1. Grade 10 Academic (MPM2D) Unit 2: Analytic Geometry Analytic Geometry – Word Problems 2
Mr. Choi
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2. Properties of Triangles
Medians:
line from vertex to the midpoint of the opposite side
Midpoint:
point on a line that divides it into 2 equal parts
to draw a median  find mid point of opposite side and connect to vertex
Altitude:
shows the height of a polygon
to draw an altitude  line from vertex to opposite side so that it meets at 90°
Perpendicular bisectors:
perpendicular to a line segment and meets at its midpoint.
to draw a perpendicular bisector  find midpoint, then measure a right angle, draw line.
Angle bisectors:
The (interior) bisector of an angle, also called the internal angle bisector is the line or line segment that divides the angle into two equal parts 
Centroid:
where all 3 medians meet
divides each median in the ratio 1: 2
it is the centre of mass
Orthocentre:
where all 3 altitudes meet
Circumcentre:
where all 3 perpendicular bisectors meet centre of the circle that passes through the vertices of the triangle
the circle is called circumcircle or circumscribed circle
Incentre:
where all 3 angle bisectors meet centre of the circle that meets each side once
the circle is called incircle or inscribed circle
Analytic Geometry – Problem Solving 2
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3. Quadrilaterals If you know the coordinates of the vertices of a quadrilateral, You can determine the length and slopes of the sides and use the information to classify the quadrilateral as - Parallelogram (both pairs of opposite sides are parallel), - Rectangle (parallelogram and adjacent sides are perpendicular), - Square (rectangle and all sides are same length), - Rhombus (both pairs of opposite sides are parallel and all sides are the same length).
Analytic Geometry – Problem Solving 2
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4. When triangles and quadrilaterals are drawn on a coordinate plane, they are constructed from line segments. The lengths and slopes of the line segments can be used to classify or define specific types of triangles and quadrilaterals.  
Analytic Geometry – Problem Solving 2
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5. Example 1 .
The plan shows the positions of three entrances to a parking lot. A tall mast light is to be erected to illuminate the three entrances, A (-8, 14),
B (-4, 8), and C (18, 10).
a) Where should the light be located to illuminate each entrance equally?
b) If the closest power line to the parking lot runs along a straight line containing points (0, 4) and (12, 10).
At what point should the lighting contractors connect to the power line, and how much cable will they need to reach the lamp? Each unit on the plan equals to 1 m.
Analytic Geometry – Problem Solving 2
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6. Example 1 - Perpendicular Bisector of AB.
a) Where should the light be located to illuminate each entrance equally?
For AB: A (-8, 14) B (-4, 8)
Midpoint:
Substitute midpoint (-6,11)
The light should be located the same distance from each entrance. This point is the centre of the circle that contains the three entrance points 
circumcentre
Analytic Geometry – Problem Solving 2
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7. Example 1 – Perpendicular Bisector of BC.
a) Where should the light be located to illuminate each entrance equally?
For BC: B (-4, 8) C (18, 10)
Midpoint:
Substitute midpoint (7, 9)
The light should be located the same distance from each entrance. This point is the centre of the circle that contains the three entrance points 
circumcentre
Analytic Geometry – Problem Solving 2
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8. Example 1 – Point of Intersection.
a) Where should the light be located to illuminate each entrance equally?
Point of intersection
(6, 19)
Multiply both sides by 3
circumcentre
Analytic Geometry – Problem Solving 2
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9. Example 1 – Conclusion circumcentre.
a) Where should the light be located to illuminate each entrance equally?
Therefore if the light is placed at (6, 19), it will be about the same distance from each entrance and illuminate each equally.
(6, 19)
circumcentre
Analytic Geometry – Problem Solving 2
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10. Example 1b .
b) If the closest power line to the parking lot runs along a straight line containing points (0, 4) and (12, 10).
At what point should the lighting contractors connect to the power line, and how much cable will they need to reach the lamp? Each unit on the plan equals to 1 m.
Analytic Geometry – Problem Solving 2
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11. Example 1b: Equation of the Power line.
Recall: The light is at (6, 19). The shortest distance from the light to the power line is the perpendicular distance shown in the sketch.
You want to find the point of intersection of the power line and the perpendicular line
from (6, 19).
(6, 19)
(0, 4)
(12, 10)
Find the equations of the lines that contain these two line segments.
The slope of the power line is:
The y – intercept is at (0, 4).
The equation of the power line is
Analytic Geometry – Problem Solving 2
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12. Example 1b: Shortest Distance Equation.
Recall: The light is at (6, 19). The shortest distance from the light to the power line is the perpendicular distance shown in the sketch.
Recall: Power line is:
(6, 19)
(0, 4)
(12, 10)
Use the point (6, 19)
(10.8, 9.4)
At the point where the perpendicular
from (6, 19) meets the power line,
Multiply both sides by 2
Therefore the contractor should connect into the power line at point (10.8, 9.4).
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Analytic Geometry – Problem Solving 2

13. Example 1b: Distance .
Recall: The light is at (6, 19). The shortest distance from the light to the power line is the perpendicular distance shown in the sketch.
Recall:
M (6, 19)
(0, 4)
(12, 10)
The length of the cable required:
M(6, 19) D (10.8, 9.4)
D (10.8, 9.4)
Therefore the contractor will need about 10.7 m of cable to connect the light to the power line.
Recall: 1 unit = 1m
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Analytic Geometry – Problem Solving 2

14. Homework Work sheet: Day 1: #1 – 13 (odd) Day 2: Even Text: Day 1
P182 #2, 3, 5 —10
P183 #11odd, 13, 17a, 18, 20
Day 2
Read P. 200 Examples 1 & 2
P196 #22
P203 #3—7, 10, 11
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Analytic Geometry – Problem Solving 2
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15. End of lesson Analytic Geometry – Problem Solving 2
© 2017 E. Choi – MPM2D - All Rights Reserved