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Ch.6 Linear Relations
Notes
Skill #1 Find the slope(steepness) of a line given a graph.
(page334 and sheet 6A side 1)
Pick any two points with integer coordinates,(where the line goes through the
corners of the grid), make a right triangle using the line as the hypotenuse, and
determine the slope = rise/run
Method: Use the formula: Slope =
See 4 examples page 334
Solutions:
a) rise = 6 or -6 = 3
run 10
-10
5
(ROC)
rise =
run
Change in y = ∆y
Change in x
∆x
b) rise = -10 or 10 = -2
run 5
-5
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Solutions:
1a) rise = 9 or -9 = 3
run
6
-6 2
1b) rise = -3 or 3 = -1
run
9
-9
3
Skill #2 Find the slope of a line, without a graph, if given
two points:A(x1 , y1 ) and B(x2 , y2) (like page337 and sheet 6-A side 2)
Now draw slanted line with coordinates of 2 endpoints, and derive slope formula:
Method: Use the formula:
Slope of line AB = y2 - y1 = rise
x2 - x1
run
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a) Find the slope of the line containing points C(-5, 3) and D(2,1)
Soln: First label the coordinates as:
slope = y2 - y1
x2 - x1
= 1- 3 = 1- 3
2 - -5
2+ 5
C(-5,-3) and D(2,1)
= -2
7
(x1 , y1)
(x2 , y2)
b) Find the slope of the line containing points E(4,-5) and F(8,6)
Soln: First label the coordinates as:
E(4,-5) and D(8,6)
(x1 , y1)
slope =
(x2 , y2)
y2 - y1 = 6 - -5 = 11
x2 - x1 8 - 4
4
Practise: Sheet 6A (both sides)
Answers to sheet 6A-side 2:
Find the slope of the line containing the following points:
a) (3 , 7) and ( 9 , 5 )
5 - 7 = -2 = -1
9- 3
6
3
d) ( 1, 2 ) and ( 3, 5 ).
5- 2 = 3
3- 1
2
b) ( -2 , 3 ) and ( 4 , 1 )
1 - 3 = -2 = -1
4 - -2
6
3
e) ( -6, 5 ) and ( 4, -9)
-9 - 5 = -14 = -7
4 - -6
10
5
c) ( -5 , -2 ) and ( -1 ,-7 )
-7 - -2 = -5
-1 - -5
4
f) ( 8 ,-2 ) and ( 5 , 4 )
4 - -2 = 6 = -2
5- 8
-3
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g) ( 4, -3 ) and ( -8, 1 ).
1 - -3 = 4 = -1
-8 - 4 -12
3
h) ( -6, 7 ) and ( -3, -2)
-2 - 7 = -9 = -3
-3 - -6
3
i) ( 5 ,2 ) and ( 4 , 0 )
0 - 2 = -2 = 2
4- 5
-1
j) ( 2, 3 ) and x-int of 4
k) ( -6, 5 ) and y-int of -2
l) x-int 3 and y-int -7
Use ( 2, 3 ) and( 4, 0 )
Use ( -6, 5 ) and (0, -2 )
Use ( 3, 0 ) and ( 0, -7 )
0 - 3 = -3
4- 2
2
-2 - 5 = -7
0 - -6
6
-7 - 0 = -7 = 7
0- 3
-3
3
m) ( 9 ,-2 ) and ( 5 , 6 )
n) ( 3 , 7 ) and x- intercept 6
o)
p)
y- intercept -4 and ( -5 ,-3 )
x-int -2 and y- int -5
Skill #3 Graphing lines given point(s) and slope
a) Graph the line containing the point A(-2,1) with slope of 3. Name the
coordinates of two other points on the line. (Like sheet 6B)
Step 1: Plot the given point (x, y) and place a star* there for “start”
Step 2: Start at the given point and use the slope to rise up or down ,
run right or left then plot a second point. If slope is not a fraction,
put 1 on the bottom to make the run = 1
Keep repeating until you fill the full grid, and connect the dots.
Area of possible confusion:
Slope = rise
run
but a point is (run , rise)
When calculating slope, rise goes first (on top)------rise up, run over,
but when writing coordinates, run goes first (in brackets)
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b) Graph the line containing the points A(3,2) and B(7,8) and name
the coordinates of 2 other points on the line
Method: Plot the points then calculate the slope using the formula:
y2 - y1 = rise
x2 - x1
run
Now start at one point and use the slope to rise, run and plot a third point as
done in part a). Repeat to find other points to fill the graph.
Skill #4 Find coordinates of other points without graphing.
a)
If a line has slope -4 and passes through the point (2,5), find the
3
coordinates of 3 other points on the line.
+3 -4
+3 -4
(2, 5)  (5, 1)
(5 , 1) 
+3 -4
(8 ,-3)
(8 , -3)  (11 ,-7)
Another way:
+3 -4
+6 -8
(2, 5) = (5, 1)
(2 , 5) = (8 ,-3)
+9 -12
(2 , 5) = (11 ,-7)
Sol: (2 + 3 ,5 - 4) = (5,9)
(5 + 3 ,9 - 4) = (8,-3)
(8 + 3 ,13 -4) = (11,-7)
b) Find the missing coordinate(s): (-2,4), (1, y ), (4 , 6), ( 7, a), ( 10,b), ( c,11)
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Solution:
Use the given points (-2,4) and (4 , 6) to get slope:
E(-2,4) and D(4,6)
(x1 , y1)
slope =
(x2 , y2)
y2 - y1 = 6 - 4 = 2 = 1
x2 - x1 4 -- 2
6 3
Now add/subtract multiples of 3 to the x-values and
multiples of 1 to the y-values
means
∆y = 1 or -1
∆x
3
-3
Solution:
(-2,4), (1, y ), (4 , 6), ( 7 , a ), ( 10 , b ), ( c , 11 )
Answers: y= 5, a = 7, b = 8, c = 19
Solution:
Solution:
(-2,4), (1, y ), (4, 6), ( 7 , a ), ( 10 , b ), ( c , 11 )
(-2,4), (1, 5 ), ( 4, 6 ), ( 7 , 7), ( 10 , 8 ), ( 19 , 11 )
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Skill #5 Using slope to identify parallel and perpendicular lines.
Parallel lines lines have equal slopes.
Perpendicular lines have negative reciprocal slopes.
Note: Negative reciprocals always multiply to give -1
Examples of negative reciprocals:
Number
negative
reciprocal
-1
7
7 x -1 = -1
7
-1
5
5 =5
1
-1 x 5= -1
5
-2
1
2
7
Product
-2 x 1 = -1
2
Ex. Classify these lines as parallel, perpendicular, or neither.
( Like # 9 on review see page 346 of text)
a)
Are these lines parallel, perpendicular or neither?:
Line EF passes through E(-7,2) and F(-2,10)
Line GH passes through G(-3,-4) and H(5, 1)
Step 1: slope EF =
y2 - y1 = 10 - 2 = 8
x2 - x1 -2 - -7
5
Step 2: slope GH =
1 - -4 = 5
5 - -3
8
Step 3: Comparing slopes
8
5
and 5
8
shows that they are not equal or negative reciprocals,
so the lines are not parallel or perpendicular.
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Solution in more detail:
Step 1: Calculate slopes of EF and GH
Soln: First label the EF coordinates as:
slope EF =
y2 - y1 = 10 - 2 = 8
x2 - x1 -2 - -7
-2 + 7
and label the GH coordinates as:
slope GH =
y2 - y1 = 1- - 4 = 5
x2 - x1 5 - -3
5 +3
(x1 , y1)
(x2 , y2)
E(-7,2) and F(-2,10)
= 8
5
(x1 , y1)
(x2 , y2)
G(-3,-4) and H(5, 1)
= 5
8
Step 2: Compare slopes to see if they are equal or if they are negative
reciprocals.
Answer: Since = 8
and 5 are not equal nor are they negative
5
8
reciprocals the lines are neither parallel or perpendicular.
Skill #6 Use slopes of parallel and perp. lines to answer various questions like:
identify a right triangle, parallelogram, rectangle, trapezoid ,etc.
Refer to sheet 6C and add some 3204 notes here.
a) Determine the slope of the line that is perpendicular
to the line through E(2,3) and F(-4, -1)
Step 1: slope EF =
-1 - 3 = -4 = 2
-4 - 2
-6 3
Step 2: Required slope is the negative reciprocal of
2
3
which is -3
2
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c) Is parallelogram ABCD a rectangle? ( Like # 11 on review see
page 348 of text)
Solution: Show that one angle, say angle B,
is a right angle(perpendicular lines).
So show slopes of AB and BC are negative reciprocals.
Step 1: slope AB =
Step 2: slope BC =
rise = -4
run
1
rise = 2 = 1
run
8 4
Since these slopes are negative reciprocals, angle B is a right angle so ABCD
is a rectangle. Note: -4 x 1 = -1
1 4
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More detailed explanation from the text book:
d) What is the value of k such that the line passing through
and
is parallel to the line
?
A.
B.
C.
D.
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Solution: The slope of the required line is -4(number multiplied by x) -4
Slope formula: =
k - -5 =
2- 4
k +5
-2
k +5 = -4
-2
Cross multiply to get : k
k
k
k
e)
+ 5 = (-2)(-4)
+5=8
=8–5
=3
In the graph below, LM is represented by the equation
If NP is parallel to LM, what is the equation of NP?
A.
B.
C.
D.
.
L
N
f)
Is quadrilateral ABCD a parallelogram? Justify your answer.
y
M
8
7
6
5
4
3
2
1
- 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1- 1
-2
-3
-4
-5
-6
-7
-8
-9
P
1
2
3
4
5
6
7
8
x
[3 points]
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y
5
4
B
3
2
A
-5
1
-4
-3
-2
-1
1
2
C
3
4
5
x
-1
D
-2
-3
-4
-5
(Sheet 6D) Big discovery!!!!:
If you graph y = 3x + 5 (using a table of values)
you will discover the line has a slope of 3 and a y–intercept of 5.
This pattern is always true and leads to the following:
The slope-intercept form of a linear equation is y = mx + b where
m = slope (mountain) and
b = y-intercept (begins on y-axis)
Note: the slope is the number in front of x (multiplied)
and the y –intercept is the constant number (without a variable).
Skill #7 Various questions regarding equations in slope - intercept form: (“y = mx + b”)
7a) State the slope and y-intercept of the following:
x
+5
4
Slope = ______
Y- intercept = ________
i) y = 3x -7
ii) y =
Ans: Slope = ______
Y- intercept = ______
Answers:
x
+5
4
i) y = 3x -7
ii) y =
Ans: Slope = 3
1
4
Y- intercept = 5
Y- intercept = -7
Slope =
iii) y = -x + 8
Slope = ______
Y- intercept = ______
iii) y = -x + 8
Slope = -1
Y- intercept = 8
iv) Given y = 4x + 9 , state slope and y-intercept Ans: slope(m) = 4
v) Given y = 5 + 3x , state slope and y-intercept Ans: slope(m) = 3
and y-intercept(b) = 9
and y-intercept(b) = 5
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7b) Write the equation in slope-intercept form of these lines:
i) having slope 2 and y- intercept -7 Ans: y = 2x - 7
ii) having y-intercept 6 and slope -5 Ans: y = -5x + 6
Example: More practice Skills #7 a and 7b: (Like p.362-363)
Equation in slopeintercept form:
y = mx + b
1. y = 3x-7
1
2. y = - x  3
2
3. y = x + 5
4. y = 6 + 8x
5. y = 4
6.
7.
8.
9.
10.
Slope
(m)
Y- intercept
(b)
y = 7x
x
y=
-2
5
y = - 4 - 6x
4
2/3
1
-6
Answers:
Equation in slopeintercept form:
y = mx + b
1. y = 3x-7
1
2. y = - x  3
2
3. y = x + 5
4. y = 6 + 8x
5. y = 4
6.
y = 7x
x
7. y =
-2
5
8. y = - 4 - 6x
9. y = 4x - 1
2
10. y = x  6
3
Slope
(m)
Y- intercept
(b)
3
-1/2
-7
3
1
8
0
5
6
4
7
1/5
0
-2
-6
-4
4
2/3
1
-6
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Note : If number multipied in front of x is missing, it’s understood to be 1 (#3)
If constant number added to x is missing, it’s understood to be 0. (#6)
Skill#7c: Use the graph to determine the slope, y-intercept and equation in
slope-intercept of the following lines:
Skill#7d:Given equation( in slope-intercept form) ,graph the line: (Use
skill #7 and 4)
i) y = 3x + 4
ii) y = x - 5
Step 1: Look at the equation and state the slope m=_____ and y- int. b = ____
Step 2: Plot the y-intercept (0, 4) and place a star* there for “start”
Step 3: Start at (0, 4) and use the slope
3
to rise 3 and run 1 then plot a second
1
point. Note: since slope is not a fraction, put 1 on the bottom to make the run = 1
Keep repeating until you draw the line in both directions filling the grid, and
connect the dots.
Section 6.5
The slope formula m = y - y1
can be rearranged
x - x1
by cross- multiplyingto produce a linear equation
in slope-point form: y - y1 = m(x - x1)
where m is the slope and (x1,y1) is a point on the line.
Skill #8 Various questions regarding slope- point form:
8a) State the slope and a point of these equations:
i) y - 3 = -7(x + 5)
Ans: Slope = ______
point = ______
ii) y + 4 =2(x - 6)
Slope = ______
point= ________
iii) y = -3(x + 2)
Slope = ______
point = ______
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Answers:
Ans: slope(m) = -7,
point = (-5, 3)
Ans: slope(m) = 2
point = (6, -4)
Ans: slope(m) = -3
point = (-2, 0)
8b)Write the equation of the line:
i) having slope 2 and a point (1, -5) Ans: y + 5 = 2(x - 1)
ii) having slope -3 and a point (-2, 8) Ans: y - 8 = -3(x + 2)
Example: More practice Skills #8a and 8b: Like page Like #4 and 5 on page 372-373
Equation in slopepoint form:
y - y1 = m(x - x1)
1. y - 5 = -4(x - 1)
2. y + 7 = 3(x - 8)
3. y -2 = -5(x + 4)
4. y + 8 = 7(x - 6)
1
5.
y + 3 = - x  7 
4
6. y - 4 = 2(x - 5)
7.
8.
y -6 = 5(x + 1)
Slope
m
Point on the
line (x1, y1)
-4
3
-5
7
-1/4
(1, 5)
(8, -7)
(-4,2)
(6,-8)
(-3, 7)
2
4
5
(5, 4)
(2, 3)
(-1, 6)
Slope
m
-4
3
-5
7
-1/4
Point on the
line (x,y)
(1, 5)
(8, -7)
(-4,2)
(6,-8)
(7, -3)
2
4
2
(5, 4)
(2, 3)
(-1, -6)
Answers:
Equation in slopepoint form
1. y - 5 = -4(x - 1)
2. y + 7 = 3(x - 8)
3. y - 2 = -5(x + 4)
4. y + 8 = 7(x - 6)
1
5.
y + 3 = - x  7 
4
6. y - 4 = 2(x - 5)
7. y - 3 = 4(x - 2)
8. y + 6 = 2(x + 1)
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Skill#8c: Use the graph to determine the slope, point and equation in slopepoint form of the following lines:
Skill#8d: Given equation( in slope-point form) , graph the line: (Use
skill #7 and 4)
i) y - 3 = -2(x + 5)
ii)
y + 1 =3(x - 2)
4
iii) y = -3(x + 2)
p.372-374
# 4 - 9, 12, 14, 19 - 23
Skill #8e: Change from slope-point form to slope-intercept form
and state slope and y-intercept:
y + 2 = 3 (x + 1)
4
4(y + 2) = 4[ 3(x + 1)]
(Multiply both sides by LCD 0f 4)
4
4(y + 2) = 3(x + 1)
Divide denominator into LCD
4y + 8 = 3x + 3
Multiply brackets(Distributive property
4y = 3x + 3 - 8
Move all terms to the right except “y” term
4y = 3x - 5
Simplify right side
y
Note
3x
4
slope m= 3
4
Another way:
y + 2 = 3 (x + 1)
4

5
4
Divide all terms by coefficient of “y”
and y-intercept b -5
4
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y = 3 (x + 1) - 2
4
Solve for y by subtracting 2 from both sides
y =3x+3 -2
4
4
remove brackets using distributive property
y =3x+3 -8
4
4 4
get common denominator
y =3x -5
4
4
add numerators
Note
slope m= 3
4
and y-intercept b -5
4
Skill #8f Write the equation of a line(in slope-point form) that is
a) parallel to
b)
y=2x-5
3
perpendicular to
ANSWERS:
a)
Answer:
b)
Answer:
and passes through (4, -7)
y=2x-5
3
y + 7 = 2(x - 4)
3
y + 7 = -3(x - 4)
2
and passes through (4, -7)
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Detailed explanations:
a)
parallel to
Answer:
y=2x-5
3
and passes through (4, -7)
y + 7 = 2(x - 4)
3
since parallel lines have equal slopes
m= 2 and point (x,y) = (4, -7)
3
Step 1:
Step 2:
Write equation in slope-point form: y - y1 = m(x - x1)
y - - 7 = 2(x - 4)
3
y + 7 = 2(x - 4)
3
b) is perpendicular to
Answer:
y=2x-5
3
Nand passes through (4, -7)
y + 7 = -3(x - 4)
2
Detailed explanation:
Step 1:
m= -3 and point (x,y) = (4, -7)
2
Step 2:
Write equation in slope-point form: y - y1 = m(x - x1)
y - - 7 = -3(x - 4)
2
y + 7 = -3(x - 4)
2
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y - y1 = m(x - x1)
Skill #8g) Write the equation of the line(in slope-point form)
that passes through two points:
i) (2, -6) and (-3, 4)
ii) (-4, 7) and (5, -2)
i)Step 1: slope = y2 - y1
= 4 - -6 = 10 = -2
x2 - x1 -3 - 2
-5
Step 2: Equation: y - y1 = m(x - x1)
y - - 6 = -2( x - 2)
y + 6 = -2( x - 2)
y - 4 = 2( x + 3)
using (2, -6)
using (2, -6) OR
using (-3, 4)
y - y1 = m(x - x1)
ii) Write the equation of the line(in slope-point form) that
passes through (-4, 7) and (5, -2)
Step 1: slope = y2 - y1
= -2 -7 = -9
x2 - x1 5 - -4
9
= -1
Equation:
using (-4, 7)
y - 7 = -1( x + 4) or
y + 2 = -1( x - 5)
(5, -2)
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Section 6.6
A third form a linear equation is general form: Ax + By + C = 0
Skill#9a Change the following from general to slope –intercept form
then state slope and y-intercept. (Like #’s 16,19 on review sheet
& ex.3 page 381)
Method: Moving all terms to the right side except the “y” term.
Then divide all terms by coefficient of “y” and simplify to get “y = mx + b” format
We must rearrange the equation and solve for y to
a) 3x - 4y - 12 = 0
b) 8x + 3y - 24 = 0
Soln:
3y = -8x + 24
-4y = -3x + 12
c)
5x - y - 14 = 0
-y = -5x + 14
y
 3x 12

4 4
y
 8 x 24

3
3
y
y
3
x3
4
y
8
x 8
3
y  5x  14
Slope: ? 3/4
y-int.: ? -3
Slope: ? -8/3
y-int.: ? 8
 5 x 14

1 1
Slope: ?
y-int.: ?
5
-14
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Summary :Three forms of a linear equation:
Equation Formula
1. Slope -intercept form:
y = mx + b
2. Slope -point form:
y - y1 = m(x -x1)
Sample
equations
y = 7x + 2
Point (x,y) Slope
(0,2)
y-int.
7
y-3 = 5(x - 4)
(4,3)
5
2x + 5y + 4 = 0
3. General Form:
Ax + By + C = 0
Skill#9b
Change the following to general form Ax + By + C = 0(Like # 18 on
review sheet & ex.1 page 379)
Method: Bring all terms to left side, change signs, and make right side = 0. Put x term first,
y next and constant # last(you may need to add like terms) In the last step, when
the x term is negative, change signs on all terms.
If there is a fraction, multiply all terms by LCD
i)
y = -2x + 4
ii)
y = 3x - 6
5
2x + y - 4 = 0
5y =5 (3x) - 6(5)
5
5y = 3x – 30
y + 2 = 5x – 35
-3x + 5y + 30 = 0
- 5x + y + 37 = 0
(Group 2 Wed)
iii)
y + 2 = 5(x – 7)
y + 2 - 5x + 35 = 0
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iv)
y + 4 = 2(x – 5)
3
3(y + 4) = 3[ 2(x – 5)]
3
(Multiply both sides by LCD of 3)
3(y + 4) = 2(x – 5)
Divide denominator 3 into LCD 3 to get answer of 1
3y + 12 = 2x – 10
Multiply brackets(Distributive property)
3y + 12 - 2x + 10 = 0
Move all terms to the left and put 0 on the right side.
- 2x + 3y + 22 = 0
Put x term first, y next and constant # last(Add like
terms 12 + 10 = 22)
When x term is negative, change signs on all terms.
2x - 3y - 22 = 0
ANSWERS:
a)
y = -2x + 4
y + 2x - 4 =0
2x + y - 4 =0
b)
y = 3x - 6
5
5y = 5 (3x) - 6(5)
5
c)
y + 2 = 5(x – 7)
y + 2 = 5x – 35
5y = 3x – 30
y + 2 = 5x – 35
5y - 3x + 30 = 0
y + 2 - 5x + 35 = 0
-3x + 5y + 30 = 0
3x - 5y - 30 = 0
- 5x + y + 37 = 0
5x - y - 37 = 0
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iv)
y + 4 = 2(x – 5)
3
3(y + 4) = 3[ 2(x – 5)]
3
3y + 12 = 2(x – 5)
3y + 12 = 2x – 10
3y + 12 - 2x + 10 = 0
2x - 3y - 22 = 0
Skill #10 Graph(using the two- intercept method (Like #17 on review)
a) 3x - 4y -12 =0
b) 8x + 3y - 24 = 0
Find y-int: (Let x = 0)
3(0)- 4y - 12 = 0
- 4y -12 =0
- 4y = 12
- 4y = 12
- 4 -4
y = -3
1– 8
2
3
4
5
6
7
8
1
2
3
4
5
6
7
1– 8
2
3
4
5
6
7
8
1
2
3
4
5
6
7
coordinates:  ( 0 ,-3 )
Find x-int: (Let y = 0)
3x – 4(0) - 12 =0
3x
- 12 =0
3x = 12
3x = 12
3
3
x=4
( 4 ,0)
y
8
7
6
5
4
3
2
1
– 8– 7– 6– 5– 4– 3– 2– 1
– 1
– 2
– 3
– 4
– 5
– 6
– 7
– 8
1
2
3
4
5
6
7
8
x
c)
4x + 7y - 28 =0
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And now for the graphs 
The two intercepts are 
( 0, -3 )
( 4, 0 )
y
8– 8
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1– 8
2
3
4
5
6
7
8
1
2
3
4
5
6
7
( 0, 8 )
( 3, 0 )
y
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
– 8– 7– 6– 5– 4– 3– 2– 1
– 1
1
2
3
4
5
6
7
8
x
8– 8
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1– 8
2
3
4
5
6
7
8
1
2
3
4
5
6
7
– 8– 7– 6– 5– 4– 3– 2– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
( 0, 4 )
( 7, 0 )
y
8
7
6
5
4
3
2
1
1
2
3
4
5
6
7
8
x
– 8– 7– 6– 5– 4– 3– 2– 1
– 1
1
2
3
4
5
6
– 2
– 3
– 4
– 5
– 6
– 7
– 8
Using short cuts:
a) 3x - 4y -12 =0
Step 1: Find y-int: (Let x = 0)
- 4y -12 =0
- 4y = 12
-4 -4
y = -3
The coordinates are  ( 0 , -3 )
Step 2: Find x-int:
(Let y = 0)
3x - 4y – 12 =0
3x - 12 =0
3x = 12
3
3
x=4
The coordinates are  ( 4 , 0 )
b) 8x + 3y - 24 = 0
( ,
)
c)
( , )
8x + 3y - 24 = 0
( ,
)
4x + 7y - 28 =0
4x + 7y - 28 =0
( , )
7
8
x
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Answers:
Step 1: Find y-int:
(Let x = 0)
a) 3x - 4y - 12 = 0
3(0)- 4y - 12 = 0
- 4y - 12 =0
-4y = 12
12
y=
4
y=3
The coordinates are 
( 0, 3 )
Step 2:Find x-int:
3x - 4y - 12 =0
(Let y = 0)
3x – 4(0) - 12=0
3x = 12
12
x=
3
x=4
The coordinates are 
( 4, 0 )
b) 8x + 3y - 24 = 0
8(0) + 3y - 24 = 0
3y - 24 = 0
3y = 24
24
y=
3
y=8
( 0, 8 )
( 0, 4 )
8x + 3y - 24 = 0
8x + 3(0) - 24 = 0
8x - 24 = 0
8x = 24
4x + 7y -28=0
4x + 7(0) - 28=0
4x = 28
28
x=
4
24
8
x=3
x=7
( 3, 0 )
( 7, 0 )
x=
c) ( 0, 4 ) and ( 7, 0 )
y
y
x
4(0) + 7 y -28 = 0
7y - 28 =0
7y = 28
28
y=
7
y=4
And now for the graphs: The two intercepts are 
a) ( 0, 3 ) and ( 4, 0 )
b) ( 0, 8 ) and ( 3, 0 )
y
c) 4x + 7 y - 28 =0
x
x
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Skill #11 Important facts/generalizations about slope: Real life examples of slope:
Steepness of a roof, incline on a treadmill, wheelchair ramp, etc.
(page335)
Type of line/sample graph
Slope
Sample
equation(s)
Slants up to the right
Positive + or y = 3x +7
y increases as x increases
+
Slants down to the right
Negative + or y = -3x +7
y decreases as x increases
+
Steep incline
high number
y = 9x + 4
y = -9x + 4
same high
steepness
Gentle incline
low number
y = 2x + 4
y = -2x + 4
same low
steepness
Horizontal line
0
y = 5 or y = -3
Vertical line
undefined
x = 5 or x = -3
Parallel lines
equal slopes
y = 5x + 3
y = 5x + 7
Perpendicular lines
negative reciprocal
slopes
y = 5x + 3
y = -1x + 7
5
Notes:
1. If rise and run have same signs, slope is positive, and line slants up to right.
2. If rise/run have different signs, slope is negative, & line slants down to right.
3. Direction doesn’t matter when calculating slope: G to H or H to G.
4.You can use any 2 points on the line to calculate slope.
5.Slope of horizontal line is always 0 because rise = 0 = 0
run run
6.Slope of vertical line is always undefined because rise = rise = undefined
run
0
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7.
8.
9.
Don’t just count the blocks. You must first identify the scale on each axis.
The rise is not the highest y-value. It’s the change between y-values.
When calculating slope, rise goes first (on top)------rise up, run over, but
when writing coordinates, run goes first (in brackets)
Slope = rise
but a point is (run,rise)
run
Extra notes/examples:
Skill # 18 How to find x and y intercepts from an equation.
First , lets note the intercepts on the given graph:
Notice: For every y-intercept, the x coordinate is 0
For every x-intercept, the y coordinate is 0
This leads to the following method to quickly find intercepts:
To find y-intercept, replace
x with 0
To find x-intercept, replace y with 0
Ex. Find the intercepts of the following equations then sketch the graph using the two
intercept method:
a) 3x - 4y = 12
b) 8x + 3y - 24 = 0
c) 4x + 7 y = 28
n
Sol : For y-int:
3(0)- 4y = 12
8(0) + 3y - 24 = 0
4(0) + 7 y = 28
- 4y = 12
3y - 24 = 0
7y = 28
28
12
Y=
3y = 24
y=
7
4
24
Y=3
y=
Y=4
3
Y=8
The coordinates are  ( 0, 3 )
( 0, 8 )
( 0, 4 )
a) 3x - 4y = 12
b) 8x + 3y - 24 = 0
c) 4x + 7y = 28
n
Sol : For x-int:
3x – 4(0) = 12
8x + 3(0) - 24 = 0
4x + 7(0) = 28
3x = 12
8x - 24 = 0
4x = 28
12
28
x=
8x = 24
x=
3
4
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24
8
x=3
x=4
The coordinates are 
x=
( 4, 0 )
x=7
( 3, 0 )
And now for the graphs:
The two intercepts are 
( 4, 0 )
y
( 0, 3 )
y
( 3, 0 )
( 0, 8 )
x
y
x
SKILL # Write the linear equation representing the data in the table of values.
Given a linear table of values, write the equation.
x
-5
0
5
10
15
( 7, 0 )
y
-20
-5
10
25
40
A.
B.
C.
D.
Method:determine the rate of change(slope) and the y –int, then write the equation
( 7, 0 )
( 0, 4 )
x
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ROC = ∆y = rise = y2 - y1 = 40 - 25 =15 = 3
∆x
run
x2 - x1
15 - 10 5
y- intercept = -5 (y-value when x=0)
so equation is y = 3x - 5(choice C)
Sample test/exam questions
1.
In the graph below, LM is represented by the equation
If NP is parallel to LM, what is the equation of NP?
A.
B.
C.
D.
.
L
N
2.
What is the value of k such that the line passing through
?
y
M
8
7
6
5
4
3
2
1
- 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1- 1
-2
-3
-4
-5
-6
-7
-8
-9
and
P
1
2
3
4
5
6
7
8
x
is parallel to the line
A.
B.
C.
D.
3.
Is quadrilateral ABCD a parallelogram? Justify your answer.
[3 points]
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y
5
4
B
3
2
A
-5
1
-4
-3
-2
-1
1
2
C
3
4
5
x
-1
D
-2
-3
-4
-5
4.
Which graph represents the equation
?
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y
A
.
-5
-4
-3
-2
-1
y
5
5
4
4
3
3
2
2
1
1
1
-1
2
3
4
5
x
-5
-4
-3
-2
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
1
2
3
4
5
x
B
.
y
-5
5.
-4
-3
-2
-1
y
5
5
4
4
3
3
2
2
1
1
-1
1
2
3
4
5
x
-5
-3
-2
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
A line has slope and passes through point
What is the equation of the line?
.
A.
B.
C.
D.
6.
-4
Which point is on the line
?
A.
B.
C.
D.
7. Which linear equation represents the data in the table of values?
1
2
3
4
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x
-5
0
5
10
15
y
-20
-5
10
25
40
A.
B.
C.
D.
8.
A boat travelling at 8 m/s begins to accelerate.
Its new speed, S, in metres per second, is modelled by the function
, where t is the length of time, in seconds, that it accelerates.
[3 points
a) Determine the speed of the boat at 7 seconds.
b) Determine the time it takes for the boat to reach 26 m/s.
c) What is the domain of this function?
9.
Determine the equation of the line passing through
and
in general form.
[3 points
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10. Which graph represents the solution to the system below?
y
A.
5
4
3
2
1
y
B.
1 2 3 4 5 x
-5 -4 -3 -2 -1-1
-2
-3
-4
-5
5
4
3
2
1
y
C.
5
4
3
2
1
-5 -4 -3 -2 -1-1
-2
-3
-4
-5
1 2 3 4 5 x
-5 -4 -3 -2 -1-1
-2
-3
-4
-5
y
D.
1 2 3 4 5 x
5
4
3
2
1
-5 -4 -3 -2 -1-1
-2
-3
-4
-5
1 2 3 4 5 x
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Skill # Solve word problems involving slope.(See p.338-339)
Ex. Jacob was charged $7 to travel 5 kilometers. Later that week he was
charged $12.50 to travel 18 kilometers. What is average cost per kilometer for
the trip?
Solution:
cost =
km
Average cost per km = slope = cost
km
y2 - y1 = 12.50 - 7 = 5.50 = 0.42 = $0.42
x2 - x1 18 - 5
13
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Summary of Horizontal and Vertical Lines
Horizontal Lines
0
Slope
Equation
y = # (the given y-coordinate)
Vertical Lines
undefined
x = # (the given x- coordinate)
Skill # 20 (Should have been skill 10)Write a linear equation y = mx + b
when given the graph. (Like skills 1 and 8)
Step 1: Find slope m Pick any two points where the line goes through the corners
of the grid and determine the slope = rise/run
Step 2: Find y-intercept b Look at the y- axis and write the number where the
graph intercepts the y-axis.
Step 3:
Summary:
Write the formula: y = m x + b then replace m and b with
the values you just obtained in steps 1 and 2!
1.
2.
3.
Find m
Find b
Write y = m x + b
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Step 2:
or
Use point (1, 1) to write equation: y - 1 = 2(x - 1) or
Use point (-2, -5) to write equation: y + 5 = 2(x + 2)
Like #4 and 5 on page 372
Please update your homework chart: