Geometry Essentials Study Guide
Accommodated Study Support • Unit 1
Geometry Foundations Study Guide
Formulas, Step-by-Step Problem Checklists & Visual Models
Name: ___________________________
Date: _____________ Period: _____
THE #1 TEST TRAP: "Did I finish the problem?"
Solving for x is only Step 1! Always re-read: does the question ask for x, or does it ask to find the length of a segment or measure of an angle? If so, plug x back in!
1. Essential Coordinate Formulas (Memorize & Use)
Distance Formula Length between 2 points
\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
SPED Tip: Label points \((x_1, y_1)\) and \((x_2, y_2)\) first. Remember: subtracting a negative turns into addition! \(-(-5) = +5\).
Midpoint Formula Halfway point (x, y)
\( M = \left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2} \right) \)
SPED Tip: You are simply finding the average of the x's and average of the y's! Add them, then divide each by 2.
Slope & Perpendicular Slope Opposite Reciprocal
\( m = \frac{y_2 - y_1}{x_2 - x_1} \)
\( m_\perp = -\frac{1}{m} \)
Flip & Negate: If segment slope is \(+\frac{1}{2}\), perpendicular slope is \(-\frac{2}{1} = -2\). If slope is \(-\frac{1}{3}\), perp slope is \(+3\).
Point-Slope Equation Form For line equations
\( y - y_1 = m(x - x_1) \)
Watch the Signs: If the midpoint is \((-4, 7)\), plug in to get \(y - 7 = m(x - (-4))\) which becomes \(y - 7 = m(x + 4)\).
2. Master Geometry Relationships (Setting Up Equations)
Segment Relationships
- Segment Addition Postulate:
\(\text{Part}_1 + \text{Part}_2 = \text{Whole}\) \((AB + BC = AC)\)
- Midpoint / Bisector: Cuts into 2 EQUAL halves!
\(\text{Left Half} = \text{Right Half}\) OR \(2 \times (\text{Half}) = \text{Whole}\)
Angle Relationships
- Complementary: Sum to \(90^\circ\) (forms a corner / right \(\angle\))
\(\angle 1 + \angle 2 = 90^\circ\)
- Supplementary / Linear Pair: Straight line = \(180^\circ\)
\(\angle 1 + \angle 2 = 180^\circ\)
- Vertical Angles: Across the 'X' = Equal (\(\cong\))
Unit 1 Geometry Exam Study Guide Page 1 of 4 • Formulas & Definitions
Topic Review 1
Segments, Midpoints & Distance
Step-by-Step Problem Breakdown
Model 1: Segment Midpoint Chaining (Test Q#5 & HW #4) Work Backwards
Problem: In the figure below, \(F\) is the midpoint of \(\overline{GH}\). The total length \(GH = 60\). Point \(B\) is the midpoint of \(\overline{GF}\). What is the length of \(\overline{BH}\)?
G B (midpoint) F (midpoint) H 15 15 30
Step 1: Big Half
\(F\) splits \(GH = 60\) in half:
\(GF = 30\), \(FH = 30\)
Step 2: Small Half
\(B\) splits \(GF = 30\) in half:
\(GB = 15\), \(BF = 15\)
Step 3: Combine for BH
Add \(BF + FH\):
\(BH = 15 + 30 = \mathbf{45}\)
Model 2: Coordinate Distance & Comparing (Test Q#1, Q#2) Formula Scaffold
Problem: Endpoints are \(A(-10, -5)\) and \(B(0, -15)\). Determine half the length of \(\overline{AB}\).
1 Label: \(x_1 = -10, \; y_1 = -5\) and \(x_2 = 0, \; y_2 = -15\)
2 Plug In: \(d = \sqrt{(0 - (-10))^2 + (-15 - (-5))^2} = \sqrt{(10)^2 + (-10)^2}\)
3 Simplify: \(d = \sqrt{100 + 100} = \sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2}\)
! Answer Check: Question asks for HALF the length \(\rightarrow \frac{10\sqrt{2}}{2} = \mathbf{5\sqrt{2}}\) (Choice C)
Model 3: Segment Bisector with Quadratic (Test Do Now #1 & Q#6) High-Yield Question
Problem: \(B\) is the midpoint of \(\overline{AP}\). If \(AP = x^2 - 9\) and \(BP = 4x\), determine the length of \(\overline{AB}\).
1. Set Up Equation:
Since \(B\) is the midpoint, the Whole = 2 × Half:
\(AP = 2(BP)\)
\(x^2 - 9 = 2(4x) \implies x^2 - 9 = 8x\)
2. Solve Quadratic & Test:
\(x^2 - 8x - 9 = 0 \implies (x - 9)(x + 1) = 0\)
\(x = 9\) or \(x = -1\). Length cannot be negative, so \(x = 9\)!
\(AB = BP = 4(9) = \mathbf{36}\)
Unit 1 Geometry Exam Study Guide Page 2 of 4 • Segments & Lengths
Topic Review 2
Perpendicular Bisector Recipe
Guaranteed 4-Step Method
4
What is a Perpendicular Bisector?
A line that does TWO things: (1) Passes through the MIDPOINT of a segment, and (2) crosses at a \(90^\circ\) angle with an OPPOSITE RECIPROCAL SLOPE.
1 Find the Midpoint
This is the point \((x_m, y_m)\) your new line will pass through:
\( M = \left(\frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2}\right) \)
2 Find Original Slope
Find the steepness of the original segment:
\( m_{\text{orig}} = \frac{y_2 - y_1}{x_2 - x_1} \)
3 Flip & Negate (Slope \(\perp\))
Take the opposite reciprocal of \(m_{\text{orig}}\):
Flip fraction upside-down & change sign (+ to −)
4 Write Point-Slope Form
Plug Midpoint \((x_m, y_m)\) and \(m_\perp\) into:
\( y - y_m = m_\perp (x - x_m) \)
Master Model: Endpoints \(Q(-6, 6)\) and \(R(-2, 8)\) Test Question #4
Step 1: Midpoint
\(x_m = \frac{-6 + (-2)}{2} = \frac{-8}{2} = -4\)
\(y_m = \frac{6 + 8}{2} = \frac{14}{2} = 7\)
Midpoint = \((-4, 7)\)
Step 2: Orig. Slope
\(m = \frac{8 - 6}{-2 - (-6)}\)
\(m = \frac{2}{-2 + 6} = \frac{2}{4}\)
\(m = \frac{1}{2}\)
Step 3: Perp. Slope
Flip \(\frac{1}{2} \to \frac{2}{1}\)
Change \(+ \to -\)
\(m_\perp = -2\)
Step 4: Equation
Point \((-4, 7)\), \(m = -2\)
\(y - 7 = -2(x + 4)\)
✓ Matches Choice B
Your Turn • Guided Practice: Endpoints \((2, 5)\) and \((8, 3)\) (HW #5)
1. Midpoint:
\((\frac{2+8}{2}, \frac{5+3}{2}) = (\_\_, \_\_)\)
2. Orig Slope:
\(m = \frac{3-5}{8-2} = \frac{-2}{6} = \_\_\_\)
3. Perp Slope:
\(m_\perp = \text{flip } -\frac{1}{3} = \_\_\_\)
4. Equation:
\(y - \_\_ = \_\_(x - \_\_)\)
Unit 1 Geometry Exam Study Guide Page 3 of 4 • Perpendicular Bisectors
Topic Review 3
Angle Pairs & Triangle Algebra
Relationships & Quadratic Checks
Complementary \((90^\circ)\)
Make a "Corner" (\(90^\circ\))
\(\angle 1 + \angle 2 = 90^\circ\)
Supplementary \((180^\circ)\)
Make a "Straight Line"
\(\angle 1 + \angle 2 = 180^\circ\)
Vertical Angles \((\cong)\)
Across an "X" = Equal
\(\angle \text{Top} = \angle \text{Bottom}\)
Model 4: Intersecting Lines (Find Product \(x \cdot y\)) Test Q#8 & HW #1
Two lines intersect forming vertical angles and linear pairs: Top-left is \((y + 20)^\circ\), bottom-left is \((4y - 15)^\circ\), and bottom-right is \((x + 15)^\circ\). What is the product \(x \cdot y\)?
1. Solve for \(y\) (Linear Pair)
Adjacent angles on a line add to \(180^\circ\):
\(y + 20 + 4y - 15 = 180\)
\(5y + 5 = 180 \implies 5y = 175\)
\(\mathbf{y = 35}\)
2. Solve for \(x\) (Vertical)
Top-left = Bottom-right:
\((y + 20) = (x + 15)\)
\(35 + 20 = x + 15\)
\(55 = x + 15\)
\(\mathbf{x = 40}\)
3. Final Question Check!
Question asks for the product \(x \cdot y\):
\(40 \times 35 = \mathbf{1400}\)
(Select Choice C)
Model 5: Right Triangle Acute Angles & Factoring Check Critical Concept
In a right triangle, the two acute angles must add up to \(90^\circ\).
Given: \(m\angle X = x^2\) and \(m\angle Y = 4x + 13\). Which value of \(x\) must we use?
Step 1: Set Sum = \(90^\circ\)
\(x^2 + (4x + 13) = 90\)
\(x^2 + 4x - 77 = 0\)
\((x + 11)(x - 7) = 0\)
Potential solutions: \(x = -11\) or \(x = 7\)
Step 2: TEST FOR REAL ANGLES!
If \(x = -11\): \(m\angle Y = 4(-11) + 13 = -31^\circ\) ✗ Impossible! (Negative angle)
If \(x = 7\): \(m\angle Y = 4(7) + 13 = 41^\circ\) & \(m\angle X = 7^2 = 49^\circ\) ✓ Valid!
We must choose \(\mathbf{x = 7}\) (Choice E).
Final Test-Taking Checklist for Tomorrow
Did I label \((x_1, y_1)\) and \((x_2, y_2)\) clearly?
Did I flip the fraction AND change the sign for \(m_\perp\)?
Did I plug \(x\) back in if asked for a length or angle?
Are all my calculated angle measures positive and realistic?
Unit 1 Geometry Exam Study Guide Page 4 of 4 • Angles & Checklist
