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Placing Figures in the Coordinate PlanePowerPoint Presentation

Placing Figures in the Coordinate Plane

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Placing Figures in the Coordinate Plane

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Lesson 6-6

Check Skills You’ll Need

(For help, go to Lesson 6-1.)

Draw a quadrilateral with the given vertices. Then determine the

most precise name for each quadrilateral.

1.H(–5, 0), E(–3, 2), A(3, 2), T(5, 0)

2.S(0, 0), A(4, 0), N(3, 2), D(–1, 2)

3.R(0, 0), A(5, 5), I(8, 4), N(7, 1)

4.W(–3, 0), I(0, 3), N(3, 0), D(0, –3)

Check Skills You’ll Need

6-6

Lesson 6-6

Check Skills You’ll Need

1.The slope of EA and the slope of HT are both 0,

so EA || HT. Since EH and AT are not parallel, HEAT is a trapezoid.

HE = 8 and TA = 8.

The legs of the trapezoid are congruent, so HEAT is an isosceles trapezoid.

2.DN and SA are both horizontal, so their slopes

are both 0. Since their slopes are the same,

DN || SA. The slope of DS and the slope of NA

are both – = –2, so DS || NA.

Since both pairs of opposite sides are parallel, the quadrilateral is

a parallelogram.

Solutions

2

1

6-6

Lesson 6-6

Check Skills You’ll Need

Solutions (continued)

3.RA = 50,

AI = 10,

IN = 10, and

NR = 50

Since two pairs of adjacent sides are congruent, the quadrilateral is a kite.

4.WI = IN = ND = DW = 18. The slope of IN and the slope of DW are both = –1. The slope of WI and the slope of ND are both = 1. Since the product of 1 and –1 is –1 WIIN, INND, NDDW, and DWWI.

–3

3

3

3

Since all sides are congruent and all angles are right angles, the quadrilateral is a square.

6-6

m B = 45, m C = m D = 135

Lesson 6-5

Lesson Quiz

Use isosceles trapezoid ABCD for Exercises 1 and 2.

1.If m A = 45, find m B, m C, and m D.

2.If AC = 3x – 16 and BD = 10x – 86, find x.

10

Use kite GHIJ for Exercises 3–6.

3.Find m 1.

4.Find m 2.

5.Find m 3.

6.Find m 4.

90

9

81

40

6-6

Lesson 6-6

Notes

You have used coordinate geometry to find the midpoint of a line segment and to find the distance between two points. Coordinate geometry can also be used to prove conjectures.

A coordinate proof is a style of proof that uses coordinate geometry and algebra. The first step of a coordinate proof is to position the given figure in the plane. You can use any position, but some strategies can make the steps of the proof simpler.

6-6

Lesson 6-6

Notes

6-6

Lesson 6-6

Notes

Once the figure is placed in the coordinate plane, you can use slope, the coordinates of the vertices, the Distance Formula, or the Midpoint Formula to prove statements about the figure.

6-6

Lesson 6-6

Notes

A coordinate proof can also be used to prove that a certain relationship is always true.

You can prove that a statement is true for all special quadrilaterals without knowing the side lengths.

To do this, assign variables as the coordinates of the vertices.

6-6

Lesson 6-6

Notes

If a coordinate proof requires calculations with fractions, choose coordinates that make the calculations simpler.

For example, use multiples of 2 when you are to find coordinates of a midpoint. Once you have assigned the coordinates of the vertices, the procedure for the proof is the same, except that your calculations will involve variables.

6-6

Caution!

Do not use both axes when positioning a figure unless you know the figure has a right angle.

Lesson 6-6

Notes

6-6

Lesson 6-6

Additional Examples

Real-World Connection

An art class creates T-shirt designs by drawing quadrilaterals and connecting their midpoints to form other quadrilaterals.Show that TWVU is a parallelogram by proving both pairs of opposite sides congruent.

If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram by Theorem 6-6.

You can prove that TWVU is a parallelogram by showing that TW = VU and WV = TU. Use the Distance Formula.

6-6

TW = (a + c – a)2 + (b + d – b)2 =

VU = (c + e – e)2 + (d – 0)2 =

WV = ((a + c) – (c + e))2 + (b + d – d)2 =

TU = (a – e)2 + (b – 0)2 =

(a – e)2 + b2

(a – e)2 + b2

c2 + d2

c2 + d2

Lesson 6-6

Additional Examples

(continued)

Use the coordinates T(a, b), W(a + c, b + d), V(c + e, d), and U(e, 0).

Because TW = VU and WV = TU, TWVU is a parallelogram.

Quick Check

6-6

Because AB || CO and CO is horizontal, AB is also horizontal. So point B has the same second coordinate, q, as point A.

Lesson 6-6

Additional Examples

Naming Coordinates

Use the properties of parallelogram OCBA to find the missing coordinates. Do not use any new variables.

The vertex O is the origin with coordinates O(0, 0).

Because point A is p units to the left of point O, point B is also p units to the left of point C because OCBA is a parallelogram. So the first coordinate of point B is – p – x.

The missing coordinates are O(0, 0) and B(–p – x, q).

Quick Check

6-6

b

a

–

b

2

c + a

2

b

2

a

2

midpoint:, ;

slope:

midpoint:, ;

slope:

c + a

b

Lesson 6-6

Lesson Quiz

Find the missing coordinates of each figure.

2. rhombus

3. rectangle

1. parallelogram

M (b, c + a)

M(2a, 0), D(a, –b)

A(0, b), D(a, 0)

Find the coordinates of the midpoint and the slope.

4.OM in Exercise 15. AD in Exercise 26. AD in Exercise 3

midpoint: (a, 0);

slope: undefined

6-6

Lesson 6-7

Notes

The midsegment of a trapezoid is the segment that joins the midpoints of the nonparallel opposite sides.

6-7

Lesson 6-7

Notes

6-7

In a trapezoid, only one pair of sides is parallel.

In TRAP, TP || RA . Because TP lies on the

horizontal x-axis, RA also must be horizontal.

Lesson 6-7

Additional Examples

Naming Coordinates

Examine trapezoid TRAP. Explain why you can assign the same y-coordinate to points R and A.

The y-coordinates of all points on a horizontal line are the same, so points R and A have the same y-coordinates.

Quick Check

6-7

The quadrilateral XYZW formed by connecting the midpoints of ABCD is shown below.

Lesson 6-7

Additional Examples

Using Coordinate Proof

Use coordinate geometry to prove that the

quadrilateral formed by connecting the midpoints of rhombus ABCD is a rectangle.

From Lesson 6-6, you know that XYZW is a parallelogram.

If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle by Theorem 6-14.

6-7

( –2a)2 + (2b)2 = 4a2 + 4b2

XZ = (–a – a)2 + (b – (–b))2 =

( –2a)2 + (–2b)2 = 4a2 + 4b2

YW = (–a – a)2 + (– b – b)2 =

Lesson 6-7

Additional Examples

Quick Check

(continued)

To show that XYZW is a rectangle, find the lengths of its diagonals,

and then compare them to show that they are equal.

Because the diagonals are congruent, parallelogram XYZW is a rectangle.

6-7

Lesson 6-7

Lesson Quiz

Use the diagram for Exercises 1–5.

1. Point M is the midpoint of AC. Find its coordinates.

2. Point N is the midpoint of BC. Find its coordinates.

(a + c, d)

(b + c, d)

3.Explain how you know that MN || AB.

4.Show that MN = AB.

5.What theorem do Exercises 1–4 prove?

Both have slope 0, so they are parallel.

1

2

The Distance Formula finds MN = b – a and AB = 2b – 2a.

Triangle Midsegment Theorem

6-7