Post on 08-Aug-2018
Chapter 9 Coordinate Geometry
What is coordinate geometry?
DefinitionCoordinate geometry is a system of geometry where the position ofpoints on the plane is described using an ordered pair of numbers.
So basically, we are using algebra and distances to prove geometricconcepts, describe and identify polygons, etc.
What is coordinate geometry?
DefinitionCoordinate geometry is a system of geometry where the position ofpoints on the plane is described using an ordered pair of numbers.
So basically, we are using algebra and distances to prove geometricconcepts, describe and identify polygons, etc.
Some formulas we need
The Distance Formula
•(a, b)
P
•(c, d)
Q
The distance between points P and Q is given by
D =√
(c− a)2 + (d − b)2
Justification of Distance Formula
•(a, b)
P
•(c, d)
Q
•(c− a, b)
d1 = c− a
d2 = d − b
Justification of Distance Formula
•(a, b)
P
•(c, d)
Q
•(c− a, b)
d1 = c− a
d2 = d − b
Justification of Distance Formula
•(a, b)
P
•(c, d)
Q
•(c− a, b)
d1 = c− a
d2 = d − b
Some formulas we need
The Midpoint Formula
•(a, b)
P
•(c, d)
Q
•M
The midpoint of the line segment PQ is(a + c
2,
b + d2
)
Verification of Midpoint Formula
•(a, b)
P
•(c, d)
Q
•Md1
d2
What can we use to show that d1 = d2?
Verification of Midpoint Formula
•(a, b)
P
•(c, d)
Q
•Md1
d2
What can we use to show that d1 = d2?
Verification of Midpoint Formula
d1 =
√(a + c
2− a)2
+
(b + d
2− b)2
=
√(a + c
2− 2a
2
)2
+
(b + d
2− 2b
2
)2
=
√(c− a
2
)2
+
(d − b
2
)2
=12
√(c− a)2 + (d − b)2
Verification of Midpoint Formula
d1 =
√(a + c
2− a)2
+
(b + d
2− b)2
=
√(a + c
2− 2a
2
)2
+
(b + d
2− 2b
2
)2
=
√(c− a
2
)2
+
(d − b
2
)2
=12
√(c− a)2 + (d − b)2
Verification of Midpoint Formula
d1 =
√(a + c
2− a)2
+
(b + d
2− b)2
=
√(a + c
2− 2a
2
)2
+
(b + d
2− 2b
2
)2
=
√(c− a
2
)2
+
(d − b
2
)2
=12
√(c− a)2 + (d − b)2
Verification of Midpoint Formula
d1 =
√(a + c
2− a)2
+
(b + d
2− b)2
=
√(a + c
2− 2a
2
)2
+
(b + d
2− 2b
2
)2
=
√(c− a
2
)2
+
(d − b
2
)2
=12
√(c− a)2 + (d − b)2
Verification of Midpoint Formula
d2 =
√(c− a + c
2
)2
+
(d − b + d
2
)2
=
√(2c2− a + c
2
)2
+
(2d2− b + d
2
)2
=
√(c− a
2
)2
+
(d − b
2
)2
=12
√(c− a)2 + (d − b)2
Verification of Midpoint Formula
d2 =
√(c− a + c
2
)2
+
(d − b + d
2
)2
=
√(2c2− a + c
2
)2
+
(2d2− b + d
2
)2
=
√(c− a
2
)2
+
(d − b
2
)2
=12
√(c− a)2 + (d − b)2
Verification of Midpoint Formula
d2 =
√(c− a + c
2
)2
+
(d − b + d
2
)2
=
√(2c2− a + c
2
)2
+
(2d2− b + d
2
)2
=
√(c− a
2
)2
+
(d − b
2
)2
=12
√(c− a)2 + (d − b)2
Verification of Midpoint Formula
d2 =
√(c− a + c
2
)2
+
(d − b + d
2
)2
=
√(2c2− a + c
2
)2
+
(2d2− b + d
2
)2
=
√(c− a
2
)2
+
(d − b
2
)2
=12
√(c− a)2 + (d − b)2
Some formulas we need
SlopeThe slope of a line is given by
∆y∆x
=y2 − y1
x2 − x1
Parallel LinesIf two lines are parallel, then they are in the same plane and have thesame slope.
Perpendicular LinesIf two lines are perpendicular, then their slopes are negativereciprocals of each other; that is, if the slopes of the two lines are m1and m2, then m1m2 = −1.
Some formulas we need
SlopeThe slope of a line is given by
∆y∆x
=y2 − y1
x2 − x1
Parallel LinesIf two lines are parallel, then they are in the same plane and have thesame slope.
Perpendicular LinesIf two lines are perpendicular, then their slopes are negativereciprocals of each other; that is, if the slopes of the two lines are m1and m2, then m1m2 = −1.
Some formulas we need
SlopeThe slope of a line is given by
∆y∆x
=y2 − y1
x2 − x1
Parallel LinesIf two lines are parallel, then
they are in the same plane and have thesame slope.
Perpendicular LinesIf two lines are perpendicular, then their slopes are negativereciprocals of each other; that is, if the slopes of the two lines are m1and m2, then m1m2 = −1.
Some formulas we need
SlopeThe slope of a line is given by
∆y∆x
=y2 − y1
x2 − x1
Parallel LinesIf two lines are parallel, then they are in the same plane and have thesame slope.
Perpendicular LinesIf two lines are perpendicular, then their slopes are negativereciprocals of each other; that is, if the slopes of the two lines are m1and m2, then m1m2 = −1.
Some formulas we need
SlopeThe slope of a line is given by
∆y∆x
=y2 − y1
x2 − x1
Parallel LinesIf two lines are parallel, then they are in the same plane and have thesame slope.
Perpendicular LinesIf two lines are perpendicular, then
their slopes are negativereciprocals of each other; that is, if the slopes of the two lines are m1and m2, then m1m2 = −1.
Some formulas we need
SlopeThe slope of a line is given by
∆y∆x
=y2 − y1
x2 − x1
Parallel LinesIf two lines are parallel, then they are in the same plane and have thesame slope.
Perpendicular LinesIf two lines are perpendicular, then their slopes are negativereciprocals of each other; that is, if the slopes of the two lines are m1and m2, then m1m2 = −1.
Justification of Slopes of Perpendicular Lines
Suppose the first line is y = m1x and the second line is y = m2x
Justification of Slopes of Perpendicular Lines
At the point where x = 1, each line has risen (or fallen) exactly thevalue of the slope. So, at x = 1, the y-coordinate of l1 is m1 and they-coordinate of l2 is m2, where m1 and m2 are the slopes of lines l1and l2 respectively.
Justification of Slopes of Perpendicular Lines
These two lines are only perpendicular if our triangle satisfies thePythagorean Theorem, which says that in any right triangle the sum ofthe squares of the legs is equal to the square of the hypotenuse. Wegenerally express this as a2 + b2 = c2.
In order to apply this, we need to find the distance between thevertices of the triangle and then substitute those values into therelation above.
Justification of Slopes of Perpendicular Lines
These two lines are only perpendicular if our triangle satisfies thePythagorean Theorem, which says that in any right triangle the sum ofthe squares of the legs is equal to the square of the hypotenuse. Wegenerally express this as a2 + b2 = c2.
In order to apply this, we need to find the distance between thevertices of the triangle and then substitute those values into therelation above.
Justification of Slopes of Perpendicular Lines
Distance between (0, 0) and (1,m1):a2 = (
√(1− 0)2 + (m1 − 0)2)2 = 1 + m2
1
Distance between (0, 0) and (1,m2):b2 = (
√(1− 0)2 + (m2 − 0)2)2 = 1 + m2
2
Distance between (1,m1) and (1,m2):c2 = (
√(1− 1)2 + (m2 − m1)2)2 = (m2 − m1)2
Justification of Slopes of Perpendicular Lines
Distance between (0, 0) and (1,m1):a2 = (
√(1− 0)2 + (m1 − 0)2)2 = 1 + m2
1
Distance between (0, 0) and (1,m2):b2 = (
√(1− 0)2 + (m2 − 0)2)2 = 1 + m2
2
Distance between (1,m1) and (1,m2):c2 = (
√(1− 1)2 + (m2 − m1)2)2 = (m2 − m1)2
Justification of Slopes of Perpendicular Lines
Distance between (0, 0) and (1,m1):a2 = (
√(1− 0)2 + (m1 − 0)2)2 = 1 + m2
1
Distance between (0, 0) and (1,m2):b2 = (
√(1− 0)2 + (m2 − 0)2)2 = 1 + m2
2
Distance between (1,m1) and (1,m2):c2 = (
√(1− 1)2 + (m2 − m1)2)2 = (m2 − m1)2
Justification of Slopes of Perpendicular Lines
This gives the following :
a2 + b2 = c2
1 + m21 + 1 + m2
2 = (m2 − m1)2
2 + m21 + m2
2 = m22 − 2m1m2 + m2
1
2 = −2m1m2
− 1m1
= m2
That is, the slopes are negative reciprocals of each other.
Justification of Slopes of Perpendicular Lines
This gives the following :
a2 + b2 = c2
1 + m21 + 1 + m2
2 = (m2 − m1)2
2 + m21 + m2
2 = m22 − 2m1m2 + m2
1
2 = −2m1m2
− 1m1
= m2
That is, the slopes are negative reciprocals of each other.
Justification of Slopes of Perpendicular Lines
This gives the following :
a2 + b2 = c2
1 + m21 + 1 + m2
2 = (m2 − m1)2
2 + m21 + m2
2 = m22 − 2m1m2 + m2
1
2 = −2m1m2
− 1m1
= m2
That is, the slopes are negative reciprocals of each other.
Justification of Slopes of Perpendicular Lines
This gives the following :
a2 + b2 = c2
1 + m21 + 1 + m2
2 = (m2 − m1)2
2 + m21 + m2
2 = m22 − 2m1m2 + m2
1
2 = −2m1m2
− 1m1
= m2
That is, the slopes are negative reciprocals of each other.
Justification of Slopes of Perpendicular Lines
This gives the following :
a2 + b2 = c2
1 + m21 + 1 + m2
2 = (m2 − m1)2
2 + m21 + m2
2 = m22 − 2m1m2 + m2
1
2 = −2m1m2
− 1m1
= m2
That is, the slopes are negative reciprocals of each other.
Equations of Lines
Slope intercept form:
y = mx + b
Point-slope form: y− y1 = m(x− x1)
Point-point form: y− y1 = y2−y1x2−x1
(x− x1)
Equations of Lines
Slope intercept form: y = mx + b
Point-slope form: y− y1 = m(x− x1)
Point-point form: y− y1 = y2−y1x2−x1
(x− x1)
Equations of Lines
Slope intercept form: y = mx + b
Point-slope form: y− y1 = m(x− x1)
Point-point form: y− y1 = y2−y1x2−x1
(x− x1)
Equations of Lines
Slope intercept form: y = mx + b
Point-slope form: y− y1 = m(x− x1)
Point-point form: y− y1 = y2−y1x2−x1
(x− x1)
Triangles and Parallel Lines
ExampleProve that the segment joining the midpoints of opposite sides of anytriangle is parallel to the third side.
What do we want to do first?
A C
B
• •M1 M2
Triangles and Parallel Lines
ExampleProve that the segment joining the midpoints of opposite sides of anytriangle is parallel to the third side.
What do we want to do first?
A C
B
• •M1 M2
Triangles and Parallel Lines
ExampleProve that the segment joining the midpoints of opposite sides of anytriangle is parallel to the third side.
What do we want to do first?
A C
B
• •M1 M2
Triangles and Parallel Lines
How should we assign coordinates to the points?
A (0, 0) C (2e, 0)
B (2c, 2d)
• •M1 (c, d) M2 (c + e, d)
Where does c + e come from?
Triangles and Parallel Lines
How should we assign coordinates to the points?
A (0, 0) C (2e, 0)
B (2c, 2d)
• •M1 (c, d) M2 (c + e, d)
Where does c + e come from?
Triangles and Parallel Lines
How should we assign coordinates to the points?
A (0, 0) C (2e, 0)
B (2c, 2d)
• •M1 (c, d) M2 (c + e, d)
Where does c + e come from?
Triangles and Parallel Lines
m1 =d − d
c + e− c=
0e
= 0
m2 =0− 02e− 0
=02e
= 0
Since the lines are coplanar and have the same slope, they are parallel.
Triangles and Parallel Lines
m1 =d − d
c + e− c=
0e
= 0
m2 =0− 02e− 0
=02e
= 0
Since the lines are coplanar and have the same slope, they are parallel.
Triangles and Parallel Lines
m1 =d − d
c + e− c=
0e
= 0
m2 =0− 02e− 0
=02e
= 0
Since the lines are coplanar and have the same slope, they are parallel.
Diagonals of a Parallelogram
ExampleProve that the diagonals of a parallelogram bisect each other.
A B
D C
How do we want to label the vertices?
Diagonals of a Parallelogram
ExampleProve that the diagonals of a parallelogram bisect each other.
A B
D C
How do we want to label the vertices?
Diagonals of a Parallelogram
ExampleProve that the diagonals of a parallelogram bisect each other.
A B
D C
How do we want to label the vertices?
Diagonals of a Parallelogram
A (0, 0) B (2a, 0)
D (2b, 2c) C (2a + 2b, 2c)
Now what do we need to find in order to prove that the diagonalsbisect each other?
Diagonals of a Parallelogram
A (0, 0) B (2a, 0)
D (2b, 2c) C (2a + 2b, 2c)
Now what do we need to find in order to prove that the diagonalsbisect each other?
Diagonals of a Parallelogram
Midpoint1 =
(2a + 2b + 0
2,
2c + 02
)= (a + b, c)
Midpoint2 =
(2a + 2b
2,
2c2
)= (a + b, c)
Since the midpoint of each line is the same point, the diagonals bisecteach other.
Diagonals of a Parallelogram
Midpoint1 =
(2a + 2b + 0
2,
2c + 02
)= (a + b, c)
Midpoint2 =
(2a + 2b
2,
2c2
)= (a + b, c)
Since the midpoint of each line is the same point, the diagonals bisecteach other.
Diagonals of a Parallelogram
Midpoint1 =
(2a + 2b + 0
2,
2c + 02
)= (a + b, c)
Midpoint2 =
(2a + 2b
2,
2c2
)= (a + b, c)
Since the midpoint of each line is the same point, the diagonals bisecteach other.
Rectangle?
Using what we have, what additional property could be used to showthat the parallelogram is a rectangle?
If a parallelogram is a rectangle, then the diagonals are of equallength.
AC =√
(2a + 2b)2 + (2c)2
BD =√
(2a− 2b)2 + (2c)2
Rectangle?
Using what we have, what additional property could be used to showthat the parallelogram is a rectangle?
If a parallelogram is a rectangle, then the diagonals are of equallength.
AC =√
(2a + 2b)2 + (2c)2
BD =√
(2a− 2b)2 + (2c)2
Rectangle?
Using what we have, what additional property could be used to showthat the parallelogram is a rectangle?
If a parallelogram is a rectangle, then the diagonals are of equallength.
AC =
√(2a + 2b)2 + (2c)2
BD =√
(2a− 2b)2 + (2c)2
Rectangle?
Using what we have, what additional property could be used to showthat the parallelogram is a rectangle?
If a parallelogram is a rectangle, then the diagonals are of equallength.
AC =√
(2a + 2b)2 + (2c)2
BD =
√(2a− 2b)2 + (2c)2
Rectangle?
Using what we have, what additional property could be used to showthat the parallelogram is a rectangle?
If a parallelogram is a rectangle, then the diagonals are of equallength.
AC =√
(2a + 2b)2 + (2c)2
BD =√
(2a− 2b)2 + (2c)2
Rectangle?
If AC = BD then we have
(2a + 2b)2 + (2c)2 = (2a− 2b)2 + (2c)2
4a2 + 8ab + 4b2 = 4a2 − 8ab + 4b2
16ab = 0
This means either a = 0 or b = 0. a 6= 0 since A 6= B.
If b = 0 then D is on the y-axis, so ... AB is horizontal and AD isvertical.
Rectangle?
If AC = BD then we have
(2a + 2b)2 + (2c)2 = (2a− 2b)2 + (2c)2
4a2 + 8ab + 4b2 = 4a2 − 8ab + 4b2
16ab = 0
This means either a = 0 or b = 0. a 6= 0 since A 6= B.
If b = 0 then D is on the y-axis, so ... AB is horizontal and AD isvertical.
Rectangle?
If AC = BD then we have
(2a + 2b)2 + (2c)2 = (2a− 2b)2 + (2c)2
4a2 + 8ab + 4b2 = 4a2 − 8ab + 4b2
16ab = 0
This means either a = 0 or b = 0. a 6= 0 since A 6= B.
If b = 0 then D is on the y-axis, so ... AB is horizontal and AD isvertical.
Rectangle?
If AC = BD then we have
(2a + 2b)2 + (2c)2 = (2a− 2b)2 + (2c)2
4a2 + 8ab + 4b2 = 4a2 − 8ab + 4b2
16ab = 0
This means either a = 0 or b = 0.
a 6= 0 since A 6= B.
If b = 0 then D is on the y-axis, so ... AB is horizontal and AD isvertical.
Rectangle?
If AC = BD then we have
(2a + 2b)2 + (2c)2 = (2a− 2b)2 + (2c)2
4a2 + 8ab + 4b2 = 4a2 − 8ab + 4b2
16ab = 0
This means either a = 0 or b = 0. a 6= 0 since A 6= B.
If b = 0 then D is on the y-axis, so ... AB is horizontal and AD isvertical.
Rectangle?
If AC = BD then we have
(2a + 2b)2 + (2c)2 = (2a− 2b)2 + (2c)2
4a2 + 8ab + 4b2 = 4a2 − 8ab + 4b2
16ab = 0
This means either a = 0 or b = 0. a 6= 0 since A 6= B.
If b = 0 then D is on the y-axis, so ...
AB is horizontal and AD isvertical.
Rectangle?
If AC = BD then we have
(2a + 2b)2 + (2c)2 = (2a− 2b)2 + (2c)2
4a2 + 8ab + 4b2 = 4a2 − 8ab + 4b2
16ab = 0
This means either a = 0 or b = 0. a 6= 0 since A 6= B.
If b = 0 then D is on the y-axis, so ... AB is horizontal and AD isvertical.
Squares
ExampleProve that the diagonals of a square are perpendicular bisectors ofeach other.
A B
D C
How do we want to label the vertices?
Squares
ExampleProve that the diagonals of a square are perpendicular bisectors ofeach other.
A B
D C
How do we want to label the vertices?
Squares
ExampleProve that the diagonals of a square are perpendicular bisectors ofeach other.
A B
D C
How do we want to label the vertices?
Squares
ExampleProve that the diagonals of a square are perpendicular bisectors ofeach other.
A (0, 0)
B (2a, 0)
D (0, 2a) C (2a, 2a)
Squares
ExampleProve that the diagonals of a square are perpendicular bisectors ofeach other.
A (0, 0) B (2a, 0)
D (0, 2a) C (2a, 2a)
Squares
ExampleProve that the diagonals of a square are perpendicular bisectors ofeach other.
A (0, 0) B (2a, 0)
D (0, 2a) C (2a, 2a)
Squares
First, we need to show the diagonals bisect each other. To do so, wewill show that the midpoint of each diagonal is the same point.
mid1 =
(2a + 0
2,
2a + 02
)= (a, a)
mid2 =
(0 + 2a
2,
2a + 02
)= (a, a)
Now that we know know the diagonals bisect each other, we need toshow that the lines are perpendicular. We will do so by finding theslopes and showing that they are negative reciprocals of each other.
m1 =2a− 02a− 0
= 1
m2 =2a− 00− 2a
= −1
Since m1m2 = −1, the diagonals are perpendicular.
Squares
First, we need to show the diagonals bisect each other. To do so, wewill show that the midpoint of each diagonal is the same point.
mid1 =
(2a + 0
2,
2a + 02
)= (a, a)
mid2 =
(0 + 2a
2,
2a + 02
)= (a, a)
Now that we know know the diagonals bisect each other, we need toshow that the lines are perpendicular. We will do so by finding theslopes and showing that they are negative reciprocals of each other.
m1 =2a− 02a− 0
= 1
m2 =2a− 00− 2a
= −1
Since m1m2 = −1, the diagonals are perpendicular.
Squares
First, we need to show the diagonals bisect each other. To do so, wewill show that the midpoint of each diagonal is the same point.
mid1 =
(2a + 0
2,
2a + 02
)= (a, a)
mid2 =
(0 + 2a
2,
2a + 02
)= (a, a)
Now that we know know the diagonals bisect each other, we need toshow that the lines are perpendicular. We will do so by finding theslopes and showing that they are negative reciprocals of each other.
m1 =2a− 02a− 0
= 1
m2 =2a− 00− 2a
= −1
Since m1m2 = −1, the diagonals are perpendicular.
Squares
First, we need to show the diagonals bisect each other. To do so, wewill show that the midpoint of each diagonal is the same point.
mid1 =
(2a + 0
2,
2a + 02
)= (a, a)
mid2 =
(0 + 2a
2,
2a + 02
)= (a, a)
Now that we know know the diagonals bisect each other, we need toshow that the lines are perpendicular. We will do so by finding theslopes and showing that they are negative reciprocals of each other.
m1 =2a− 02a− 0
= 1
m2 =2a− 00− 2a
= −1
Since m1m2 = −1, the diagonals are perpendicular.
Squares
First, we need to show the diagonals bisect each other. To do so, wewill show that the midpoint of each diagonal is the same point.
mid1 =
(2a + 0
2,
2a + 02
)= (a, a)
mid2 =
(0 + 2a
2,
2a + 02
)= (a, a)
Now that we know know the diagonals bisect each other, we need toshow that the lines are perpendicular. We will do so by finding theslopes and showing that they are negative reciprocals of each other.
m1 =2a− 02a− 0
= 1
m2 =2a− 00− 2a
= −1
Since m1m2 = −1, the diagonals are perpendicular.
Squares
First, we need to show the diagonals bisect each other. To do so, wewill show that the midpoint of each diagonal is the same point.
mid1 =
(2a + 0
2,
2a + 02
)= (a, a)
mid2 =
(0 + 2a
2,
2a + 02
)= (a, a)
Now that we know know the diagonals bisect each other, we need toshow that the lines are perpendicular. We will do so by finding theslopes and showing that they are negative reciprocals of each other.
m1 =2a− 02a− 0
= 1
m2 =2a− 00− 2a
= −1
Since m1m2 = −1, the diagonals are perpendicular.
Squares
First, we need to show the diagonals bisect each other. To do so, wewill show that the midpoint of each diagonal is the same point.
mid1 =
(2a + 0
2,
2a + 02
)= (a, a)
mid2 =
(0 + 2a
2,
2a + 02
)= (a, a)
Now that we know know the diagonals bisect each other, we need toshow that the lines are perpendicular. We will do so by finding theslopes and showing that they are negative reciprocals of each other.
m1 =2a− 02a− 0
= 1
m2 =2a− 00− 2a
= −1
Since m1m2 = −1, the diagonals are perpendicular.
Squares
First, we need to show the diagonals bisect each other. To do so, wewill show that the midpoint of each diagonal is the same point.
mid1 =
(2a + 0
2,
2a + 02
)= (a, a)
mid2 =
(0 + 2a
2,
2a + 02
)= (a, a)
Now that we know know the diagonals bisect each other, we need toshow that the lines are perpendicular. We will do so by finding theslopes and showing that they are negative reciprocals of each other.
m1 =2a− 02a− 0
= 1
m2 =2a− 00− 2a
= −1
Since m1m2 = −1, the diagonals are perpendicular.
Isosceles Triangles
ExampleProve that is the midpoint of one side of a rectangle is joined to theendpoints of the opposite side, an isosceles triangle is formed.
(0, 0) (2a, 0)
(2a, 2b)(0, 2b) (a, 2b)
d1 d2
Isosceles Triangles
ExampleProve that is the midpoint of one side of a rectangle is joined to theendpoints of the opposite side, an isosceles triangle is formed.
(0, 0) (2a, 0)
(2a, 2b)(0, 2b) (a, 2b)
d1 d2
Isosceles Triangles
ExampleProve that is the midpoint of one side of a rectangle is joined to theendpoints of the opposite side, an isosceles triangle is formed.
(0, 0)
(2a, 0)
(2a, 2b)(0, 2b) (a, 2b)
d1 d2
Isosceles Triangles
ExampleProve that is the midpoint of one side of a rectangle is joined to theendpoints of the opposite side, an isosceles triangle is formed.
(0, 0) (2a, 0)
(2a, 2b)
(0, 2b) (a, 2b)
d1 d2
Isosceles Triangles
ExampleProve that is the midpoint of one side of a rectangle is joined to theendpoints of the opposite side, an isosceles triangle is formed.
(0, 0) (2a, 0)
(2a, 2b)(0, 2b)
(a, 2b)
d1 d2
Isosceles Triangles
ExampleProve that is the midpoint of one side of a rectangle is joined to theendpoints of the opposite side, an isosceles triangle is formed.
(0, 0) (2a, 0)
(2a, 2b)(0, 2b)
(a, 2b)
d1 d2
Isosceles Triangles
ExampleProve that is the midpoint of one side of a rectangle is joined to theendpoints of the opposite side, an isosceles triangle is formed.
(0, 0) (2a, 0)
(2a, 2b)(0, 2b) (a, 2b)
d1 d2
Isosceles Triangles
ExampleProve that is the midpoint of one side of a rectangle is joined to theendpoints of the opposite side, an isosceles triangle is formed.
(0, 0) (2a, 0)
(2a, 2b)(0, 2b) (a, 2b)
d1 d2
Isosceles Triangles
We need only to show that d1 = d2 to show that an isosceles triangleis formed in this manner.
d1 =√
(2b− 0)2 + (a− 0)2
=√
a2 + 4b2
d2 =√
(a− 2a)2 + (2b− 0)2
=√
a2 + 4b2
Since the distances are equal, we have an isosceles triangle.
Isosceles Triangles
We need only to show that d1 = d2 to show that an isosceles triangleis formed in this manner.
d1 =√
(2b− 0)2 + (a− 0)2
=√
a2 + 4b2
d2 =√
(a− 2a)2 + (2b− 0)2
=√
a2 + 4b2
Since the distances are equal, we have an isosceles triangle.
Isosceles Triangles
We need only to show that d1 = d2 to show that an isosceles triangleis formed in this manner.
d1 =√
(2b− 0)2 + (a− 0)2
=√
a2 + 4b2
d2 =√
(a− 2a)2 + (2b− 0)2
=√
a2 + 4b2
Since the distances are equal, we have an isosceles triangle.
Isosceles Triangles
We need only to show that d1 = d2 to show that an isosceles triangleis formed in this manner.
d1 =√
(2b− 0)2 + (a− 0)2
=√
a2 + 4b2
d2 =√
(a− 2a)2 + (2b− 0)2
=√
a2 + 4b2
Since the distances are equal, we have an isosceles triangle.
Isosceles Triangles
We need only to show that d1 = d2 to show that an isosceles triangleis formed in this manner.
d1 =√
(2b− 0)2 + (a− 0)2
=√
a2 + 4b2
d2 =√
(a− 2a)2 + (2b− 0)2
=√
a2 + 4b2
Since the distances are equal, we have an isosceles triangle.
Isosceles Triangles
We need only to show that d1 = d2 to show that an isosceles triangleis formed in this manner.
d1 =√
(2b− 0)2 + (a− 0)2
=√
a2 + 4b2
d2 =√
(a− 2a)2 + (2b− 0)2
=√
a2 + 4b2
Since the distances are equal, we have an isosceles triangle.
Right Triangles
ExampleProve that the midpoint of the hypotenuse of a right triangle isequidistant from all three vertices.
(0, 0) (2a, 0)
(0, 2b)
•(a, b)
d1
d2
d3
Right Triangles
ExampleProve that the midpoint of the hypotenuse of a right triangle isequidistant from all three vertices.
(0, 0) (2a, 0)
(0, 2b)
•(a, b)
d1
d2
d3
Right Triangles
ExampleProve that the midpoint of the hypotenuse of a right triangle isequidistant from all three vertices.
(0, 0)
(2a, 0)
(0, 2b)
•(a, b)
d1
d2
d3
Right Triangles
ExampleProve that the midpoint of the hypotenuse of a right triangle isequidistant from all three vertices.
(0, 0) (2a, 0)
(0, 2b)
•(a, b)
d1
d2
d3
Right Triangles
ExampleProve that the midpoint of the hypotenuse of a right triangle isequidistant from all three vertices.
(0, 0) (2a, 0)
(0, 2b)
•(a, b)
d1
d2
d3
Right Triangles
ExampleProve that the midpoint of the hypotenuse of a right triangle isequidistant from all three vertices.
(0, 0) (2a, 0)
(0, 2b)
•(a, b)
d1
d2
d3
Right Triangles
We know by definition that d2 = d3, so it is pointless to prove thesesegments have equal measure. We do need to find the length of one ofthem, however, as we need something to compare to d1.
d1 =√
(a− 0)2 + (2b− b)2 =√
a2 + b2
d2 =√
(a− 2a)2 + (b− 0)2 =√
a2 + b2
Since d1 = d2, we have that the midpoint of the hypotenuse isequidistant from all three vertices.
Right Triangles
We know by definition that d2 = d3, so it is pointless to prove thesesegments have equal measure. We do need to find the length of one ofthem, however, as we need something to compare to d1.
d1 =√
(a− 0)2 + (2b− b)2 =√
a2 + b2
d2 =√
(a− 2a)2 + (b− 0)2 =√
a2 + b2
Since d1 = d2, we have that the midpoint of the hypotenuse isequidistant from all three vertices.
Right Triangles
We know by definition that d2 = d3, so it is pointless to prove thesesegments have equal measure. We do need to find the length of one ofthem, however, as we need something to compare to d1.
d1 =√
(a− 0)2 + (2b− b)2 =√
a2 + b2
d2 =√
(a− 2a)2 + (b− 0)2 =√
a2 + b2
Since d1 = d2, we have that the midpoint of the hypotenuse isequidistant from all three vertices.
Right Triangles
We know by definition that d2 = d3, so it is pointless to prove thesesegments have equal measure. We do need to find the length of one ofthem, however, as we need something to compare to d1.
d1 =√
(a− 0)2 + (2b− b)2 =√
a2 + b2
d2 =√
(a− 2a)2 + (b− 0)2 =√
a2 + b2
Since d1 = d2, we have that the midpoint of the hypotenuse isequidistant from all three vertices.
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0) (2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0) (2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0) (2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0) (2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0)
(2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0) (2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0) (2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0) (2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0) (2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0) (2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0) (2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
ExampleProve that the segments that join the midpoints of a quadrilateral forma parallelogram.
•
•
•
•
(0, 0) (2a, 0)
(2c, 2d)
(2e, 2f )
(a, 0)
(a + c, d)
(c + e, d + f )
(e, f )
Another Parallelogram Proof
To show that the inside quadrilateral is a parallelogram, we need toshow that opposite sides have the same slope. Working clockwise, wehave
m1 =d + f − fc + e− e
=dc
m2 =d − (d + f )
a + c− (c + e)=−f
a− e
m3 =0− d
a− (a + c)=−d−c
=dc
m4 =f − 0e− a
=−f
a− e
Since we have opposite sides parallel, the quadrilateral is aparallelogram.
Another Parallelogram Proof
To show that the inside quadrilateral is a parallelogram, we need toshow that opposite sides have the same slope. Working clockwise, wehave
m1 =d + f − fc + e− e
=dc
m2 =d − (d + f )
a + c− (c + e)=−f
a− e
m3 =0− d
a− (a + c)=−d−c
=dc
m4 =f − 0e− a
=−f
a− e
Since we have opposite sides parallel, the quadrilateral is aparallelogram.
Another Parallelogram Proof
To show that the inside quadrilateral is a parallelogram, we need toshow that opposite sides have the same slope. Working clockwise, wehave
m1 =d + f − fc + e− e
=dc
m2 =d − (d + f )
a + c− (c + e)=−f
a− e
m3 =0− d
a− (a + c)=−d−c
=dc
m4 =f − 0e− a
=−f
a− e
Since we have opposite sides parallel, the quadrilateral is aparallelogram.
Another Parallelogram Proof
To show that the inside quadrilateral is a parallelogram, we need toshow that opposite sides have the same slope. Working clockwise, wehave
m1 =d + f − fc + e− e
=dc
m2 =d − (d + f )
a + c− (c + e)=−f
a− e
m3 =0− d
a− (a + c)=−d−c
=dc
m4 =f − 0e− a
=−f
a− e
Since we have opposite sides parallel, the quadrilateral is aparallelogram.
Another Parallelogram Proof
To show that the inside quadrilateral is a parallelogram, we need toshow that opposite sides have the same slope. Working clockwise, wehave
m1 =d + f − fc + e− e
=dc
m2 =d − (d + f )
a + c− (c + e)=−f
a− e
m3 =0− d
a− (a + c)=−d−c
=dc
m4 =f − 0e− a
=−f
a− e
Since we have opposite sides parallel, the quadrilateral is aparallelogram.
Medians
ExampleProve that the three medians of a triangle are concurrent at the pointthat is two-thirds of the distance from any vertex to the midpoint ofthe opposite side.
DefinitionThe median of a triangle is the line segment joining a vertex to themidpoint of the opposite side.
Medians
ExampleProve that the three medians of a triangle are concurrent at the pointthat is two-thirds of the distance from any vertex to the midpoint ofthe opposite side.
DefinitionThe median of a triangle is the line segment joining a vertex to themidpoint of the opposite side.
Medians
A (0, 0) B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2(a + b, c)M3(b, c)
Approach?
Medians
A (0, 0)
B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2(a + b, c)M3(b, c)
Approach?
Medians
A (0, 0) B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2(a + b, c)M3(b, c)
Approach?
Medians
A (0, 0) B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2(a + b, c)M3(b, c)
Approach?
Medians
A (0, 0) B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2(a + b, c)M3(b, c)
Approach?
Medians
A (0, 0) B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2(a + b, c)M3(b, c)
Approach?
Medians
A (0, 0) B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2
(a + b, c)M3(b, c)
Approach?
Medians
A (0, 0) B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2(a + b, c)
M3(b, c)
Approach?
Medians
A (0, 0) B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2(a + b, c)M3
(b, c)
Approach?
Medians
A (0, 0) B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2(a + b, c)M3(b, c)
Approach?
Medians
A (0, 0) B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2(a + b, c)M3(b, c)
Approach?
Medians
A (0, 0) B (2a, 0)
C (2b, 2c)
M1
(a, 0)
M2(a + b, c)M3(b, c)
Approach?
Medians
First, we’ll start with AM2. What is the slope?
m1 =c
a + bSince the y-intercept is the origin, we have
y1 =c
a + bx
Medians
First, we’ll start with AM2. What is the slope?
m1 =c
a + b
Since the y-intercept is the origin, we have
y1 =c
a + bx
Medians
First, we’ll start with AM2. What is the slope?
m1 =c
a + bSince the y-intercept is the origin, we have
y1 =c
a + bx
Medians
First, we’ll start with AM2. What is the slope?
m1 =c
a + bSince the y-intercept is the origin, we have
y1 =c
a + bx
Medians
We need the equation for another line. WLOG, we’ll use BM3.
What is the slope?
m3 =c
b− 2a
Now we need the y-intercept.
0 =c
b− 2a(2a) + b′
0 =2ac
b− 2a+ b′
b′ =2ac
b− 2aSo,
y3 =c
b− 2ax− 2ac
b− 2a
Medians
We need the equation for another line. WLOG, we’ll use BM3.
What is the slope?
m3 =c
b− 2a
Now we need the y-intercept.
0 =c
b− 2a(2a) + b′
0 =2ac
b− 2a+ b′
b′ =2ac
b− 2aSo,
y3 =c
b− 2ax− 2ac
b− 2a
Medians
We need the equation for another line. WLOG, we’ll use BM3.
What is the slope?
m3 =c
b− 2a
Now we need the y-intercept.
0 =c
b− 2a(2a) + b′
0 =2ac
b− 2a+ b′
b′ =2ac
b− 2aSo,
y3 =c
b− 2ax− 2ac
b− 2a
Medians
We need the equation for another line. WLOG, we’ll use BM3.
What is the slope?
m3 =c
b− 2a
Now we need the y-intercept.
0 =c
b− 2a(2a) + b′
0 =2ac
b− 2a+ b′
b′ =2ac
b− 2aSo,
y3 =c
b− 2ax− 2ac
b− 2a
Medians
We need the equation for another line. WLOG, we’ll use BM3.
What is the slope?
m3 =c
b− 2a
Now we need the y-intercept.
0 =c
b− 2a(2a) + b′
0 =2ac
b− 2a+ b′
b′ =2ac
b− 2aSo,
y3 =c
b− 2ax− 2ac
b− 2a
Medians
We need the equation for another line. WLOG, we’ll use BM3.
What is the slope?
m3 =c
b− 2a
Now we need the y-intercept.
0 =c
b− 2a(2a) + b′
0 =2ac
b− 2a+ b′
b′ =2ac
b− 2aSo,
y3 =c
b− 2ax− 2ac
b− 2a
Medians
We need the equation for another line. WLOG, we’ll use BM3.
What is the slope?
m3 =c
b− 2a
Now we need the y-intercept.
0 =c
b− 2a(2a) + b′
0 =2ac
b− 2a+ b′
b′ =2ac
b− 2a
So,
y3 =c
b− 2ax− 2ac
b− 2a
Medians
We need the equation for another line. WLOG, we’ll use BM3.
What is the slope?
m3 =c
b− 2a
Now we need the y-intercept.
0 =c
b− 2a(2a) + b′
0 =2ac
b− 2a+ b′
b′ =2ac
b− 2aSo,
y3 =c
b− 2ax− 2ac
b− 2a
Medians
Now we solve ...
ca + b
x =c
b− 2a− 2ac
b− 2a
bcx− 2acx = acx + bcx− 2a2c− 2abc
− 2acx = acx− 2a2c− 2abc
− 3acx = −2a2c− 2abc
− 3acx = −2ac(a + b)
x =23
(a + b)
Medians
Now we solve ...
ca + b
x =c
b− 2a− 2ac
b− 2abcx− 2acx = acx + bcx− 2a2c− 2abc
− 2acx = acx− 2a2c− 2abc
− 3acx = −2a2c− 2abc
− 3acx = −2ac(a + b)
x =23
(a + b)
Medians
Now we solve ...
ca + b
x =c
b− 2a− 2ac
b− 2abcx− 2acx = acx + bcx− 2a2c− 2abc
− 2acx = acx− 2a2c− 2abc
− 3acx = −2a2c− 2abc
− 3acx = −2ac(a + b)
x =23
(a + b)
Medians
Now we solve ...
ca + b
x =c
b− 2a− 2ac
b− 2abcx− 2acx = acx + bcx− 2a2c− 2abc
− 2acx = acx− 2a2c− 2abc
− 3acx = −2a2c− 2abc
− 3acx = −2ac(a + b)
x =23
(a + b)
Medians
Now we solve ...
ca + b
x =c
b− 2a− 2ac
b− 2abcx− 2acx = acx + bcx− 2a2c− 2abc
− 2acx = acx− 2a2c− 2abc
− 3acx = −2a2c− 2abc
− 3acx = −2ac(a + b)
x =23
(a + b)
Medians
Now we solve ...
ca + b
x =c
b− 2a− 2ac
b− 2abcx− 2acx = acx + bcx− 2a2c− 2abc
− 2acx = acx− 2a2c− 2abc
− 3acx = −2a2c− 2abc
− 3acx = −2ac(a + b)
x =23
(a + b)
Medians
x =23
(a + b)
What does this do for us?
y =c
a + b
(23
(a + b)
)y =
23
c
This gives us the coordinate 23 of the way from the origin to the
endpoint of the the associated median.
Medians
x =23
(a + b)
What does this do for us?
y =c
a + b
(23
(a + b)
)
y =23
c
This gives us the coordinate 23 of the way from the origin to the
endpoint of the the associated median.
Medians
x =23
(a + b)
What does this do for us?
y =c
a + b
(23
(a + b)
)y =
23
c
This gives us the coordinate 23 of the way from the origin to the
endpoint of the the associated median.
Medians
x =23
(a + b)
What does this do for us?
y =c
a + b
(23
(a + b)
)y =
23
c
This gives us the coordinate 23 of the way from the origin to the
endpoint of the the associated median.