Intro

Up until René Descartes conceived of the ordered plane, we proved geometry theorems using logic, axioms, propositions and deductions. We wrote formal proofs with statements and justifications for those statements -- and the whole thing was quite impossible for many to master -- since it demanded pure logic and precise reference to create an invincible formal proof.

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Now, however, thanks to Descartes, we can assign variables to represent points, equations to represent lines and curves in two dimensions and three dimensions -- so now we can prove geometry theorems with algebra. Such proofs are called synthetic proofs.

| intro | synthetic proofs | word problems | practice | solutions |

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Synthetic Proofs

The main difference between a formal proof and a synthetic proof for a geometry deduction or proposition is there is no need to justify the algebraic statements of a synthetic proof whereas every statement in a formal proof must be justified by an axiom or proposition. Since the coordinates of the lines and points determine the variables in the algebra, one need only look at the graph or diagram in question to justify the statements.

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Therefore, it is essential to make a diagram or graph when we want to prove things synthetically.

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In lesson 3.4 of this MathRoom -- example 4 shows how to use numerical coordinates to prove the theorem that the line joining the midpoints of 2 sides of a triangle is parallel to the third side and equal to one half its length. Now, for an example of synthetic proof, we'll prove this is true for any triangle -- not the specific one in the example. I'll leave the second part of the proof -- to show the line is half the length of the base -- as an exercise.

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As mentioned previously, a good approach is to use the axes and the origin as points in the figure whenever possible. It helps to keep the number of variables to a minimum since some of the coordinates will be 0.

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The slope of MN = , The slope of OA = 0

So the lines are parallel.

Now use this diagram to prove that MN = ½ OA

by finding the length of each in terms of a and b.

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| intro | synthetic proofs | word problems | practice | solutions |

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Word Problems with Points, Distances and Linear Equations

In lesson 2 - point of division -- practice question 3, we did a word problem where we had to find the distance between the disabled cargo boat and the repair boat. If you need a reminder, check out the solution to this question.

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This kind of question comes up time and again in city planning -- determining the path for sewers, highways, and utility lines.

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The best, and in my opinion, the only way to deal with these questions is with a visual image -- a diagram. A well-labeled diagram can make things clear and precise allowing us to solve the problem easily. Without the picture, our brains can't grasp the concepts nearly as easily as with the picture -- so make a picture if you can -- especially when the question involves rectilinear figures or street maps. If our variables are defined in the diagram, we often omit the "let" statement -- for its obvious what the variables represent. However, it's best to always define our variables with a "let" statement.

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| intro | synthetic proofs | word problems | practice | solutions |

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Practice

1) In a Cartesian coordinate graph of a provincial park, a hiking trail is represented by the linear equation . The ranger's lookout tower is situated at the point P(26, 36). The scale on the graph is in meters.

How far from the trail is the lookout tower? (Round your answer to 1 decimal place.)

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2) Mario is moving from Québec City to Rimouski. The moving truck with all his possessions travels at an average speed of 75 km/hr. Mario leaves Québec City in his car, 15 minutes after the departure of the moving truck. He travels at an average speed of 90 km/hr. Both Mario and the truck do not stop on the road to Rimouski.

How far from Quebec City will Mario overtake the moving truck?

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3) The graph represents a drawing of a public park in the village of Sans Cerveau.

How long is the hedge AP?

diagram

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4) The city of Lottalumps is planning a new road called Lumpy Lane. On the Cartesian graph of the setup, (scaled in kilometers), the beginning of Lumpy Lane is situated at point P(6, 5).

The city planners want to join the new road at a right angle with Chunky St., which runs
from A(3, – 4) to B( – 1, 8).

a) What is the equation for Lumpy Lane?

b) If it ends at the intersection with Chunky St., how long is Lumpy Lane?

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| intro | synthetic proofs | word problems | practice | solutions |

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Solutions

1) The linear equation in General Form is x + 2y - 68 = 0

Now we use the formula for distance from a point to a line.

The distance from this line to the lookout tower at P(26, 36) is:

The lookout tower is 13.4 meters from the trail.

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2) The moving truck travels at an average speed of 75 km/hr.

Let t = the time in hours that the truck's been on the road

Let d = the distance the truck is from Québec City when Mario overtakes it.

then d = 75t

Mario leaves 15 minutes later at 90 km/hr -- 15 minutes = 0. 25 hr.

so (t - 0.25) = Mario's time on the road in hours

and d = 90(t - 0.25) is Mario's distance from Québec when he meets the truck.

Since they must be the same distance for overtaking to occur

75t = 90(t - 0.25)

15t = 22.5 so t = 1.5 hours.

after an hour and a half, the truck will have covered 75(1.5) = 112.5 km.

Mario will overtake the truck in 1.5 hours, 112.5 km. from Québec City.

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3) Looking at the diagram,

The key here is to find the y-coordinate of P. That will give us the length of the hedge.

S is the y-intercept of y = , so S is at (0, 27)

RQ = the y-value at Q, so RQ = = 51

So, Q is at (60, 51)

Since SP is 1/3 of PQ, and the trapezoids OSPA & OSQR are similar,

SP is ¼ SQ

so, we find the coordinates of P by moving S, ¼ the distance between SQ

This puts P at (15, 33)

The length of the hedge is 33 meters.

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4) Make your own diagram here -- I'm tired -- and this one's so easy. Use graph paper.

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a) If we find the slope of Chunky St. -- the line through A(3, - 4) and B(-1, 8)

We'll know the slope of Lumpy Lane -- since it is perpendicular to AB.

We have a point on the line -- and that's all we need for the equation of the line.

slope AB = -3

so slope of Chunky St. =

P(6, 5) is on the line, so its equation is:

(y - 5) = (x - 6) or x + 3y - 21 = 0

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b) Now we need the perpendicular distance between P(6, 5) and the line AB.

So we need the equation of AB.

y + 4 = -3(x - 3) which is 3x + y - 5 = 0 in General Form.

Lumpy Lane will be 5.7 km. long.

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| intro | synthetic proofs | word problems | practice | solutions |

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