Geometry Review Answer Key
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GEOMETRY IA REVIEW #1 — Master Answer Key
PAGE 1 OF 6 Topic: Parallel & Perpendicular Lines
SWBAT: Determine parallel & perpendicular line equations in coordinate geometry SECTION: DO NOW
**#1
Equation of a Parallel Line
Final: \(y = \frac{4}{3}x + 9\)
Problem: If \(a \parallel b\) and line \(b\) has the equation \(6y - 8x = 24\) and line \(a\) passes through the point \((-6, 1)\), determine the equation of line \(a\).
Step 1: Find slope of given line \(b\)
Solve \(6y - 8x = 24\) for \(y\) (Slope-Intercept Form):
\(6y = 8x + 24 \implies y = \frac{8}{6}x + 4 \implies y = \frac{4}{3}x + 4\)
Slope of line \(b\): \(m_b = \frac{4}{3}\).
Step 2: Apply parallel property & point \((-6, 1)\)
Parallel lines share identical slopes: \(m_a = m_b = \frac{4}{3}\).
Point-Slope Form: \(y - 1 = \frac{4}{3}(x + 6)\)
Slope-Intercept Form: \(y = \frac{4}{3}x + 9\)
Visual: Parallel Slopes (\(m = \frac{4}{3}\)) (-6,1) Line b Line a Both rise 4 units for every 3 units run
*#2
Perpendicular Line from Graph
Correct Choice: (c)
Problem: In the graph below \(\overline{AB}\) is graphed. What is the equation of the line that passes through \(C(1,7)\) that is perpendicular to \(\overline{AB}\)?
Plotted Lines with Right Angle (\(\perp\)) 1 2 -2 4 8 A(-2,8) B(0,4) C(1,7)
Step 1: Determine slope of line \(\overline{AB}\)
Points from graph: \(A(-2, 8)\) and \(B(0, 4)\):
\(m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 8}{0 - (-2)} = \frac{-4}{2} = -2\)
Step 2: Negative reciprocal perpendicular slope
\(m_\perp = -\frac{1}{m_{AB}} = -\frac{1}{-2} = +\frac{1}{2}\)
Using point \(C(1, 7)\) in point-slope formula \(y - y_1 = m(x - x_1)\):
\(y - 7 = \frac{1}{2}(x - 1)\)
a) \(y - 7 = -2(x - 1)\) Used parallel slope
b) \(y + 7 = -2(x + 1)\) Sign & slope errors
c) \(y - 7 = \frac{1}{2}(x - 1)\) ✓ Correct
d) \(y + 7 = \frac{1}{2}(x + 1)\) Coordinate signs flipped
Key Exam Concept: Parallel lines have equal slopes (\(m_1 = m_2\)). Perpendicular lines have negative reciprocal slopes (\(m_1 \cdot m_2 = -1\)). Geometry IA
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GEOMETRY IA REVIEW #1 — Skills Drill
PAGE 2 OF 6 Score Target: 4 / 4
**#1 Answer: (d)
Which line is perpendicular to line \(q\)?
\(q: y = \frac{1}{2}x + 5\)
1. Identify given slope: \(m_q = \frac{1}{2}\)
2. Flip & negate for \(\perp\):
\(m_\perp = -\frac{1}{1/2} = -2\)
3. Match option with slope \(-2\):
a) \(y = \frac{1}{2}x + 2\) (\(m=\frac{1}{2}\) \(\parallel\))
b) \(y = 2x + 2\) (\(m=2\))
c) \(y = -\frac{1}{2}x + 2\) (\(m=-\frac{1}{2}\))
d) \(y = -2x + 2\) ✓ Correct (\(m=-2\))
*#2 Slope: \(m = 2\)
What is the slope of a line parallel to:
\(k: 4y - 8x = 12\)
Step 1: Convert to slope-intercept form (\(y = mx + b\)):
\(4y = 8x + 12\)
\(y = \frac{8}{4}x + \frac{12}{4} \implies y = 2x + 3\)
Step 2: Determine parallel slope condition:
Parallel lines have equal slopes (\(m_\parallel = m_k\)).
PARALLEL SLOPE ANSWER: \(m_\parallel = 2\) (or \(\frac{2}{1}\))
**#3 Perpendicular Slope: \(+3\)
Perpendicular slope of the line below:
(0, -3) (-6, -1) run = +6 rise = -2
Given line slope: \(m = \frac{-2}{6} = -\frac{1}{3}\)
Perpendicular slope: \(m_\perp = -\frac{1}{-1/3} = \mathbf{3}\) (or \(\frac{3}{1}\))
#4 Parallel Graph
Graph a line parallel (\(\parallel\)) to \(\overleftrightarrow{AB}\):
A B Parallel line
Slope of \(\overleftrightarrow{AB}\): \(m = -2\).
Any line drawn with slope \(= -2\) (e.g. \(y = -2x\) through origin) is 100% correct.
Scoring Tip: On free response graphing questions, students must plot at least 2 integer grid points and draw line extending through the grid with arrows. Skills Drill Review
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GEOMETRY IA REVIEW #1 — Problem Set A
PAGE 3 OF 6 Triangle Perimeter & Distance Formula
**#1
Perimeter & Quadratic Factoring
Largest Side = 36 inches
Problem: In triangle \(ABC\) below the perimeter is \(72\text{ inches}\). What is the length of the largest side?
Triangle \(ABC\) A C B \(x^2 - 7x\) \(2x\) \(4x\)
Step 1: Set up Perimeter equation
\(AB + BC + AC = \text{Perimeter}\)
\(2x + 4x + (x^2 - 7x) = 72 \implies x^2 - x - 72 = 0\)
Step 2: Factor & solve for \(x\)
\((x - 9)(x + 8) = 0 \implies x = 9 \quad \text{or} \quad x = -8\)
Reject \(x = -8\) because side length cannot be negative: \(2(-8) = -16 < 0\). Thus, \(x = 9\).
Step 3: Calculate each side length with \(x = 9\)
\(AB = 2(9) = \mathbf{18}\) \(BC = 4(9) = \mathbf{36}\) \(AC = 9^2 - 7(9) = \mathbf{18}\)
Largest side is \(BC = 36\text{ inches}\). (Check: \(18 + 36 + 18 = 72\text{ in}\))
#2
Distance Formula Modeling
Bruce lives closer
Problem: Abby lives at point \(A(5, -2)\) and Bruce lives at point \(B(-2, 0)\). Who lives closer to the school at point \(S(0, 4)\)?
Locations on Coordinate Plane S(0,4) A(5,-2) B(-2,0) d ≈ 7.81 d ≈ 4.47
Abby to School (\(AS\))
\(A(5, -2)\) to \(S(0, 4)\):
\(d = \sqrt{(0 - 5)^2 + (4 - (-2))^2}\)
\(d = \sqrt{(-5)^2 + (6)^2}\)
\(d = \sqrt{25 + 36} = \mathbf{\sqrt{61} \approx 7.81}\)
Bruce to School (\(BS\))
\(B(-2, 0)\) to \(S(0, 4)\):
\(d = \sqrt{(0 - (-2))^2 + (4 - 0)^2}\)
\(d = \sqrt{(2)^2 + (4)^2}\)
\(d = \sqrt{4 + 16} = \mathbf{\sqrt{20} \approx 4.47}\)
Conclusion: Since \(\sqrt{20} < \sqrt{61}\) (\(4.47 < 7.81\)), Bruce lives closer to the school by \(\approx 3.34\) units.
Distance Formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Always watch negative coordinates carefully when subtracting! Problem Set A (Part 1)
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GEOMETRY IA REVIEW #1 — Problem Set A (Cont.)
PAGE 4 OF 6 Perpendicular Bisectors & Midpoints
**#5
Perpendicular Bisector Equation
Choice 2: \(y - 1 = -2(x - 1)\)
Problem: \(\overleftrightarrow{AB}\) is the perpendicular bisector of \(\overline{QP}\). The endpoints of \(\overline{QP}\) are \(Q(3,2)\) and \(P(-1, 0)\). Which is the equation of \(\overleftrightarrow{AB}\)?
1. Find Midpoint \(M\) of \(\overline{QP}\)
\(M = \left(\frac{3 + (-1)}{2}, \frac{2 + 0}{2}\right) = \mathbf{(1, 1)}\)
2. Find Slope of \(\overline{QP}\)
\(m_{QP} = \frac{0 - 2}{-1 - 3} = \frac{-2}{-4} = \frac{1}{2}\)
3. Take Opposite Reciprocal Slope & Apply Point-Slope Form
Perpendicular slope \(m_\perp = -\frac{1}{1/2} = \mathbf{-2}\). Through midpoint \((x_1, y_1) = \mathbf{(1, 1)}\):
\(y - 1 = -2(x - 1)\)
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\(y+1=-2(x+1)\)
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\(y-1=-2(x-1)\) ✓
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\(y-11=\frac{1}{2}(x+4)\)
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\(y+11=\frac{1}{2}(x-4)\)
P(-1,0) Q(3,2) M(1,1) Line AB (⊥ bisector)
#6
Segment Addition & Multiple Midpoints
Length of \(BH = 45\)
Problem: In the figure, \(F\) is the midpoint of \(\overline{GH}\). If \(GH = 60\), and point \(B\) is the midpoint of \(\overline{GF}\), what is the length of \(\overline{BH}\)?
G B F (midpoint of GH) H GB = 15 BF = 15 FH = 30 BH = 15 + 30 = 45
1. Since \(F\) is midpoint of \(GH = 60\):
\(GF = FH = \frac{60}{2} = 30\)
2. Since \(B\) is midpoint of \(GF = 30\):
\(GB = BF = \frac{30}{2} = 15\)
3. Calculate length of \(\overline{BH}\):
\(BH = BF + FH = 15 + 30 = \mathbf{45}\)
*#7
Half the Length of a Segment
Half Length = \(5\sqrt{2} \approx 7.07\)
Problem: The segment \(AB\) has endpoints \(A(-10, -5)\) and \(B(0, -15)\). Determine half the length of \(\overline{AB}\).
Method 1: Total Length Divided by 2
\(AB = \sqrt{(0 - (-10))^2 + (-15 - (-5))^2}\)
\(AB = \sqrt{(10)^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200}\)
\(AB = \sqrt{100 \cdot 2} = 10\sqrt{2}\)
\(\text{Half length} = \frac{AB}{2} = \frac{10\sqrt{2}}{2} = \mathbf{5\sqrt{2}}\) (or \(\approx 7.07\))
Method 2: Distance from Midpoint to Endpoint
Midpoint of \(AB\):
\(M = \left(\frac{-10+0}{2}, \frac{-5-15}{2}\right) = (-5, -10)\)
\(AM = \sqrt{(-5 - (-10))^2 + (-10 - (-5))^2}\)
\(AM = \sqrt{5^2 + (-5)^2} = \sqrt{50} = \mathbf{5\sqrt{2}}\)
Both methods verify the exact same result.
Common Pitfall: Always simplify radicals fully! \(\sqrt{200} = 10\sqrt{2}\), so half is \(5\sqrt{2}\) (do NOT divide 200 inside the radical by 2). Problem Set A (Part 2)
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GEOMETRY IA REVIEW #1 — Problem Set B
PAGE 5 OF 6 Writing Perpendicular Lines & System Classification
#8
Equation of Perpendicular Line Through Given Point
\(y - 6 = \frac{5}{2}(x + 2)\)
Problem: Line \(m\) is represented by the equation \(y = -\frac{2}{5}x + 4\). Line \(n\) intersects line \(m\) at the point \((-2, 6)\) and is perpendicular to it. Write an equation for line \(n\).
Step 1: Identify slope of line \(m\) and find perpendicular slope
Line \(m\): \(y = -\frac{2}{5}x + 4 \implies m_m = -\frac{2}{5}\).
Opposite reciprocal slope for perpendicular line \(n\):
\(m_n = -\frac{1}{-2/5} = +\frac{5}{2}\)
Step 2: Write Equation in Point-Slope & Slope-Intercept Forms
Passes through \((x_1, y_1) = (-2, 6)\) with slope \(m = \frac{5}{2}\):
Point-Slope Form:
\(y - 6 = \frac{5}{2}(x + 2)\)
Slope-Intercept Form:
\(y = \frac{5}{2}x + 11\)
Perpendicular at \((-2,6)\) Line m Line n (-2,6)
#9
Comparing Multiple Linear Equations
Choice 4: II and III
Problem: Consider the linear equations below: \(s: -4y = -3x + 16\) | \(t: 6y + 8x + 36 = 0\) | \(r: 6x - 8 = 8y\) Which of the following statements is true? \(I: s \parallel t\) | \(II: s \parallel r\) | \(III: s \perp t\)
Line \(s\)
\(-4y = -3x + 16\)
\(y = \frac{-3}{-4}x + \frac{16}{-4}\)
\(y = \frac{3}{4}x - 4\)
Slope \(m_s = \frac{3}{4}\)
Line \(t\)
\(6y = -8x - 36\)
\(y = -\frac{8}{6}x - 6\)
\(y = -\frac{4}{3}x - 6\)
Slope \(m_t = -\frac{4}{3}\)
Line \(r\)
\(8y = 6x - 8\)
\(y = \frac{6}{8}x - 1\)
\(y = \frac{3}{4}x - 1\)
Slope \(m_r = \frac{3}{4}\)
Statement I: \(s \parallel t\) \(\implies m_s = \frac{3}{4} \neq -\frac{4}{3} = m_t\) FALSE (Slopes not equal)
Statement II: \(s \parallel r\) \(\implies m_s = \frac{3}{4} = m_r = \frac{3}{4}\) (different y-intercepts: \(-4 \neq -1\)) TRUE (Equal slopes \(\parallel\))
Statement III: \(s \perp t\) \(\implies m_s \cdot m_t = \left(\frac{3}{4}\right)\left(-\frac{4}{3}\right) = -1\) TRUE (Opposite reciprocals \(\perp\))
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I only
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I and II
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I and III
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II and III
Strategic Advice: When testing statements with multiple linear equations, isolate \(y\) for all equations first so slopes and intercepts can be compared instantly. Problem Set B (Part 1)
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GEOMETRY IA REVIEW #1 — Problem Set B (Cont.)
PAGE 6 OF 6 Error Analysis & Coordinate Graphing
*#10
Student Error Analysis
Correct Student: Student B
Problem: \(g\) is perpendicular to \(k\) and the equation of \(g\) is \(4y + x = 12\). If \(k\) passes through the point \((4, -1)\), determine the equation of \(k\).
Student A
\(y = mx + b\)
\(-1 = -\frac{1}{4}(4) + b\)
Student B ✓
\(y = mx + b\)
\(-1 = 4(4) + b\)
Student C
\(y = mx + b\)
\(-1 = 1(4) + b\)
Analysis of Line \(g\) & Required Slope of \(k\):
\(4y = -x + 12 \implies y = -\frac{1}{4}x + 3\). Thus \(m_g = -\frac{1}{4}\). Since \(k \perp g\), the perpendicular slope is \(m_k = -\frac{1}{-1/4} = \mathbf{4}\).
Who is correct?
Student B used the correct perpendicular slope \(m = 4\) with point \((4, -1)\): \(-1 = 4(4) + b \implies b = -17\). Equation: \(y = 4x - 17\).
What is incorrect about the others?
Student A used \(m = -\frac{1}{4}\) (parallel slope instead of perpendicular). Student C incorrectly assumed \(m = 1\) from \(x\)'s coefficient.
**#11
Perpendicular Lines Coordinate Graphing
Comprehensive Free-Response
Problem: Lines \(n\) and \(p\) are perpendicular. The equation of line \(n\) is \(6y - 2x = 12\).
Accurate Graph of Lines \(n\) and \(p\) x y (0,2) line n (-2,-2) (0,-8) line p (-3,1)
a) Graph line \(n\) and label:
Solve \(6y - 2x = 12 \implies 6y = 2x + 12 \implies y = \frac{1}{3}x + 2\).
y-intercept: \((0, 2)\), Slope \(m_n = \frac{1}{3}\) (up 1, right 3).
b) Point-slope equation for line \(p\):
Passes through \((-2, -2)\). Since \(p \perp n\), perpendicular slope is \(m_p = -\frac{1}{1/3} = \mathbf{-3}\).
\(y - (-2) = -3(x - (-2)) \implies \mathbf{y + 2 = -3(x + 2)}\)
c) Slope-intercept form & graph line \(p\):
\(y + 2 = -3x - 6 \implies \mathbf{y = -3x - 8}\)
Plot y-intercept \((0, -8)\), point \((-2, -2)\), and intersection point \((-3, 1)\) with slope \(-3\).
Final Review Check: Lines \(n\) and \(p\) intersect perpendicularly at \((-3, 1)\), proving \(\left(\frac{1}{3}\right)(-3) = -1\). Good luck on the IA! End of IA Review #1
