Here are the steps to solve these problems: 1. Find the slopes of the two lines: m1 = (8-2)/(5--2) = 6/3 = 2 (slope of r) m2 = (7-0)/(-8--2) = 7/-6 = -1 (slope of s) The slopes are negative reciprocals, so r ⊥ s. 2. The slopes are m1 = 2 and m2 = -1/2. Since m1 × m2 = -1, the lines are perpendicular. 3. The given line has slope 3. The perpendicular line will have slope -1/3. Plug into the point-slope form

3. Gradient measures the steepness of a slope.
• Step 1: Measure the rise (difference in height
between 2 points)
• Step 2 : Measure the run (the distance
between 2 points). Make sure that you
convert the scale into metres
• Both the rise and run need to be expressed in
metres.

8. • Say the rise is 42 metres and the run is 600
metres.
• 42(rise)/600(run) - Formula (divide the top by
itself and the bottom by the top)
• Divide 42 by itself = 1
• Divide 600 by 42 = 14.3

9. • The answer can be expressed in three ways:
• a) As a statement 1 in 14.3
• b) As a ratio 1: 14.3
• c) As a representative fraction 1/14.3

11. A theorem to find the length of sides of right triangles
• What do the variables stand for?
a = the Y, vertical side of the triangle
b = the X, horizontal side of the triangle
c = the hypotenuse of the triangle
• What type of triangle do we use the theorem for?
-Right angled triangles.

12. • Draw a right triangle with two sides labeled with numbers
x
32+42=x2 3
9+16=x2
x=25 4
(The opposite of x2 is )
X=5

13. The distance formula is a mathematical formula used to
measure how far apart two points are from one
another.
• What steps do you follow to use the distance
formula?
Label the points.
Put them in the distance formula.
Do the math.
(x2 – x1)2 + (y2 – y1)2

14. • List two points
(3,12)(9,5)
(3-9)2+(12-5)2
-62+72
36+42
(Square root)78
8.83 is the answer

15. •The midpoint of a segment is the POINT M.
•The midpoint is a dot with a coordinate (x, y).
•M = ( [x₁ + x₂]/2, [y₁ + y₂]/2 )
•Take the x coordinates, add, divide by 2 = new x
coordinate.
•Take the y coordinates, add, divide by 2 = new y
coordinate.
•M = ( x, y )

16. •M = ( [x₁ + x₂]/2, [y₁ + y₂]/2 )
•Find the midpoint between:
• G(-3, 2) and H(7, -2)
• ( [-3 + 7]/2, [2 + -2]/2 )
• ( [4]/2, [0]/2 )
• ( 2, 0 ) ← Midpoint between G and H

17. •Midpoint between A(2, 5) and B(8, 1):
•Midpoint between P(-4, -2) and Q(2, 3):

19. Perpendicular Lines Postulate:
• Two non-vertical lines are perpendicular if and only
if the product of their slopes is -1.
Vertical and horizontal lines are perpendicular.
• l1⊥l2 if and only if
m1∙m2 = -1
• That is, m2 = -1/m1,
The slopes are
negative reciprocals
of each other.

20. Theorem: Perpendicular to Parallel Lines:
• In a plane, if a line is perpendicular to
one of two parallel lines, then it is
perpendicular to the other.
and
Then

21. Theorem: Two Perpendiculars:
• If two coplanar lines are each
perpendicular to the same line, then
they are parallel to each other.

23. If 2 perpendicular lines have gradients m1
and m2 then m2 is the negative reciprocal
of m1.
E.g. If line a has a gradient of 3 then line b
must have a gradient of -3 if both lines are
perpendicular to each other.

24. Given: l ll m and l ⊥ n
Prove: m ⊥ n
Statement Reason
1 l ll m, l ⊥ n Given
m∠1 = 90o
2 ∠1 is a right angle Definition of ⊥ lines
m∠2 = m∠1
3 Definition of a right angle
m∠2 = 90o
4 Corresponding angles postulate
5 Substitution property of equality
6 ∠2 is a right angle Definition of a right angle
7 m⊥n Definition of ⊥ lines

25. 1. Line r contains the points (-2,2) and (5,8).
Line s contains the points (-8,7) and (-2,0).
Is r ⊥ s?

26. 1. Given the equation of line v is
and line w is
Is v ⊥ w?

27. Given the line
3.Find the equation of the line passing through (
6,1) and perpendicular to the given line.
4. Find the equation of the line passing through
( 6,1) and parallel to the given line.