Perpendicular lines, gradients, IB SL Mathematics
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Transcript of 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.
• 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
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
32+42=x2
9+16=x2
x=25(The opposite of x2 is )X=5
4
3
x
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 Postulate:
• l1⊥l2 if and only if m1∙m2 = -1
• That is, m2 = -1/m1, The slopes are negative reciprocals of each other.
• Two non-vertical lines are perpendicular if and only if the product of their slopes is -1.Vertical and horizontal lines are perpendicular.
• In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.
Theorem: Perpendicular to Parallel Lines:
and Then
• If two coplanar lines are each perpendicular to the same line, then they are parallel to each other.
Theorem: Two Perpendiculars:
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.
Statement Reason
1 l ll m, l ⊥ n Given
2 ∠1 is a right angle Definition of lines⊥
3 m∠1 = 90o Definition of a right angle
4 m 2 ∠ = m∠1 Corresponding angles postulate
5 m∠2 = 90o Substitution property of equality
6 ∠2 is a right angle Definition of a right angle
7 m ⊥ n Definition of lines⊥
Given: l ll m and l ⊥ n Prove: m ⊥ n
1. Line r contains the points (-2,2) and (5,8). Line s contains the points (-8,7) and (-2,0). Is r ⊥ s?
2. Given the equation of line v isand 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.