4.7 Triangles and Coordinate Proof

Geometry

Mrs. Spitz (this slide is not ours… The citation is the end slide)

Fall 2004

Objectives:

1. Place geometric figures in a coordinate plane.

2. Write a coordinate proof.

Assignment:

pp. 246-249 #1-28 all Quiz after this section – similar quiz on pg.

250. Chapter 4 Review – pgs. 252-254 #1-18 Extra Credit –Algebra Review –pgs. 258-259

#1-66 all – SHOW ALL WORK. Test next week before break. If you intend to

leave early . . . Take the test before you leave. Binders are due that day.

Placing Figures in a Coordinate Plane

So far, you have studied two-column proofs, paragraph proofs, and flow proofs. A COORDINATE PROOF involves placing geometric figures in a coordinate plane. Then you can use the Distance Formula (no, you never get away from using this) and the Midpoint Formula, as well as postulate and theorems to prove statements about figures.

Ex. 1: Placing a Rectangle in a Coordinate Plane

Place a 2-unit by 6-unit rectangle in a coordinate plane.

SOLUTION: Choose a placement that makes finding distance easy (along the origin) as seen to the right.

Ex. 1: Placing a Rectangle in a Coordinate Plane

One vertex is at the origin, and three of the vertices have at least one coordinate that is 0.

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Ex. 1: Placing a Rectangle in a Coordinate Plane

One side is centered at the origin, and the x-coordinates are opposites.

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Note:

Once a figure has been placed in a coordinate plane, you can use the Distance Formula or the Midpoint Formula to measure distances or locate points

Ex. 2: Using the Distance Formula

A right triangle has legs of 5 units and 12 units. Place the triangle in a coordinate plane. Label the coordinates of the vertices and find the length of the hypotenuse.

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Ex. 2: Using the Distance Formula

One possible placement is shown. Notice that one leg is vertical and the other leg is horizontal, which assures that the legs meet as right angles. Points on the same vertical segment have the same x-coordinate, and points on the same horizontal segment have the same y-coordinate.

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Ex. 2: Using the Distance Formula

You can use the Distance Formula to find the length of the hypotenuse.

d = √(x2 – x1)2 + (y2 – y1)2

= √(12-0)2 + (5-0)2

= √169

= 13

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Ex. 3 Using the Midpoint Formula

In the diagram, ∆MLN ≅ ∆KLN). Find the coordinates of point L.

Solution: Because the triangles are congruent, it follows that ML ≅ KL. So, point L must be the midpoint of MK. This means you can use the Midpoint Formula to find the coordinates of point L.

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Ex. 3 Using the Midpoint Formula

L (x, y) = x1 + x2, y1 +y2

2 2 Midpoint Formula

=160+0 , 0+160

2 2 Substitute values

= (80, 80) Simplify.

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Writing Coordinate Proofs

Once a figure is placed in a coordinate plane, you may be able to prove statements about the figure.

Ex. 4: Writing a Plan for a Coordinate Proof

Write a plan to prove that SQ bisects PSR. Given: Coordinates of vertices of ∆PQS and ∆RQS. Prove SQ bisects PSR. Plan for proof: Use the Distance Formula to find the

side lengths of ∆PQS and ∆RQS. Then use the SSS Congruence Postulate to show that ∆PQS ≅ ∆RQS. Finally, use the fact that corresponding parts of congruent triangles are congruent (CPCTC) to conclude that PSQ ≅RSQ, which implies that SQ bisects PSR.

Ex. 4: Writing a Plan for a Coordinate Proof

Given: Coordinates of vertices of ∆PQS and ∆RQS.

Prove SQ bisects PSR.

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NOTE:

The coordinate proof in Example 4 applies to a specific triangle. When you want to prove a statement about a more general set of figures, it is helpful to use variables as coordinates.

For instance, you can use variable coordinates to duplicate the proof in Example 4. Once this is done, you can conclude that SQ bisects PSR for any triangle whose coordinates fit the given pattern.

No coordinates – just variables

x

(-h, 0) (0, 0) (h, 0)

(0, k)

y

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P R

Ex. 5: Using Variables as Coordinates Right ∆QBC has leg lengths

of h units and k units. You can find the coordinates of points B and C by considering how the triangle is placed in a coordinate plane.

Point B is h units horizontally from the origin (0, 0), so its coordinates are (h, 0). Point C is h units horizontally from the origin and k units vertically from the origin, so its coordinates are (h, k). You can use the Distance Formula to find the length of the hypotenuse QC.

k units

h units

C (h, k)

B (h, 0)Q (0, 0)

hypotenuse

Ex. 5: Using Variables as Coordinates

OC = √(x2 – x1)2 + (y2 – y1)2

= √(h-0)2 + (k - 0)2

= √h2 + k2

k units

h units

C (h, k)

B (h, 0)Q (0, 0)

hypotenuse

Ex. 5 Writing a Coordinate Proof

Given: Coordinates of figure OTUV

Prove ∆OUT ∆UVO Coordinate proof:

Segments OV and UT have the same length.

OV = √(h-0)2 + (0 - 0)2=h UT = √(m+h-m)2 + (k - k)2=h

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T (m, k) U (m+h, k)

V (h, 0)

Ex. 5 Writing a Coordinate Proof

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T (m, k) U (m+h, k)

V (h, 0)

Horizontal segments UT and OV each have a slope of 0, which implies they are parallel. Segment OU intersects UT and OV to form congruent alternate interior angles TUO and VOU. Because OU OU, you can apply the SAS Congruence Postulate to conclude that ∆OUT ∆UVO.

Resources

This is not my Powerpoint. I did not make it nor did I have any help making it. I obtained it from:

http://rsms.rcsnc.org/UserFiles/Servers/Server_4793539/File/Geometry%204.7%20Triangles%20and%20Coordinate%20Proof.pdf