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Section 2-10
Perpendiculars and Distance
Essential Questions
n How do you find the distance between a point and
a line?
n How do you find the distance between parallel
lines?
Vocabulary
1. Equidistant:
2. Distance Between a Point and a Line:
3. Distance Between Parallel Lines:
Vocabulary
1. Equidistant: The distance between any two lines as
measured along a perpendicular is the same; this
occurs with parallel lines
2. Distance Between a Point and a Line:
3. Distance Between Parallel Lines:
Vocabulary
1. Equidistant: The distance between any two lines as
measured along a perpendicular is the same; this
occurs with parallel lines
2. Distance Between a Point and a Line: The length
of the segment perpendicular to the line, with the
endpoints being that point and a point on the line
3. Distance Between Parallel Lines:
Vocabulary
1. Equidistant: The distance between any two lines as
measured along a perpendicular is the same; this
occurs with parallel lines
2. Distance Between a Point and a Line: The length
of the segment perpendicular to the line, with the
endpoints being that point and a point on the line
3. Distance Between Parallel Lines: The length of the
segment perpendicular to the two parallel lines with
the endpoints on either of the parallel lines
Postulates & Theorems
1. Perpendicular Postulate:
2. Two Lines Equidistant from a Third:
Postulates & Theorems
1. Perpendicular Postulate: If given a line and a point
not on the line, then there exists exactly one line
through the point that is perpendicular to the given
line
2. Two Lines Equidistant from a Third:
Postulates & Theorems
1. Perpendicular Postulate: If given a line and a point
not on the line, then there exists exactly one line
through the point that is perpendicular to the given
line
2. Two Lines Equidistant from a Third: In a plane, if
two lines are each equidistant from a third line,
then the two lines are parallel to each other
Steps to find the Distance
from a Point to a Line
Steps to find the Distance
from a Point to a Line
1. Find the equation of the original line
Steps to find the Distance
from a Point to a Line
1. Find the equation of the original line
2. Find the equation of the perpendicular line through the
other point
Steps to find the Distance
from a Point to a Line
1. Find the equation of the original line
2. Find the equation of the perpendicular line through the
other point
3. Solve the system of these two equations.
Steps to find the Distance
from a Point to a Line
1. Find the equation of the original line
2. Find the equation of the perpendicular line through the
other point
3. Solve the system of these two equations.
4. Use the distance formula utilizing this point on the line
and the point not on the line.
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
= −1
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
= −1 T(0, 0)
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
= −1 T(0, 0)
y = mx + b
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
= −1 T(0, 0)
y = −x
y = mx + b
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
= −1 T(0, 0)
y = −x
2. Find the equation of the perpendicular line through the
other point
y = mx + b
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
= −1 T(0, 0)
y = −x
2. Find the equation of the perpendicular line through the
other point
m = 1
y = mx + b
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
= −1 T(0, 0)
y = −x
2. Find the equation of the perpendicular line through the
other point
m = 1
V(1, 5)
y = mx + b
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
= −1 T(0, 0)
y = −x
2. Find the equation of the perpendicular line through the
other point
m = 1
V(1, 5)
y − y1
= m(x − x1
)
y = mx + b
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
= −1 T(0, 0)
y = −x
2. Find the equation of the perpendicular line through the
other point
m = 1
V(1, 5)
y − y1
= m(x − x1
)
y = mx + b
y − 5 = 1(x −1)
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
= −1 T(0, 0)
y = −x
2. Find the equation of the perpendicular line through the
other point
m = 1
V(1, 5)
y − y1
= m(x − x1
)
y = mx + b
y − 5 = 1(x −1)
y − 5 = x −1
Example 1
The line a contains the points T(0, 0) and U(−5, 5).
Find the distance between line a and the point V(1, 5).
1. Find the equation of the original line
m =
0 − 5
0 + 5
=
−5
5
= −1 T(0, 0)
y = −x
2. Find the equation of the perpendicular line through the
other point
m = 1
V(1, 5)
y − y1
= m(x − x1
)
y = mx + b
y − 5 = 1(x −1)
y − 5 = x −1
y = x + 4
Example 1
3. Solve the system of these two equations.
Example 1
3. Solve the system of these two equations.
y = −x
y = x + 4
⎧
⎨
⎩
Example 1
3. Solve the system of these two equations.
y = −x
y = x + 4
⎧
⎨
⎩
−x = x + 4
Example 1
3. Solve the system of these two equations.
y = −x
y = x + 4
⎧
⎨
⎩
−x = x + 4
−2x = 4
Example 1
3. Solve the system of these two equations.
y = −x
y = x + 4
⎧
⎨
⎩
−x = x + 4
−2x = 4
x = −2
Example 1
3. Solve the system of these two equations.
y = −x
y = x + 4
⎧
⎨
⎩
−x = x + 4
−2x = 4
x = −2
y = −(−2)
Example 1
3. Solve the system of these two equations.
y = −x
y = x + 4
⎧
⎨
⎩
−x = x + 4
−2x = 4
x = −2
y = −(−2) = 2
Example 1
3. Solve the system of these two equations.
y = −x
y = x + 4
⎧
⎨
⎩
−x = x + 4
−2x = 4
x = −2
y = −(−2) = 2
2 = −2+ 4
Example 1
3. Solve the system of these two equations.
y = −x
y = x + 4
⎧
⎨
⎩
−x = x + 4
−2x = 4
x = −2
y = −(−2) = 2
2 = −2+ 4
(−2,2)
Example 1
4. Use the distance formula utilizing this point on the line
and the point not on the line.
Example 1
4. Use the distance formula utilizing this point on the line
and the point not on the line.
(1, 5), (−2, 2)
Example 1
4. Use the distance formula utilizing this point on the line
and the point not on the line.
(1, 5), (−2, 2)
d = (x2
− x1
)2
+(y2
− y1
)2
Example 1
4. Use the distance formula utilizing this point on the line
and the point not on the line.
(1, 5), (−2, 2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (−2−1)2
+(2− 5)2
Example 1
4. Use the distance formula utilizing this point on the line
and the point not on the line.
(1, 5), (−2, 2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (−2−1)2
+(2− 5)2
= (−3)2
+(−3)2
Example 1
4. Use the distance formula utilizing this point on the line
and the point not on the line.
(1, 5), (−2, 2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (−2−1)2
+(2− 5)2
= (−3)2
+(−3)2
= 9+ 9
Example 1
4. Use the distance formula utilizing this point on the line
and the point not on the line.
(1, 5), (−2, 2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (−2−1)2
+(2− 5)2
= (−3)2
+(−3)2
= 9+ 9 = 18
Example 1
4. Use the distance formula utilizing this point on the line
and the point not on the line.
(1, 5), (−2, 2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (−2−1)2
+(2− 5)2
= (−3)2
+(−3)2
= 9+ 9 = 18 ≈ 4.24
Example 1
4. Use the distance formula utilizing this point on the line
and the point not on the line.
(1, 5), (−2, 2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (−2−1)2
+(2− 5)2
= (−3)2
+(−3)2
= 9+ 9 = 18 ≈ 4.24 units
Example 2
Find the distance between the parallel lines m and n with
the following equations.
y = 2x + 3 y = 2x −1
Example 2
Find the distance between the parallel lines m and n with
the following equations.
y = 2x + 3 y = 2x −1
1. Find the equation of the perpendicular line.
Example 2
Find the distance between the parallel lines m and n with
the following equations.
y = 2x + 3 y = 2x −1
1. Find the equation of the perpendicular line.
y = mx + b
Example 2
Find the distance between the parallel lines m and n with
the following equations.
y = 2x + 3 y = 2x −1
1. Find the equation of the perpendicular line.
y = mx + b
m = −
1
2
,(0,3)
Example 2
Find the distance between the parallel lines m and n with
the following equations.
y = 2x + 3 y = 2x −1
1. Find the equation of the perpendicular line.
y = mx + b
y = −
1
2
x + 3
m = −
1
2
,(0,3)
Example 2
2. Find the intersection of the perpendicular line and the
other parallel line using a system.
Example 2
2. Find the intersection of the perpendicular line and the
other parallel line using a system.
y = 2x −1
y = −
1
2
x + 3
⎧
⎨
⎪
⎩
⎪
Example 2
2. Find the intersection of the perpendicular line and the
other parallel line using a system.
y = 2x −1
y = −
1
2
x + 3
⎧
⎨
⎪
⎩
⎪
2x −1= −
1
2
x + 3
Example 2
2. Find the intersection of the perpendicular line and the
other parallel line using a system.
y = 2x −1
y = −
1
2
x + 3
⎧
⎨
⎪
⎩
⎪
2x −1= −
1
2
x + 3
5
2
x = 4
Example 2
2. Find the intersection of the perpendicular line and the
other parallel line using a system.
y = 2x −1
y = −
1
2
x + 3
⎧
⎨
⎪
⎩
⎪
2x −1= −
1
2
x + 3
5
2
x = 4 x = 1.6
Example 2
2. Find the intersection of the perpendicular line and the
other parallel line using a system.
y = 2x −1
y = −
1
2
x + 3
⎧
⎨
⎪
⎩
⎪
2x −1= −
1
2
x + 3
5
2
x = 4 x = 1.6
y = 2(1.6)−1
Example 2
2. Find the intersection of the perpendicular line and the
other parallel line using a system.
y = 2x −1
y = −
1
2
x + 3
⎧
⎨
⎪
⎩
⎪
2x −1= −
1
2
x + 3
5
2
x = 4 x = 1.6
y = 2(1.6)−1
y = 2.2
Example 2
2. Find the intersection of the perpendicular line and the
other parallel line using a system.
y = 2x −1
y = −
1
2
x + 3
⎧
⎨
⎪
⎩
⎪
2x −1= −
1
2
x + 3
5
2
x = 4 x = 1.6
y = 2(1.6)−1
y = 2.2
y = −
1
2
(1.6)−1
Example 2
2. Find the intersection of the perpendicular line and the
other parallel line using a system.
y = 2x −1
y = −
1
2
x + 3
⎧
⎨
⎪
⎩
⎪
2x −1= −
1
2
x + 3
5
2
x = 4 x = 1.6
y = 2(1.6)−1
y = 2.2
y = −
1
2
(1.6)−1
y = 2.2
Example 2
2. Find the intersection of the perpendicular line and the
other parallel line using a system.
y = 2x −1
y = −
1
2
x + 3
⎧
⎨
⎪
⎩
⎪
2x −1= −
1
2
x + 3
5
2
x = 4 x = 1.6
y = 2(1.6)−1
y = 2.2
y = −
1
2
(1.6)−1
y = 2.2
(1.6, 2.2)
Example 2
3. Use the new point and original y-intercept you chose in
step 2 in the distance formula.
Example 2
3. Use the new point and original y-intercept you chose in
step 2 in the distance formula.
(0, 3), (1.6, 2.2)
Example 2
3. Use the new point and original y-intercept you chose in
step 2 in the distance formula.
(0, 3), (1.6, 2.2)
d = (x2
− x1
)2
+(y2
− y1
)2
Example 2
3. Use the new point and original y-intercept you chose in
step 2 in the distance formula.
(0, 3), (1.6, 2.2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (1.6 − 0)2
+(2.2− 3)2
Example 2
3. Use the new point and original y-intercept you chose in
step 2 in the distance formula.
(0, 3), (1.6, 2.2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (1.6 − 0)2
+(2.2− 3)2
= (1.6)2
+(−0.8)2
Example 2
3. Use the new point and original y-intercept you chose in
step 2 in the distance formula.
(0, 3), (1.6, 2.2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (1.6 − 0)2
+(2.2− 3)2
= (1.6)2
+(−0.8)2
= 2.56 +.64
Example 2
3. Use the new point and original y-intercept you chose in
step 2 in the distance formula.
(0, 3), (1.6, 2.2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (1.6 − 0)2
+(2.2− 3)2
= (1.6)2
+(−0.8)2
= 2.56 +.64
= 3.2
Example 2
3. Use the new point and original y-intercept you chose in
step 2 in the distance formula.
(0, 3), (1.6, 2.2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (1.6 − 0)2
+(2.2− 3)2
= (1.6)2
+(−0.8)2
= 2.56 +.64
= 3.2 ≈1.79
Example 2
3. Use the new point and original y-intercept you chose in
step 2 in the distance formula.
(0, 3), (1.6, 2.2)
d = (x2
− x1
)2
+(y2
− y1
)2
= (1.6 − 0)2
+(2.2− 3)2
= (1.6)2
+(−0.8)2
= 2.56 +.64
= 3.2 ≈1.79 units
Example 3
Line h contains the points E(2, 4) and F(5, 1). Find
the distance between line h and the point G(1, 1).
Example 3
Line h contains the points E(2, 4) and F(5, 1). Find
the distance between line h and the point G(1, 1).
Solution:
Example 3
Line h contains the points E(2, 4) and F(5, 1). Find
the distance between line h and the point G(1, 1).
Solution:
d = 8
Example 3
Line h contains the points E(2, 4) and F(5, 1). Find
the distance between line h and the point G(1, 1).
Solution:
d = 8 ≈ 2.83
Example 3
Line h contains the points E(2, 4) and F(5, 1). Find
the distance between line h and the point G(1, 1).
Solution:
d = 8 ≈ 2.83 units

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Geometry Section 2-10

  • 2. Essential Questions n How do you find the distance between a point and a line? n How do you find the distance between parallel lines?
  • 3. Vocabulary 1. Equidistant: 2. Distance Between a Point and a Line: 3. Distance Between Parallel Lines:
  • 4. Vocabulary 1. Equidistant: The distance between any two lines as measured along a perpendicular is the same; this occurs with parallel lines 2. Distance Between a Point and a Line: 3. Distance Between Parallel Lines:
  • 5. Vocabulary 1. Equidistant: The distance between any two lines as measured along a perpendicular is the same; this occurs with parallel lines 2. Distance Between a Point and a Line: The length of the segment perpendicular to the line, with the endpoints being that point and a point on the line 3. Distance Between Parallel Lines:
  • 6. Vocabulary 1. Equidistant: The distance between any two lines as measured along a perpendicular is the same; this occurs with parallel lines 2. Distance Between a Point and a Line: The length of the segment perpendicular to the line, with the endpoints being that point and a point on the line 3. Distance Between Parallel Lines: The length of the segment perpendicular to the two parallel lines with the endpoints on either of the parallel lines
  • 7. Postulates & Theorems 1. Perpendicular Postulate: 2. Two Lines Equidistant from a Third:
  • 8. Postulates & Theorems 1. Perpendicular Postulate: If given a line and a point not on the line, then there exists exactly one line through the point that is perpendicular to the given line 2. Two Lines Equidistant from a Third:
  • 9. Postulates & Theorems 1. Perpendicular Postulate: If given a line and a point not on the line, then there exists exactly one line through the point that is perpendicular to the given line 2. Two Lines Equidistant from a Third: In a plane, if two lines are each equidistant from a third line, then the two lines are parallel to each other
  • 10. Steps to find the Distance from a Point to a Line
  • 11. Steps to find the Distance from a Point to a Line 1. Find the equation of the original line
  • 12. Steps to find the Distance from a Point to a Line 1. Find the equation of the original line 2. Find the equation of the perpendicular line through the other point
  • 13. Steps to find the Distance from a Point to a Line 1. Find the equation of the original line 2. Find the equation of the perpendicular line through the other point 3. Solve the system of these two equations.
  • 14. Steps to find the Distance from a Point to a Line 1. Find the equation of the original line 2. Find the equation of the perpendicular line through the other point 3. Solve the system of these two equations. 4. Use the distance formula utilizing this point on the line and the point not on the line.
  • 15. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5).
  • 16. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line
  • 17. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5
  • 18. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5
  • 19. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5 = −1
  • 20. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5 = −1 T(0, 0)
  • 21. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5 = −1 T(0, 0) y = mx + b
  • 22. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5 = −1 T(0, 0) y = −x y = mx + b
  • 23. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5 = −1 T(0, 0) y = −x 2. Find the equation of the perpendicular line through the other point y = mx + b
  • 24. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5 = −1 T(0, 0) y = −x 2. Find the equation of the perpendicular line through the other point m = 1 y = mx + b
  • 25. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5 = −1 T(0, 0) y = −x 2. Find the equation of the perpendicular line through the other point m = 1 V(1, 5) y = mx + b
  • 26. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5 = −1 T(0, 0) y = −x 2. Find the equation of the perpendicular line through the other point m = 1 V(1, 5) y − y1 = m(x − x1 ) y = mx + b
  • 27. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5 = −1 T(0, 0) y = −x 2. Find the equation of the perpendicular line through the other point m = 1 V(1, 5) y − y1 = m(x − x1 ) y = mx + b y − 5 = 1(x −1)
  • 28. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5 = −1 T(0, 0) y = −x 2. Find the equation of the perpendicular line through the other point m = 1 V(1, 5) y − y1 = m(x − x1 ) y = mx + b y − 5 = 1(x −1) y − 5 = x −1
  • 29. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line m = 0 − 5 0 + 5 = −5 5 = −1 T(0, 0) y = −x 2. Find the equation of the perpendicular line through the other point m = 1 V(1, 5) y − y1 = m(x − x1 ) y = mx + b y − 5 = 1(x −1) y − 5 = x −1 y = x + 4
  • 30. Example 1 3. Solve the system of these two equations.
  • 31. Example 1 3. Solve the system of these two equations. y = −x y = x + 4 ⎧ ⎨ ⎩
  • 32. Example 1 3. Solve the system of these two equations. y = −x y = x + 4 ⎧ ⎨ ⎩ −x = x + 4
  • 33. Example 1 3. Solve the system of these two equations. y = −x y = x + 4 ⎧ ⎨ ⎩ −x = x + 4 −2x = 4
  • 34. Example 1 3. Solve the system of these two equations. y = −x y = x + 4 ⎧ ⎨ ⎩ −x = x + 4 −2x = 4 x = −2
  • 35. Example 1 3. Solve the system of these two equations. y = −x y = x + 4 ⎧ ⎨ ⎩ −x = x + 4 −2x = 4 x = −2 y = −(−2)
  • 36. Example 1 3. Solve the system of these two equations. y = −x y = x + 4 ⎧ ⎨ ⎩ −x = x + 4 −2x = 4 x = −2 y = −(−2) = 2
  • 37. Example 1 3. Solve the system of these two equations. y = −x y = x + 4 ⎧ ⎨ ⎩ −x = x + 4 −2x = 4 x = −2 y = −(−2) = 2 2 = −2+ 4
  • 38. Example 1 3. Solve the system of these two equations. y = −x y = x + 4 ⎧ ⎨ ⎩ −x = x + 4 −2x = 4 x = −2 y = −(−2) = 2 2 = −2+ 4 (−2,2)
  • 39. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line.
  • 40. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2)
  • 41. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 )2 +(y2 − y1 )2
  • 42. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 )2 +(y2 − y1 )2 = (−2−1)2 +(2− 5)2
  • 43. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 )2 +(y2 − y1 )2 = (−2−1)2 +(2− 5)2 = (−3)2 +(−3)2
  • 44. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 )2 +(y2 − y1 )2 = (−2−1)2 +(2− 5)2 = (−3)2 +(−3)2 = 9+ 9
  • 45. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 )2 +(y2 − y1 )2 = (−2−1)2 +(2− 5)2 = (−3)2 +(−3)2 = 9+ 9 = 18
  • 46. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 )2 +(y2 − y1 )2 = (−2−1)2 +(2− 5)2 = (−3)2 +(−3)2 = 9+ 9 = 18 ≈ 4.24
  • 47. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 )2 +(y2 − y1 )2 = (−2−1)2 +(2− 5)2 = (−3)2 +(−3)2 = 9+ 9 = 18 ≈ 4.24 units
  • 48. Example 2 Find the distance between the parallel lines m and n with the following equations. y = 2x + 3 y = 2x −1
  • 49. Example 2 Find the distance between the parallel lines m and n with the following equations. y = 2x + 3 y = 2x −1 1. Find the equation of the perpendicular line.
  • 50. Example 2 Find the distance between the parallel lines m and n with the following equations. y = 2x + 3 y = 2x −1 1. Find the equation of the perpendicular line. y = mx + b
  • 51. Example 2 Find the distance between the parallel lines m and n with the following equations. y = 2x + 3 y = 2x −1 1. Find the equation of the perpendicular line. y = mx + b m = − 1 2 ,(0,3)
  • 52. Example 2 Find the distance between the parallel lines m and n with the following equations. y = 2x + 3 y = 2x −1 1. Find the equation of the perpendicular line. y = mx + b y = − 1 2 x + 3 m = − 1 2 ,(0,3)
  • 53. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system.
  • 54. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. y = 2x −1 y = − 1 2 x + 3 ⎧ ⎨ ⎪ ⎩ ⎪
  • 55. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. y = 2x −1 y = − 1 2 x + 3 ⎧ ⎨ ⎪ ⎩ ⎪ 2x −1= − 1 2 x + 3
  • 56. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. y = 2x −1 y = − 1 2 x + 3 ⎧ ⎨ ⎪ ⎩ ⎪ 2x −1= − 1 2 x + 3 5 2 x = 4
  • 57. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. y = 2x −1 y = − 1 2 x + 3 ⎧ ⎨ ⎪ ⎩ ⎪ 2x −1= − 1 2 x + 3 5 2 x = 4 x = 1.6
  • 58. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. y = 2x −1 y = − 1 2 x + 3 ⎧ ⎨ ⎪ ⎩ ⎪ 2x −1= − 1 2 x + 3 5 2 x = 4 x = 1.6 y = 2(1.6)−1
  • 59. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. y = 2x −1 y = − 1 2 x + 3 ⎧ ⎨ ⎪ ⎩ ⎪ 2x −1= − 1 2 x + 3 5 2 x = 4 x = 1.6 y = 2(1.6)−1 y = 2.2
  • 60. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. y = 2x −1 y = − 1 2 x + 3 ⎧ ⎨ ⎪ ⎩ ⎪ 2x −1= − 1 2 x + 3 5 2 x = 4 x = 1.6 y = 2(1.6)−1 y = 2.2 y = − 1 2 (1.6)−1
  • 61. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. y = 2x −1 y = − 1 2 x + 3 ⎧ ⎨ ⎪ ⎩ ⎪ 2x −1= − 1 2 x + 3 5 2 x = 4 x = 1.6 y = 2(1.6)−1 y = 2.2 y = − 1 2 (1.6)−1 y = 2.2
  • 62. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. y = 2x −1 y = − 1 2 x + 3 ⎧ ⎨ ⎪ ⎩ ⎪ 2x −1= − 1 2 x + 3 5 2 x = 4 x = 1.6 y = 2(1.6)−1 y = 2.2 y = − 1 2 (1.6)−1 y = 2.2 (1.6, 2.2)
  • 63. Example 2 3. Use the new point and original y-intercept you chose in step 2 in the distance formula.
  • 64. Example 2 3. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2)
  • 65. Example 2 3. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 )2 +(y2 − y1 )2
  • 66. Example 2 3. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 )2 +(y2 − y1 )2 = (1.6 − 0)2 +(2.2− 3)2
  • 67. Example 2 3. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 )2 +(y2 − y1 )2 = (1.6 − 0)2 +(2.2− 3)2 = (1.6)2 +(−0.8)2
  • 68. Example 2 3. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 )2 +(y2 − y1 )2 = (1.6 − 0)2 +(2.2− 3)2 = (1.6)2 +(−0.8)2 = 2.56 +.64
  • 69. Example 2 3. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 )2 +(y2 − y1 )2 = (1.6 − 0)2 +(2.2− 3)2 = (1.6)2 +(−0.8)2 = 2.56 +.64 = 3.2
  • 70. Example 2 3. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 )2 +(y2 − y1 )2 = (1.6 − 0)2 +(2.2− 3)2 = (1.6)2 +(−0.8)2 = 2.56 +.64 = 3.2 ≈1.79
  • 71. Example 2 3. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 )2 +(y2 − y1 )2 = (1.6 − 0)2 +(2.2− 3)2 = (1.6)2 +(−0.8)2 = 2.56 +.64 = 3.2 ≈1.79 units
  • 72. Example 3 Line h contains the points E(2, 4) and F(5, 1). Find the distance between line h and the point G(1, 1).
  • 73. Example 3 Line h contains the points E(2, 4) and F(5, 1). Find the distance between line h and the point G(1, 1). Solution:
  • 74. Example 3 Line h contains the points E(2, 4) and F(5, 1). Find the distance between line h and the point G(1, 1). Solution: d = 8
  • 75. Example 3 Line h contains the points E(2, 4) and F(5, 1). Find the distance between line h and the point G(1, 1). Solution: d = 8 ≈ 2.83
  • 76. Example 3 Line h contains the points E(2, 4) and F(5, 1). Find the distance between line h and the point G(1, 1). Solution: d = 8 ≈ 2.83 units