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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.
Unit 6: Geometric Algebra Name: ___________________________
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.
Unit 6: Geometric Algebra Name: ___________________________
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
Unit 6: Geometric Algebra Name: ___________________________
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?
Unit 6: Geometric Algebra Name: ___________________________
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).
Unit 6: Geometric Algebra Name: ___________________________
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
Unit 6: Geometric Algebra Name: ___________________________
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)
Unit 6: Geometric Algebra Name: ___________________________
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.
Unit 6: Geometric Algebra Name: ___________________________
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.
Unit 6: Geometric Algebra Name: ___________________________
11
b) Graph and show that quadrilateral W (-1,7), X (5,6), Y (4,-1) and Z (-2,0) is a
parallelogram.
Unit 6: Geometric Algebra Name: ___________________________
12
Blank page
Unit 6: Geometric Algebra Name: ___________________________
13
KEY
Unit 6: Geometric Algebra Name: ___________________________
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
Unit 6: Geometric Algebra Name: ___________________________
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
Unit 6: Geometric Algebra Name: ___________________________
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
Unit 6: Geometric Algebra Name: ___________________________
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.
Unit 6: Geometric Algebra Name: ___________________________
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
Unit 6: Geometric Algebra Name: ___________________________
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).
Unit 6: Geometric Algebra Name: ___________________________
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)
Unit 6: Geometric Algebra Name: ___________________________
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.
Unit 6: Geometric Algebra Name: ___________________________
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.
Unit 6: Geometric Algebra Name: ___________________________
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.