GRADE 10 (MATH 20S) MATH. UNIT F: COORDINATE GEOMETRY. CLASS
NOTES. 1. Being able to find your way about a coordinate system is rather ...
1 GRADE 10 (MATH 20S) MATH UNIT F: COORDINATE GEOMETRY CLASS NOTES 1. Being able to find your way about a coordinate system is rather important. You do it every time you picture the city ‘street grid’ in your head. You do it every time you navigate yourself across the tiles on the kitchen floor or play a game of battleship with your niece or nephew. SLOPE OF A LINE 2. A line is a special and simple ‘curve’. A line is a ‘locus’ of points all going in the same direction. Slope tells the direction a curve is going at any particular point on the curve. On a line the slope is ‘constant’ along the entire length of the line. 3. You had studied this in a previous unit when you studied how to graph the linear equation y = mx + b. 10
10
y
8
A Line
A Parabola Curve
6
6
4
4
2
2
x
-8
-6
-4
-2 -2 0
2
4
6
8
10
-10
-8
-6
-4
-2 -2 0
-4
-4
-6
-6
-8
-8
-10
-10
a line: all the points on it are changing in the same direction. It has a constant ‘slope’.
gr10math_F_UnitNotes.doc
x
0
0 -10
y
8
2
4
6
8
10
a non-linear curve: the direction the points are going changes, depends on where on the curve you are. The slope is continuously changing. Calculating the slope of these curves is a university subject: ‘Calculus’. We will study this curve lots in Grade 11.
Revised:20140831
2 SLOPE OF LINE CALCULATION (review from previous studies) 4. Pick any two points on the line call them: P1 and P2. Let P1 be at coordinate (x1, y1) and P2 at (x2, y2). Find out the change in the y’s and the change in the x’s between those two points. Divide the change in y’s by the change in x’s will give you how much the y changes for each change in x of one. slope =
change in y ∆y y 2 − y1 Rise = = = change in x ∆x x2 − x1 Run
5. You will see the ‘∆ ∆’ symbol (DELTA, the Greek ‘D’) often in science, it just means ‘change in’. And of course to measure a change in a value you have to subtract the first value from the second value. Example: If you want to measure the change in height of your son, you take his height this year and subtract his height from last year. Subtraction is the way to measure change! EXAMPLE CALCULATE SLOPE
10
P2(8, 10)
y
8
A Line
change in y ∆y y 2 − y1 10 − 2 = = = change in x ∆x x2 − x1 8−4 slope = 2
6
slope =
4
P1(4, 2)
2
x
0 -10
-8
-6
-4
-2 -2 0
6. So every time the points on this line change one step right along the x, the line goes up in y value by 2.
2
4
6
8
10
-4 -6 -8 -10
Caution: notice the x-axis and the y-axis are not ‘scaled’ the same here. The graph is a bit ‘squished’
7. You Try: Calculate the slope of the three lines on this graph. y 2 − y1 x2 − x1 y − y3 slopeB = 4 x 4 − x3 slope A =
slopeC =
=
8
y
P2(2, 7)
Line A Line B Line C
6 4 P1(0, 3)
=
2
=
= =
x
0 -6
=
P3(3, 2)
-4
-2
-2 -4
0
2
4 P4(5, -2)
6
3
SLOPE: HORIZONTAL LINES 8. Calculate the slope of the line between the points P1(–2, 4) and P2(5, 4). Write it below showing the slope formula!:
Plot the two points at the left here: 8 6 4 2 2
9. Try these points too: P3(–5, 7) and P4(2, 7)
4
6
8
10. Notice the points on the lines above have the same y value. All points on the lines will have the same y-value. The lines are Horizontal. Both have a slope of zero. So their formula given y = mx + b is just y = b. In other words, y is a constant no matter what the x is! That is a Horizontal line! think of horizon! The horizon is horizontal! 11. The equation for a horizontal line is just y = [a constant] 12. Plot the following horizontal lines on the graph to the right
13.
a.
y=8
b.
y = –2
c.
y=3
d.
y = –7.25
What is the equation for the dotted line?
8 6 4 2 2
4
6
8
4 VERTICAL LINES 14.
What is the slope of the line that contains the points P1(–5, -3) and P2(–5, 7)?
slope =
=
15. In mathematics, you can never get an answer by dividing by zero. (how can you divide something into zero bunches?? How many ‘nothings’ can you take away from ‘something’?). We 3 say that the operation of dividing by 0 is undefined. = ? would mean that 0*? = 3. We have no 0 way of having zero ‘bunches’ of something that makes a total of three. The equation for a Vertical line is just x = [a constant]. All points on that vertical line have the same x value. The x value never changes for any specific vertical line. 16. Some people like to think of vertical lines as having an ‘infinite’ slope. But infinity is not a number, it is an idea. You will be unable to graph vertical lines on a graphing tool. Graphing tools like to have a simple formula like ‘y = something’. 17. Plot and label the following vertical lines: a.
x = –8
b.
x=0
c.
x=3
d.
x = –4.25
8 6 4 2 2
4
6
8
5 PARALLEL LINES – EQUATIONS 18.
Any two lines that are parallel have the same slope. They never meet (intersect).
19.
Give one equation for a line that is parallel to : y = 3x + 2.
20. Solution. Any line with a slope of 3. So: y =3x + 9 will work!. So will y = 3x + 42. So will y = 3x – 77.432. In fact, there are an infinite number of lines that will be parallel to the given line! Parallel Lines Example: Given: y = 3x + 2 (the solid line)
8
21. What lines are parallel? Their slope?
6 4
You write a couple equations for lines that are parallel to y = 3x + 2
2 2
4
6
8
22. How many of the given parallel lines go through point (4, 5) with that slope? How many of the given parallel lines go through point (-6, -8) with that slope?
PERPENDICULAR LINES 23. We had learned that two lines that are parallel, ‘ || ’, have the same slope. What about two lines that are perpendicular. (‘Perpendicular’ , ‘⊥ ⊥’, means intersect at a 90° angle). 24. It works out that two lines that are perpendicular have a special relationship between their slopes.
6
25.
Slopes of Perpendicular lines. Say we have two lines L1 and L2: a. L1 with slope m1; and b. L2 with slope m2.
26.The relationship between the slopes of perpendicular lines is:
m2 = −1*
1 1 =− m1 m1
27. That is; the slopes of perpendicular lines are the negative reciprocals of each other. Another way to think about it is, if you multiply the two slopes together the product is –1. m1 * m2 = –1 28.
Example.
a. Problem. Find the equation of a line that is perpendicular to the line y = 2x + 2. b. Solution. The slope, m2, of the 1 1 =− m1 2 29. So any line that has a slope of –1/2 will be perpendicular to the line y = 2x + 2.
perpendicular line will be m 2 = −
8 6 4 2 2
4
6
8
c. How many lines are there that are ⊥ to y =2x + 2? _______________
Slope ‘Word’ Problem 30. Slope is sometimes called ‘pitch’ or ‘gradient’. Anyone who has ever done roofing would know that. If a roof has a pitch of ‘1.5 over one’ or ‘1.5 to 1’ then it is law that you wear a harness because it is a rather steep roof! What is the slope or ‘pitch’ of this roof? What is the ‘pitch’ of this roof?
7 EQUATION OF A LINE OF GIVEN SLOPE THROUGH A GIVEN POINT 31.
Recall, the equation of a line through a y-intercept, b, and with slope, m, is just y = mx + b.
32. How do we find the equation of a line through any given point given a slope and a point! In this case, the given point is not just a simple one like on the y-intercept! If we are given one point on a line and a direction of a line, we should be able to find all other points! Think about it! 33. Example: Find the equation of the line that has a slope of 2 and goes through the point (–3, –5). We know: y =mx + b is a ‘slope and intercept’ equation of a line. We know the m is 2 in this case. So we know that y = 2x + b. All we need is to find the b! 34. But we know that when x = –3 that y = –5! Let’s substitute in the x and y into what we know so far! y = 2x + b − 5 = 2(−3) + b so − 5 = −6 + b ∴b = 1
35.
2) The yintercept ends up being +1 in this case
8 6 4 2 2
4
6
8
1) We are given this point, and a slope
Some people like to use the ‘point & slope’ form equation:
The formula we want then is:
y = 2x + 1 36. Show your work to find the equation of the line that goes through the given point with the given slope. You might want to use the graph at the right to help to picture it the first few times.
y – y1 = m(x – x1)
8 6 4
Line A B C D E
Given Point (–3, –3) (–4, –8) (–6, –4) (–8, 6) (–17.5, 33.145)
Slope 1 1/4 2.5 -4 2
Answers: A: y = 1x + 0 or y = x; B: y = x/4 –7; C: y = 2.5x + 11; D: y = –4x – 26; E: y = 2x + 68.145
2 2
4
6
8
8 Your work to get the above solutions:
Chartrand
16
North
15
eek
Sherise Swamp
s Cr
38. Advanced Application. (not for tests) Your friend calls you on a radio. He is lost out in the bush. He knows he was walking in a constant direction exactly on a compass direction of 045° all day. He has been using a grid map as shown at the right. He knows he went exactly by Beaver Rock earlier in the day. It is getting dark. He is afraid of the evil and legendary ‘Chartrand Monster’ in the night swamp! What is the equation for the line he is following so that you can get Search and Rescue to look for him???!!!
Vau ghn ’
37.
14
13
Beaver Rock
45°°
Notice the grid system as on most maps!
12 03
04
05
06
LINE THROUGH TWO GIVEN POINTS 39.
We have learned to graph lines: a. given a slope and y-intercept (slope intercept form equation: y = mx + b) b. using an x and y intercept (standard form equation Ax + By = C) c. given a slope and any point that the line contains
40. We also know that a line can be made from just two points! Two points define a line! So how do we make a line given just two points P1(x1, y1) and P2(x2, y2).
9 ( y 2 − y1 ) . And we know how to find ( x 2 − x1 ) an equation given one point and a slope. So no problem to find a line.
42.
We know how to find the slope given two points: m =
43. Example: Find the equation of the line that goes through the two points P1(4, 4) and P2(6, 8). 8−4 = 2 . We need 6−4 the y-intercept now, b.. So pick one point, say P2(6, 8). 8=2(6) + b. So b = –4. So the equation of the line is y = 2x – 4
45.
The slope is m =
PRACTICE PROBLEMS 46. Given two points on a line calculate the equation of the line and plot it.
8 6 4
Line P1 A (0, 0)
P2 (8, 8)
B
(2, 5)
(–4, –5)
C
(5, 2)
(2, 7)
D
(8, –8)
(–5,1)
Equation
Room to write your calculations
2 2
4
6
8
10
DISTANCE BETWEEN POINTS 47. Examine the right-angle triangle at the right. Calculate the length of the hypotenuse here:
5 12 48.
What theorem did you use above to calculate the length of the hypotenuse? .
________________________ 49. What is the length of the hypotenuse in this triangle at the right? Show formula and work.
40
30
20
10
0
50. Label the ends of the hypotenuse with A and B. What is the length of the triangle hypotenuse at right? Show formula and work!
10
20
30
40
11 51.
How did we calculate the base (the change or difference in x)? _____________
How did we calculate the height (change or difference in y)?____________________ 52. Now you should be able to understand the formula for the distance between two points. Given two points P1(x1, y1) and P2 (x2, y2), the distance between the two points is:
d = ( x2 − x1 ) 2 + ( y2 − y1 ) 2 (You will always be given this formula on a test even though you should be able to logically figure it out anyway) 53. Practice. Find the distance, d, between the following pairs of points: (plot them if you feel it necessary to visualize) a.
(3, 3) & (6, 3)
d=___________
b.
(-4, -4) & (-4, 7) d= ___________
c.
(-6, -6) & (2, 5) d= _________
d.
(-8, 5) & (4, -6) d= ________
e.
(-10,-10) & (70,50) d= ______
FIND THE MID-POINT 54.
How to find the ‘mid-point’ between two points.
55. If Walter is standing next to Jackie, how do we find the average (or mean) height? The point that is mid-way between them? How do you find your ‘average’ mark after two assignments?
56. Examine ∆ABC at right. If we wanted to make a similar triangle that had a hypotenuse of length exactly half of ∆ABC how could we do it?
A 10 6
B
8
C
12
57. Mid-point formula. Given any two points P1(x1, y1) and P2 (x2, y2) the point that is midway between them is given by: Just like finding a ‘mean’ or ‘average’ of x1 + x 2 y1 + y 2 the x’s and y’s of the midpo int = , two points 2 2
58.
Practice problems. Find the mid-point between the following pairs of points. a.
(2, -2) (4, 8)
b.
(17, 12)
c.
(1.75, 2.25) (12.333, 11)
d.
(4, 4) (16, 4)
e.
(12,8) (12, –23)
(–4, –8)
