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Perpendicular lines, gradients, IB SL Mathematics
Perpendicular lines, gradients, IB SL Mathematics
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.
Perpendicular lines, gradients, IB SL Mathematics
Perpendicular lines, gradients, IB SL Mathematics
Perpendicular lines, gradients, IB SL Mathematics
Perpendicular lines, gradients, IB SL Mathematics
• 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
•   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
Perpendicular lines, gradients, IB SL Mathematics
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.
• 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
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
• 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
•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 )
•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
•Midpoint between A(2, 5) and B(8, 1):




•Midpoint between P(-4, -2) and Q(2, 3):
Perpendicular lines, gradients, IB SL Mathematics
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.
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
Theorem: Two Perpendiculars:
• If two coplanar lines are each
  perpendicular to the same line, then
  they are parallel to each other.
Perpendicular lines, gradients, IB SL Mathematics
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.
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
1. Line r contains the points (-2,2) and (5,8).
   Line s contains the points (-8,7) and (-2,0).
   Is r ⊥ s?
1. Given the equation of line v is
   and line w is
   Is v ⊥ w?
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.
Perpendicular lines, gradients, IB SL Mathematics

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Perpendicular lines, gradients, IB SL Mathematics

  • 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.