unit 6: geometric algebra - wordpress.com · 5/5/2015 · eoct prep unit 6: geometric algebra...

EOCT Prep

UNIT 6: Geometric Algebra

Checklist

MAX Scored

1 Vocabulary 25

2 Using distance and slope formulas 25

3 Finding parallel and perpendicular lines 15

4 Find a point between two points on given a segment for a

given ratio, including midpoint. 35

5 Find points of intersection between lines 10

6 Use coordinate to calculate perimeters and areas for

various geometric figures 20

7 Use distance and slopes to prove different geometric

figures 20

Totals 150

Name: _______________________________

Period: __________ Date: _____________

Unit 6: Geometric Algebra Name: ___________________________

2

Section 1. Vocabulary

Word Bank (word is used only once; not all words will be used below)

Circle Obtuse triangles Quadrilateral Square

Equilateral triangles Parallel lines Rectangle Transformation

Isosceles triangles Parallelogram Rhombus Trapezoid

Median Perpendicular lines Scalene triangles X-intercepts

Midpoint Pythagorean

Theorem Slope Y-intercepts

a) ______________ are points on a graph where y = 0.

b) The ____________ of a line measures the steepness of the line. It is often

referred to as the “rise over the run” or the “rate of change”.

c) ___________ are coplanar lines that do not intersect. The slopes of parallel

lines are equal to each other, but these lines have different ____________.

d) _______________ are two lines that intersect at a right angle. The slopes

of two perpendicular lines are opposite-sign reciprocals of each other.

e) _______________ are triangles where all three sides are equal in length.

_______________ are triangles with two equal sides, whereas

_______________ have three sides with three different lengths.


3

g) A ________________ is a polygon with four sides (edges) and four vertices.

h) A ________________ is a special quadrilateral with two sets of parallel

lines, with opposite sides congruent to each other, and the sum of the adjacent

interior angles equal to 1800.

i) A _____________ is a special parallelogram with all interior angles equal to

900.

j) A _____________ is a special quadrilaterial with all fours sides equal in

length. A ______________ is a special quadrilateral with all fours sides

equal in length AND with all interior angles equal to 900.

k) The distance formula uses the ____________________ to show the

shortance distance between two points is d = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2.

l) A ____________ is the EXACT (one and only) point that bisects a line

segment into two equal part.


4

Vocabulary (continued)

T F m. All lines have a y-intercept.

n. A line that is parallel to a given line is also perpendicular to any line

perpendicular to the given line.

o. Two parallel lines must have at least one unique y-intercept or x-intercept

p. The Beatles are also known as the “Fab Four”

q. An obtuse triangle can have two interior obtuse angles

r. The area of a right angle triangle is the product of the two legs divided

by two

s. A rhombus with at least one right angle is a square

t. If the diagonals of a quadrilateral are perpendicular AND the point of

intersection for the diagonals are the midpoint for both diagonals, then

the quadrilateral is a rhombus

u. All rhombuses are squares, but not all squares are rhombuses

v. A product of the slopes for two perpendicular lines is always -1


5

Section 2: Distance & Slopes of Lines

Use the graph to calculate or show the

following:

a) Distance, d(AE)

b) Distance, d(AF)

c) Distance, d(AJ)

d) Distance, d(HJ)

e) Is 𝐸𝐹̅̅ ̅̅ ≅ 𝐷𝐺̅̅ ̅̅ ? How do you know?

f) Slope (BC)

g) Slope (ED)

h) Is 𝐸𝐷̅̅ ̅̅ ∥ 𝐹𝐺̅̅ ̅̅ ? How do you know?

i) Is 𝐸𝐹̅̅ ̅̅ ∥ 𝐷𝐺̅̅ ̅̅ ? How do you know?

j) Is DEFG are parallelogram? How do you know?

k) Is 𝐴𝐵̅̅ ̅̅ ⊥ 𝐹𝐺̅̅ ̅̅ ? How do you know?


6

Section 3: Finding parallel & perpendicular lines

a) Find the equation of a line

perpendicular to 𝑦 =2

3𝑥 − 5

passing through the point (-3.4).

b) Find the equation of the line through

points (5,4) and (11,2), then find the

perpendicular line to the first line that

runs through point (7,-2).

c) Find a line parallel to line

𝑦 = −2𝑥 + 1 and through

the point (2,3).


7

Section 4: Midpoint and Line Segment Portions

Find the midpoints for these pairs of points.

a. (-11,8) & (-7,16) b. (11,-3) & (15,17) c. (-4.02, 6.22) & (14.26,7.88)

Find the missing endpoint for these problems.

d. Midpoint (-4,6); Endpoint (2,1) e. Midpoint (-7,-8); Endpoint (-4,-2)

Find the partitioned points for these problems.

f. A (-18,6); B (2,2).

Find the point that is ¼ of AB.

g. T (16,-8); U (-8,-12).

Find the point that splits UT in the ratio 3:1


8

Section 5: Find the points of intersection for the following problems:

a) The point of intersection between two perpendicular lines:

𝑦 = −4

3𝑥 + 8 and 𝑦 =

3

4𝑥 +

7

4

b) The point of intersection between the lines formed from the set of points:

Line 1: pts A (4,8) and B (12,-4), and Line 2: X (10,6) and Y (6,-2)


9

Section 6: Areas & Perimeters

a) If a triangle has the vertices A (-5,2), B (4,-3) and C (4,7), calculate the perimeter

and the area of the triangle.

b) If a quadrilateral has vertices of A (-2,1), B (-6,6), C (-1,6), and D (3,1), calculate

the perimeter and area of the quadrilateral.


10

Section 7: “Show Me”

a) Graph and show that quadrilateral A (2,4), B (-3,5), C (-4,0) and D (1.-1) is a square.


11

b) Graph and show that quadrilateral W (-1,7), X (5,6), Y (4,-1) and Z (-2,0) is a

parallelogram.


12

Blank page


13

KEY


14

Section 1. Vocabulary

a) ____________ are points on a graph where y = 0.

b) The __________ of a line measures the steepness of the line. It is often referred to

as the “rise over the run” or the “rate of change”.

c) ___________ are coplanar lines that do not intersect. The slopes of parallel lines are

equal to each other, but these lines have different ____________.

d) _______________ are two lines that intersect at a right angle. The slopes of two

perpendicular lines are negative reciprocals of each other.

e) __________________ are triangles where all three sides are equal in length.

__________________ are triangles with two equal sides, whereas

__________________ have three sides with three different lengths.

f) A ________________ is a polygon with four sides (edges) and four vertices.

X-Intercepts

Slope

Parallel Lines

Y-Intercepts

Perpendicular Lines

Equilateral triangles

Isosceles triangles

Scalene triangles

Quadrilateral


15

g) A ________________ is a special quadrilateral with two sets of parallel lines, with

opposite sides congruent to each other, and the sum of the adjacent interior angles equal to

1800.

h) A _____________ is a special parallelogram with all interior angles equal to 900.

i) A _____________ is a special quadrilaterial with all fours sides equal in length. A

______________ is a special quadrilateral with all fours sides equal in length AND

with all interior angles equal to 900.

j) The distance formula uses the ____________________ to show the shortance

distance between two points is d = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2.

k) A ____________ is the EXACT (the one and only) point that bisects a line

segment into two equal part.

Parallelogram

Rectangle

Rhombus

Square

Pythagorean Theorem

Midpoint


16

Vocabulary (continued)

T F

m. All lines have a y-intercept. Consider x = 1 F

n. A line that is parallel to a given line is also perpendicular to any line

perpendicular to the given line. T

o. Two parallel lines must have at least one unique y-intercept or x-intercept T

p. The Beatles are also known as the “Fab Four”. An oldie but a goodie. T

q.

An obtuse triangle can have two interior obtuse angles. Two obtuse

angles when summed together are greater than 180, but the sum of

the interior angles to a triangle is exactly 180.

F

r. The area of a right angle triangle is the product of the two legs divided

by two T

s. A rhombus with at least one right angle is a square T

t.

If the diagonals of a quadrilateral are perpendicular AND the point of

intersection for the diagonals are the midpoint for both diagonals, then

the quadrilateral is a rhombus T

u. All rhombuses are squares, but not all squares are rhombuses. All

squares are indeed rhombuses because they have 4 equal sides. F

v.

A product of the slopes for two perpendicular lines is always -1. There is

one exception . . . when the slope of the line is zero, then the

reciprocal is undefined. F


17

Section 2: Distance and slope of lines

a) Distance, d(AE)

√(5 − (−6))2 + (4 − 4)2 = 11 units

b) Distance, d(AF)

√(5 − (−7))2

+ (4 − (−3))2

= √(12)2 + (7)2 = 13.89 units

c) Distance, d(AJ) √(5 − 6)2 + (4 − (−3))2 = √(−1)2 + (7)2 = √50 = 7.07 units

d) Distance, d(HJ) √(2 − 6)2 + (−1 − (−3))2 = √(−4)2 + (2)2 = √20 = 4.47 units

e) Is 𝐸𝐹̅̅ ̅̅ ≅ 𝐷𝐺̅̅ ̅̅ ? How do you know? 𝑑(𝐷𝐺) = 𝑑(𝐸𝐹) = √50. Same length means congruent.

f) Slope (BC) = 𝑦2− 𝑦1

𝑥2−𝑥1=

2−1

(4−6)= −

1

2

g) Slope (ED) = 𝑦2− 𝑦1

𝑥2−𝑥1=

4−3

(−6−(−2))= −

1

4

h) Is 𝐸𝐷̅̅ ̅̅ ∥ 𝐹𝐺̅̅ ̅̅ ? How do you know? Slope (ED) = Slope (FG) = -¼. Because the slopes are

equal, and there are two separate y-intercepts (from diagram), 𝑬𝑫̅̅ ̅̅ ∥ 𝑭𝑮.̅̅ ̅̅ ̅

i) Is 𝐸𝐹̅̅ ̅̅ ∥ 𝐷𝐺̅̅ ̅̅ ? How do you know? Slope (EF) = Slope (DG) = +7. Because the slopes are

equal, and there are two separate y-intercepts (from diagram), 𝑬𝑫̅̅ ̅̅ ∥ 𝑭𝑮.̅̅ ̅̅ ̅

j) Is DEFG are parallelogram? How do you know? Since 𝑬𝑫̅̅ ̅̅ ∥ 𝑭𝑮̅̅ ̅̅ and 𝑬𝑭̅̅ ̅̅ ∥ 𝑫𝑮̅̅ ̅̅ , we have a

quadrilaterial with two sets of parallel lines, which by definition, is a parallelogram.

k) Is 𝐴𝐵̅̅ ̅̅ ⊥ 𝐹𝐺̅̅ ̅̅ ? How do you know? Slope (AB) = 2. Slope (BC) = - ½ . Because the

slopes of AB and BC are oppostive sign reciprocals, the two lines are perpendicular.


18

Section 3: Finding parallel & perpendicular lines

1. Find the equation of a line

perpendicular to 𝑦 =2

3𝑥 − 5

passing through the point (-3.4).

Slope of original line is 2/3.

Slope of perpendicular line is -3/2.

𝒚 = −𝟑

𝟐 𝒙 + 𝒃. 𝟒 = −

𝟑

𝟐 (−𝟑) + 𝒃. 𝒃 = −

𝟏

𝟐.

Equation: 𝒚 = −𝟑

𝟐 𝒙 −

𝟏

𝟐

2. Find the equation of the line through

points (5,4) and (11,2), then find the

perpendicular line to the first line

that runs through point (7,-2).

Slope of original line is 𝑦2− 𝑦1

𝑥2−𝑥1=

2−4

11−5= −

1

3

Equation of line: 𝒚 = −𝟏

𝟑 𝒙 +

𝟏𝟕

𝟑.

𝟐 = −𝟏

𝟑(𝟏𝟏) + 𝒃. 𝒃 =

𝟏𝟕

𝟑

Slope of perpendicular line is 3.

𝒚 = 𝟑 𝒙 + 𝒃. −𝟐 = 𝟑 (𝟕) + 𝒃. 𝒃 = −𝟐𝟑

Equation: 𝒚 = 𝟑𝒙 − 𝟐𝟑

3. Find a line parallel to

line 𝑦 = −2𝑥 + 1 and

through the point (2,3).

Slope of original line is -2

Slope of parallel line -2

𝒚 = −𝟐 𝒙 + 𝒃. 𝟑 = −𝟐 (𝟐) + 𝒃. 𝒃 = 𝟕

Equation: 𝒚 = −𝟐𝒙 + 𝟕


19

Section 4: Midpoint and Line Segment Portions

Find the midpoints for these pairs of points.

a. (-11,8) & (-7,16) b. (11,-3) & (15,17) c. (-4.02, 6.22) & (14.26,7.88)

Midpt = (−11−7

2,

8+16

2)

= (−𝟗, 𝟏𝟐)

Midpt = (11+15

2,

−3+17

2)

= (𝟏𝟑, 𝟕)

Midpt = (−4.02+14.26

2,

6.22+7.88

2)

= (𝟓. 𝟏𝟐, 𝟕. 𝟎𝟓)

Find the missing endpoint for these problems.

d. Midpoint (-4,6); Endpoint (2,1) e. Midpoint (-7,-8); Endpoint (-4,-2)

Let (A,B) be the missing endpoint.

The change in “x” from (2,1) and (-4,6) is -6

The change in “y” from (2,1) and (-4,6) is +5

A = -4 – 6 = -10, B = 6 + 5 = 11

The missing endpoint is (-10, 11).

Check: Midpt = (2−10

2,

1+11

2) = (−4,6).

Let (T,U) be the missing endpoint.

The change in “x” from (-4,-2) and (-7,-8) is -3

The change in “y” from (-4,-2) and (-7,-8) is -6

A = -7 – 3 = -10, B = -8 – 6 = -14

The missing endpoint is (-10, -14).

Check: Midpt = (−4−10

2,

−2−14

2) = (−7, −8).

Find the partitioned points for these problems.

f. A (-18,6); B (2,2).

Find the point that is ¼ of AB.

g. T (16,-8); U (-8,-12).

Find the pt. that splits UT in the ratio 3:1

Let P(x,y) be the partitioned point

The total change in “x” from A to B is +20

The total change in “y” from A to B is -4

P(x) = A(x) + ¼ (total “x” distance from A to B)

x = -18 + ¼ (20) = -18 + 5 = -13

P(y) = A(Y) + ¼ (total “y” distance from A to B)

y = 6 + ¼ (-4) = 6 – 1 = 5

P(x,y) = (-13, 5).

Let P(x,y) be the partitioned point.

The total change in “x” from U to T is +24

The total change in “y” from U to T is +4

The ratio 3:1 mean P is ¾ from U to T.

P(x,y) = (-8 + ¾ (24), -12 + ¾ (4))

= (-8+18, -12+3) = (10,-9)

P(x,y) = (10,-9).


20

Section 5: Find the points of intersection for the following problems:

a) The point of intersection between two perpendicular lines:

𝑦 = −4

3𝑥 + 8 and 𝑦 =

3

4𝑥 +

7

4

Equate the two sides of y . . . −4

3𝑥 + 8 =

3

4𝑥 +

7

4

Multiple through by 12 . . . -16x + 96 = 9x + 21

Simplify . . . 75 = 25x. x = 3.

Substitute x = 3. y = 4

The point of intersection is (x,y) = (3,4)

b) The point of intersection between the lines formed from the set of points:

Line 1: pts A (4,8) and B (12,-4), and Line 2: X (10,6) and Y (6,-2)

Line 1: m = −12

8= −

3

2. Using point (4,8),

8 = −3

2 (4) + 𝑏. 𝑏 = 14. Line 1: 𝑦 = −

3

2𝑥 + 14.

Line 2: m = 2. Using point (10,6),

6 = 2(10) + 𝑏. 𝑏 = −14. Line 2: 𝑦 = 2𝑥 − 14.

Equating the two sides of y . . . −3

2𝑥 + 14 = 2𝑥 − 14

Multiplying through by 2 . . . -3x + 28 = 4x – 28

Simplifying . . . 56 = 7x. x = 8.

Substituting x = 8. y = 2.

The point of intersection is (x,y) = (8,2)


21

Section 6: Areas & Perimeters

a) If a triangle has the vertices A (-5,2), B (4,-3) and C (4,7), calculate the perimeter and the

area of the triangle. Hint: Graph diagram.

I. Perimeter . . . sum of the bordering sides

Length (AB) = √(−5 − 4)2 + (2 + 3)2 = √106

Length (AC) = √(−5 − 4)2 + (2 − 7)2 = √106

Length (BC) = 7 + 3 = 10

Perimeter = 10 + √106 + √106 = 30.591 units.

II. Area . . . ½(B)(H).

Base = Length (BC) = 10

The slope of perpendicular AZ = negative reciprocal BC = 0.

Height is the distance AZ = 5 + 4 = 9 units.

Area = ½ (10) (9) = 45 square units.

b) If a quadrilateral has vertices of A (-2,1), B (-6,6), C (-1,6), and D (3,1), calculate the perimeter

and area of the quadrilateral. Hint: Graph diagram.

I. Perimeter . . . sum of the sides

Length (AB) = √(−6 + 2)2 + (6 − 1)2 = √41

Length (BC) = -1 – (-6) = 5

Length (CD) = √(3 + 1)2 + (1 − 6)2 = √41

Length (AD) = 3 + 2 = 5

Perimeter = 5 + 5 + 2√41 = 22.806 units.

Note: because 𝐴𝐵̅̅ ̅̅ ≅ 𝐶𝐷̅̅ ̅̅ and 𝐵𝐶̅̅ ̅̅ ≅ 𝐴𝐷 ̅̅ ̅̅ ̅, ABCD

is parallelogram.

II. With parallelogram, Area = Base x Height.

Base AD = 5 (from above)

Slope BE is perpendicular to BC, so BE is parallel to y-axis.

Length (BE) = 5 – 0 = 5

Area ABCD = (5)(5) = 25 square units.


22

Section 7: “Show Me”

c) Graph and show that quadrilateral A (2,4), B (-3,5), C (-4,0) and D (1.-1) is a square.

Show: Show that ABCD is a Square.

Required:

Lengths AB = BC = CD = AD

AB BC, BC CD, CD DA and DA AB,

which in turn shows each of the interior

angles as 90

Approach: Use the distance formula to show all

the sides are equal in length, which shows the

quadrilateral is a rhombus. Then, show the

slopes of the corresponding lines are opposite

reciprocals.

Proof:

Length (AB) = √(2 − (−3))2 + (5 − 4)2 = √26; Length (BC) = √(−4 − (−3))2 + (0 − 5)2 =

√26

Length (CD) = √(1 − (−4))2 + (−1 − 0)2 = √26; Length (DA) = √(1 − 2)2 + (−1 − 4)2 = √26

Thus, lengths AB = BC = CD = AD. So, quadrilateral ABCD is a rhombus.

Slope (AB) = .𝑦2− 𝑦1

𝑥2−𝑥1=

5−4

−3−2= −

1

5. Slope (CD) = Slope (AB) since ABCD is a rhombus.

Slope (AD) = 𝑦2− 𝑦1

𝑥2−𝑥1=

0−5

−4+3=

−5

−1= 5. Slope (BC) = Slope (AD) since ABCD is a rhombus.

Slope (AB) is opposite sign reciprocal of Slope (BC), meaning that AB BC, and B is 90o.

Slope (BC) is opposite sign reciprocal of Slope (CD), meaning that BC CD, and C is 90o.

Slope (CD) is opposite sign reciprocal of Slope (DA), meaning that CD DA, and D is 90o.

Slope (DA) is opposite sign reciprocal of Slope (AB), meaning that DA AB, and A is 90o.

So, we have a rhombus ABCD with all 4 interior angles = 90, which, by definition, is a square.


23

d) Graph and show that quadrilateral W (-1,7), X (5,6), Y (4,-1) and Z (-2,0) is a

parallelogram.

Show: Show that WXYZ is parallelogram.

Required:

Show 𝑊𝑋̅̅ ̅̅ ̅ ≅ 𝑌𝑍̅̅̅̅ and 𝑊𝑍̅̅ ̅̅ ̅ ≅ 𝑋𝑍̅̅ ̅̅ OR

Show 𝑊𝑋̅̅ ̅̅ ̅ ∥ 𝑌𝑍̅̅̅̅ and 𝑊𝑍̅̅ ̅̅ ̅ ∥ 𝑋𝑍̅̅ ̅̅

Approach(es):

(1) Use the distance formula to show

the opposite sides are equal in

length, OR

(2) Use the slope formula to show that

the opposite sides are parallel.

Proof (1):

Length (WX) = √(5 − (−1))2 + (6 − 7)2 = √37; Length (YZ) = √(−1 − 0)2 + (4 − (−2))2 = √37

Length (WZ) = √(−1 − (−2))2 + (0 − 7)2 = √50; Length (XY) = √(−1 − 6)2 + (4 − 5)2 = √50

Thus, length (WX) = length (YZ), meaning 𝑊𝑋̅̅ ̅̅ ̅ ≅ 𝑌𝑍̅̅̅̅ , & length (WZ) = length (XY), meaning

𝑊𝑍̅̅ ̅̅ ̅ ≅ 𝑋𝑍̅̅ ̅̅ .

So, quadrilateral WXYZ is, by definition, a parallelogram since opposite sides of the

parallelogram are congruent.

Proof (2):

Slope (WX) = .𝑦2− 𝑦1

𝑥2−𝑥1=

6−7

5+1= −

1

6. Slope (YZ) =

𝑦2− 𝑦1

𝑥2−𝑥1=

0+1

−2−4= −

1

6.

Slope (XY) = .𝑦2− 𝑦1

𝑥2−𝑥1=

−1−6

4−5= 7. Slope (WZ) =

𝑦2− 𝑦1

𝑥2−𝑥1=

0−7

−2+1= 7.

Thus, slope (WX) = slope (YZ), and slope (WZ) = slope (XY).

So, quadrilateral WXYZ is, by definition, a parallelogram because its opposite sides of the

parallelogram are parallel.

unit 6: geometric algebra - wordpress.com · 5/5/2015 · eoct prep unit 6: geometric algebra...

Documents