a2 + (am2)2
c12 + c22 =
a2 + (am1)2
+ a2 + (am2)2
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Linear Equations
Linear Equations
7/2/2013

33. Parallel and Perpendicular Lines
Equations of Lines
Perpendicular Lines in General (continued)
Thus
Subtracting like terms gives
Recalling that
b1 = am1 and b2 = – am2
gives x y L2 L1 m2 m1 a b1 c1 b2 c2
am1 =
– am2
b12 + 2b1b2 + b22
= b12 + 2a2 + b22
2b1b2 = 2a2
2(am1)(– am2) = 2a2
Perpendicular Lines in General
Carrying forward the equation
b12 + 2b1b2 + b22 = b12 + 2a2 + b22
we subtract like terms from both sides, giving
2b1b2 = 2a2
or simply b1b2 = a2
As shown on the diagram, b1 = am1 and b2 = –am2 , so substituting in the above equation we get
2(am1 )(–am2 ) = 2a2
This simplifies to
m1m2 = –1
Or we express as
m2 = –
The bottom line here is that the slopes of perpendicular lines are the negative reciprocals of each other.
– 2a2m1m2 = 2a2
m1m2 = –1 OR
m2 = 1 m1 –
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Linear Equations m1 1
Linear Equations
7/2/2013

34. Parallel and Perpendicular Lines
Equations of Lines
Perpendicular Lines in General (continued)
Thus
Subtracting like terms gives
Recalling that
b1 = am1 and b2 = – am2
gives x y L2 L1 m2 m1 a b1 c1 b2 c2
am1 =
– am2
b12 + 2b1b2 + b22
= b12 + 2a2 + b22
2b1b2 = 2a2
2(am1)(– am2) = 2a2
Perpendicular Lines in General
Carrying forward the equation
b12 + 2b1b2 + b22 = b12 + 2a2 + b22
we subtract like terms from both sides, giving
2b1b2 = 2a2
or simply b1b2 = a2
As shown on the diagram, b1 = am1 and b2 = –am2 , so substituting in the above equation we get
2(am1 )(–am2 ) = 2a2
This simplifies to
m1m2 = –1
Or we express as
m2 = –
The bottom line here is that the slopes of perpendicular lines are the negative reciprocals of each other.
– 2a2m1m2 = 2a2
m1m2 = –1 OR
m2 = 1 m1 –
7/2/2013
Linear Equations m1 1
Linear Equations
7/2/2013