HSG.GPE.B.4: Proving Geometric Theorems with Coordinates
In plain English: HSG.GPE.B.4 is the Common Core geometry standard that asks students to use coordinates to prove simple geometric theorems algebraically. Students use the distance, midpoint and slope formulas to prove or disprove claims, such as whether four points form a rectangle or whether a point lies on a given circle. It is usually taught in Geometry.
Use coordinates to prove simple geometric theorems algebraically. For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point (1, √3) lies on the circle centered at the origin and containing the point (0, 2).
Common Core State Standards for Mathematics · Domain: Expressing Geometric Properties with Equations (GPE) · Cluster: Use coordinates to prove simple geometric theorems algebraically Also written as HSG-GPE.B.4 or G-GPE.4 · Official standard
Students learn to prove and disprove geometric claims with algebra on the coordinate plane. They match each property to a formula: slopes for parallel and perpendicular sides, the distance formula for congruent segments and for points on a circle, and the midpoint formula for segments that bisect each other. The lesson works through both official examples of the standard: deciding whether four given points form a rectangle, and deciding whether (1, √3) lies on the circle centered at the origin through (0, 2).
Students also move from proofs about one figure with numbers to general proofs with letters for the coordinates, such as proving that the diagonals of every parallelogram bisect each other. Throughout, they write each proof as a claim, a definition, the computations, and a conclusion, and they learn that one failed condition, clearly named, disproves a claim.
Learning Objectives
By the end of this lesson, students will be able to:
Choose the slope, distance or midpoint formula that proves a given geometric property
Prove or disprove that four given points form a parallelogram, rectangle, rhombus or square
Prove or disprove that a given point lies on a circle defined by its center and one point on it
Write a general coordinate proof using letters for the coordinates of a figure placed conveniently on the axes
Write a clear conclusion that links each computation to a definition or theorem
Prior Knowledge Required
Students should already be comfortable with:
Finding the distance between two points with the Pythagorean Theorem 8.G.B.8
Computing the slope of a line through two points 8.EE.B.6
The slope criteria for parallel and perpendicular lines HSG.GPE.B.5
Definitions of parallelogram, rectangle, rhombus, square and circle HSG.CO.A.1
Post three points on a coordinate grid, A(1, 2), B(4, 6) and C(-3, 5), and ask students to connect them into a triangle.
Warm-Up Prompt
"Find the lengths AB and AC and the slopes of AB and AC. What kind of triangle is ABC? How sure are you, and what would you need to show someone who does not trust your sketch?"
Collect answers. AB = √(3² + 4²) = 5 and AC = √(4² + 3²) = 5, so the triangle is isosceles. The slopes are 4/3 and -3/4, whose product is -1, so angle A is a right angle. Many students guess "right triangle" from the sketch; ask what would convince a skeptic. The numbers, not the picture, are the proof. If time allows, ask for the midpoint of BC, (1/2, 11/2), and save it for later: midpoints are the third tool in the lesson.
Direct Instruction20 minutes
The coordinate proof toolkit. A coordinate proof turns a geometric claim into a calculation. Each geometric property matches one formula:
Geometric property and the calculation that proves it
To show that...
Compute
Condition
two segments are parallel
slope of each
equal slopes (or both vertical)
two segments are perpendicular
slope of each
product of slopes is -1 (or one horizontal, one vertical)
two segments are congruent
distance formula
equal lengths
two segments bisect each other
midpoint formula
same midpoint
a point lies on a circle
distance from the center
equals the radius
Teach a four-step write-up for every proof:
State the claim in words, for example "ABCD is a rectangle."
Choose the definition or theorem you will use: a rectangle is a parallelogram with a right angle, or a parallelogram with congruent diagonals.
Compute every slope, length or midpoint you need, and show the substitution.
Conclude with a sentence that links the numbers to the definition. To disprove a claim, one failed condition is enough, but name it.
Work through the examples below. Examples 1 and 2 are the two official examples of the standard. Use Diagram 1 with Example 1 and Diagram 2 with Example 2. For Example 4, explain why the vertices use letters: a proof with numbers shows the claim for one figure only, while letters cover every parallelogram. Placing one vertex at the origin and one side on the x-axis keeps the algebra short without losing generality.
Prove a rectangle (official example)
Prove or disprove that A(1, 1), B(5, 3), C(4, 5), D(0, 3) form a rectangle.
Equation: Slopes: AB = 1/2, BC = -2, CD = 1/2, DA = -2. Opposite sides are parallel and (1/2)(-2) = -1, so ABCD is a rectangle. Check: AC = BD = 5.
Point on a circle (official example)
Prove or disprove that (1, √3) lies on the circle centered at the origin that contains (0, 2).
Equation: Radius = 2. Distance from O to (1, √3) = √(1 + 3) = 2, so the point lies on the circle.
Disprove a rectangle
Prove or disprove that J(0, 0), K(6, 1), L(7, 5), M(1, 4) form a rectangle.
Equation: Slopes 1/6, 4, 1/6, 4: a parallelogram, but (1/6)(4) = 2/3, not -1. Diagonals √74 and √34. Not a rectangle.
General proof with letters
Prove that the diagonals of every parallelogram bisect each other, using O(0, 0), A(a, 0), B(a + b, c), C(b, c).
Equation: Midpoint of OB = ((a + b)/2, c/2) = midpoint of AC, so the diagonals bisect each other.
Classify a triangle
Prove that (-3, 1), (1, 4), (4, 0) form an isosceles right triangle.
Equation: Two sides of length 5, third side √50; slopes 3/4 and -4/3 multiply to -1, so the right angle is at (1, 4).
After Example 3, stress that a disproof names the specific condition that fails. "It does not look like a rectangle" is not a disproof; "the product of the slopes of JK and KL is 2/3, not -1, so angle K is not a right angle" is.
Guided Practice15-20 minutes
Pairs work through three proofs, one at a time. After each, one pair presents its write-up and the class checks that the conclusion names the definition used.
Quadrilateral: Prove that W(0, 0), X(5, 0), Y(8, 4), Z(3, 4) is a rhombus but not a square. (All four sides have length 5; WX is horizontal and XY has slope 4/3, so angle X is not a right angle. The diagonals have slopes 1/2 and -2, so they are perpendicular, as in every rhombus.)
Circle: Prove or disprove that (-3, 4) lies on the circle centered at (1, 1) that passes through (5, 4). (Radius = 5, and the distance from (1, 1) to (-3, 4) is also 5, so it lies on the circle.)
General proof: For the triangle with vertices (0, 0), (2a, 0), (2b, 2c), prove that the segment joining the midpoints of the two non-horizontal sides is parallel to the base and half as long. (The midpoints are (b, c) and (a + b, c): the segment is horizontal, like the base, and has length a, half of 2a.)
Circulate and listen for these errors: subtracting coordinates in a different order in the numerator and denominator of a slope, dropping the square root in the distance formula, and concluding "rectangle" after checking only side lengths.
Independent Practice15 minutes
Students write complete proofs for three claims on their own:
P(-2, -1), Q(2, 1), R(0, 5) form a right triangle. (Slopes of PQ and QR are 1/2 and -2, so angle Q is right. PQ = QR = √20, so it is also isosceles.)
(4, -2) lies on the circle centered at (-1, 1) that passes through (3, 4). (Radius 5, but the distance to (4, -2) is √34, so the claim is false.)
E(1, 0), F(4, 1), G(3, 4), H(0, 3) form a square. (All sides √10, adjacent slopes 1/3 and -3, so it is a square.)
Students who finish early prove that the diagonals of EFGH are congruent and perpendicular.
Closure5-10 minutes
Exit ticket: (1) Prove or disprove that (2, 2√3) lies on the circle centered at the origin that contains (-4, 0). (Radius 4, and 2² + (2√3)² = 4 + 12 = 16, so the distance is 4: it lies on the circle.) (2) Name the formula you would use to prove that two diagonals bisect each other, and say what you would need to find. (The midpoint formula: both diagonals must have the same midpoint.) (3) In one sentence, explain why four equal sides do not prove a square.
Differentiation Strategies
For Struggling Students
Give a one-page toolkit card with the table from Direct Instruction and the three formulas written out with a worked substitution for each
Have students record each slope as "rise over run" with the subtraction written out, such as (5 - 3)/(4 - 5), before simplifying
Start with figures that have at least one horizontal or vertical side, then move to figures with no side on a grid line
For Advanced Students
Prove that the midpoints of the sides of any quadrilateral form a parallelogram, using vertices (2a, 2b), (2c, 2d), (2e, 2f), (2g, 2h)
Find all points with integer coordinates on the circle centered at the origin with radius 5, and prove that there are exactly 12
Given three vertices of a rectangle, find the fourth, and prove the result in two different ways (slopes, then diagonals)
Assessment Guidance
What to Look For
Check that every conclusion names the property it proves and the definition it relies on, for example "opposite sides have equal slopes, so ABCD is a parallelogram." A correct list of numbers with no conclusion is not yet a proof. Watch for students who prove "rectangle" from four side lengths alone, who treat a sketch as evidence, or who prove a general theorem with one numerical example. For disproofs, the student should name the specific condition that fails.
02
Classroom Activities
3 Activities
1
Prove It or Disprove It Card Sort
20 minGroups of 3-4
Each group receives 8 claim cards. For each card, the group decides whether the claim is true, writes a short coordinate proof or a disproof on the back, and sorts the cards into two piles: "Proved" and "Disproved." Every card is decided by computation, not by the look of a sketch.
The 8 Claim Cards (with answers for the teacher)
Card 1: (2, 0), (6, 2), (4, 6), (0, 4) form a square. (Proved: all sides √20, adjacent slopes 1/2 and -2.)
Card 2: (-1, -1), (5, 1), (4, 4), (-2, 2) form a rectangle. (Proved: slopes 1/3, -3, 1/3, -3; it is not a square, since the sides are √40 and √10.)
Card 3: (0, 0), (5, 0), (7, 3), (2, 3) form a rectangle. (Disproved: a parallelogram with slopes 0 and 3/2, so no right angle.)
Card 4: (1, 1), (4, 5), (7, 1), (4, -3) form a square. (Disproved: all sides 5, but adjacent slopes 4/3 and -4/3 multiply to -16/9, so it is a rhombus only.)
Card 5: (0, 0), (8, 0), (6, 3), (2, 3) form a parallelogram. (Disproved: the horizontal sides have lengths 8 and 4, and the other two sides have slopes -3/2 and 3/2. It is an isosceles trapezoid.)
Card 6: (-2, 3), (1, -1), (5, 2) form a right triangle. (Proved: from (1, -1), the sides have slopes -4/3 and 3/4.)
Card 7: (-1, -5) lies on the circle centered at (2, -1) through (5, 3). (Proved: both distances from the center are 5.)
Card 8: (4, 2) lies on the circle centered at the origin through (3, 4). (Disproved: 4² + 2² = 20, not 25, so the point is inside the circle.)
Procedure
Groups split the cards so each student proves two, then trade and check each other's work
A card moves to a pile only when the whole group agrees with its proof or disproof
Each group presents one disproof and names the exact condition that fails
Discussion Questions
Which cards needed more than one kind of calculation? Why?
For Card 4, which single calculation is enough to disprove "square"?
Is there a card you could decide with the diagonals instead of the sides?
2
General Proof Stations
20 minGroups of 3
Three stations each hold one theorem and a blank coordinate grid. Groups choose general coordinates with letters, prove the theorem for every figure of that type, and leave their proof on chart paper for the next group to check.
Stations (with suggested coordinates)
Station A: In an isosceles triangle, the median to the base is perpendicular to the base. Use A(-a, 0), B(a, 0), C(0, b). (CA = CB = √(a² + b²), the midpoint of AB is the origin, so the median lies on the y-axis and is perpendicular to the x-axis.)
Station B: The diagonals of a rectangle are congruent. Use (0, 0), (a, 0), (a, b), (0, b). (Both diagonals have length √(a² + b²).)
Station C: Opposite sides of a parallelogram are congruent. Use O(0, 0), A(a, 0), B(a + b, c), C(b, c). (OA = CB = a and OC = AB = √(b² + c²).)
Procedure
Spend 6 minutes at each station: 4 minutes to write the proof, 2 minutes to check the previous group's proof and add a sticky note
Before writing, the group must explain why its coordinates describe every figure of that type, not only one
Challenge Variation
Groups redo Station C with the parallelogram placed anywhere, with vertices (p, q), (p + a, q), (p + a + b, q + c), (p + b, q + c), and compare how much longer the algebra is.
3
Proof Critique Gallery
15 minPairs
Four flawed student proofs are posted around the room. Pairs find the flaw in each, decide whether the claim is actually true, and write a corrected proof or disproof.
The Four Flawed Proofs
Proof 1: "A(0, 0), B(4, 0), C(5, 3), D(1, 3) is a rectangle because AB = CD = 4 and BC = DA = √10." (Flaw: equal opposite sides prove only a parallelogram. The diagonals are √34 and √18, so it is not a rectangle.)
Proof 2: "The point (3, 3) is on the circle centered at the origin through (0, 4), because it is on the circle in my sketch." (Flaw: a sketch is not a proof. 3² + 3² = 18, not 16, so the point is outside the circle.)
Proof 3: "The triangle with vertices (0, 0), (3, 2), (5, 5) has a right angle at (3, 2) because the slopes 2/3 and 3/2 are reciprocals." (Flaw: perpendicular slopes are negative reciprocals. The product is 1, so there is no right angle there.)
Proof 4: "The diagonals of every square are perpendicular. Proof: for (0, 0), (2, 0), (2, 2), (0, 2), the diagonals have slopes 1 and -1." (Flaw: one example is not a general proof. Fix: use (0, 0), (a, 0), (a, a), (0, a); the diagonals have slopes 1 and -1 for every a > 0.)
Procedure
Pairs spend 3 minutes at each proof and write the flaw and the fix on a sticky note
Close with a whole-class list: "What a coordinate proof must include"
Modification for Distance Learning
Post the four proofs on a shared slide deck, one per slide. Pairs work in breakout rooms and type the flaw and the fix in the speaker notes.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Proving That Four Points Form a Rectangle
The official rectangle example with A(1, 1), B(5, 3), C(4, 5), D(0, 3). Equal slopes on opposite sides prove a parallelogram, and slopes whose product is -1 prove a right angle, so ABCD is a rectangle. The equal diagonals AC = BD = 5 give a second check. The grid is drawn to scale.
Diagram 2: Proving That a Point Lies on a Circle
The official circle example. The circle centered at the origin through (0, 2) has radius 2. The dashed right triangle with legs 1 and √3 shows that the distance from the origin to (1, √3) is √(1 + 3) = 2, so the point lies on the circle. The grid is drawn to scale.
04
Homework Assignment
~30 min
HSG.GPE.B.4 Homework: Coordinate Proofs
Directions: Write a complete proof or disproof for each problem: state the claim, name the definition or theorem you use, show every slope, distance or midpoint calculation, and finish with a concluding sentence. Sketch each figure on graph paper, but remember that the sketch is not the proof.
Part 1: Quadrilaterals and Triangles (Problems 1-3)
Prove or disprove that K(-3, 1), L(1, -2), M(4, 2), N(0, 5) form a rectangle. If it is a rectangle, decide whether it is also a square and prove your answer.
Prove or disprove that P(0, 1), Q(7, 2), R(8, 6), S(1, 5) form a rectangle. Then name the most specific type of quadrilateral that PQRS is, and prove it.
A right triangle has vertices O(0, 0), P(2a, 0) and Q(0, 2b), with a > 0 and b > 0. Prove that the midpoint of the hypotenuse PQ is the same distance from all three vertices.
Part 2: Points on Circles (Problems 4-5)
A circle is centered at the origin and contains the point (0, -4). Prove or disprove that the point (-√7, 3) lies on the circle.
A circle has center (2, 3) and passes through (-4, 11). Prove or disprove that each of the points (10, 9) and (-5, -3) lies on the circle. For any point that is not on the circle, say whether it is inside or outside.
Part 3: Two Claims about One Triangle (Problem 6)
Triangle ABC has vertices A(-4, -1), B(2, -3) and C(0, 3). Claim 1: the triangle is isosceles. Claim 2: the triangle is a right triangle. Prove or disprove each claim.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Choice of Tool
Uses the formula that matches each property (slope, distance or midpoint)
Right tools, but one property checked with the wrong formula
Tools missing or unrelated to the claim
Computation
All slopes, lengths and midpoints correct, with substitutions shown
One or two arithmetic errors
Many errors or no work shown
Logic and Conclusion
Conclusion names the definition used; disproofs name the failed condition
Correct verdict, but the link to a definition is missing
No conclusion, or a verdict based on the sketch
General Proof (Problem 3)
Letters used correctly; the proof covers every right triangle
Correct idea with an algebra error
Only a numerical example given
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
To prove with coordinates that quadrilateral ABCD is a parallelogram, which of these is enough?
Answer: A
A parallelogram is a quadrilateral with both pairs of opposite sides parallel, and parallel segments have equal slopes. Choice B proves only that angle B is a right angle. Choice C describes a property of rectangles, and a quadrilateral with equal diagonals need not be a parallelogram.
Question 2 of 20 · Multiple Choice
In quadrilateral PQRS, side PQ has slope 3/5 and side QR has slope -5/3. What can you conclude about angle Q?
Answer: C
(3/5)(-5/3) = -1, so PQ is perpendicular to QR and angle Q is a right angle. Choice A confuses the perpendicular condition with the parallel one, which needs equal slopes. Choice D is wrong because negative reciprocal slopes always have opposite signs.
Question 3 of 20 · Multiple Choice
What is the distance between (-2, 4) and (4, -4)?
Answer: B
The distance is √((4 - (-2))² + (-4 - 4)²) = √(36 + 64) = √100 = 10. Choice A comes from computing 4 - 2 instead of 4 - (-2). Choice C adds the legs 6 + 8 instead of using the Pythagorean Theorem. Choice D forgets the square root.
Question 4 of 20 · Multiple Choice
A circle is centered at the origin and passes through (0, 5). Which point lies on the circle?
Answer: B
The radius is 5, so a point is on the circle when x² + y² = 25. For (√7, 3√2): 7 + 18 = 25. Choice A gives 13 and choice C gives 20 (both inside). Choice D gives 50: its coordinates are each 5, but its distance from the origin is 5√2.
Question 5 of 20 · Multiple Choice
Quadrilateral ABCD has vertices A(-2, 1), B(2, -1), C(5, 5) and D(1, 7). Which description is proved by the coordinates?
Answer: D
The slopes of AB, BC, CD and DA are -1/2, 2, -1/2 and 2. Opposite sides are parallel and (-1/2)(2) = -1, so ABCD is a rectangle. AB = √20 and BC = √45, so the sides are not all equal and it is not a square. Choice C misses the right angles; choice A assumes equal sides without checking them.
Question 6 of 20 · Multiple Choice
Which computation proves that the diagonals of a quadrilateral bisect each other?
Answer: A
Two segments bisect each other when they cut each other in half, which means they share a midpoint. Choice B tests congruent diagonals, and choice C tests perpendicular diagonals. Choice D cannot happen for diagonals that cross, since equal slopes would make them parallel or the same line.
Question 7 of 20 · Multiple Choice
A circle is centered at the origin and contains the point (0, 2). Which point does NOT lie on the circle?
Answer: D
The radius is 2, so a point is on the circle when x² + y² = 4. For (1, √2): 1 + 2 = 3, so its distance from the origin is √3, not 2. The other three points give 1 + 3 = 4, 2 + 2 = 4 and 3 + 1 = 4. A common error is to square only the whole numbers and not the radicals.
Question 8 of 20 · Multiple Choice
Which statement about the triangle with vertices (0, 0), (4, 0) and (2, 5) is proved by its coordinates?
Answer: B
The sides are 4, √(4 + 25) = √29 and √29, so exactly two sides are equal: isosceles. It is not right: 4² + (√29)² = 45, not 29, so the Pythagorean relation fails for every choice of hypotenuse. Choice A assumes all sides equal, and choice D misses the two equal sides.
Question 9 of 20 · Multiple Choice
A student wants to prove that, in every rectangle, the two segments joining the midpoints of opposite sides are perpendicular. Which vertices should she use?
Answer: C
Choice C describes every rectangle with one vertex at the origin and sides on the axes, for any a > 0 and b > 0, and the factors of 2 keep the midpoints free of fractions. Choice A proves the claim for one rectangle only. Choice B describes only squares. Choice D is not a rectangle unless b = a. With C, the midpoints are (a, 0) and (a, 2b), a vertical segment, and (0, b) and (2a, b), a horizontal segment.
Question 10 of 20 · Multiple Choice
Parallelogram OABC has vertices O(0, 0), A(6, 0), B(8, 4) and C(2, 4). At which point do its diagonals intersect?
Answer: A
The diagonals of a parallelogram bisect each other, so they meet at their common midpoint. The midpoint of OB is ((0 + 8)/2, (0 + 4)/2) = (4, 2), and the midpoint of AC is ((6 + 2)/2, (0 + 4)/2) = (4, 2). Choice B forgets to halve the y-coordinates. Choice C averages the x-coordinates of O and A instead of the endpoints of a diagonal.
Question 11 of 20 · Multiple Choice
Which fact, shown with coordinates, proves that a quadrilateral is a rectangle?
Answer: C
A parallelogram with congruent diagonals is a rectangle. Choice A proves only a parallelogram. Choice B is not enough for a general quadrilateral: a kite or a trapezoid can have one right angle. Choice D describes rhombuses and kites, not rectangles.
Question 12 of 20 · Multiple Choice
A circle has center (-2, 1) and passes through (1, 5). Which point lies on the circle?
Answer: D
The radius is √(3² + 4²) = 5. The distance from (-2, 1) to (3, 1) is 5, so it lies on the circle. The other distances are 3, √17 and 4, so those points are inside. Choice A is a tempting guess because it shares a coordinate with each given point, but its distance is only 3.
Question 13 of 20 · Multiple Choice
A student shows that all four sides of quadrilateral WXYZ have length 5 and writes "WXYZ is a square." What is wrong with the proof?
Answer: B
Every rhombus has four equal sides, and a rhombus is a square only if it has a right angle. Checking that two adjacent sides have slopes with product -1 (or that the diagonals are congruent) completes the proof. Choice C is impossible for diagonals that cross.
Question 14 of 20 · Multiple Choice
Triangle ABC has vertices A(0, 0), B(8, 0) and C(2, 6). What is the length of the segment joining the midpoints of AC and BC?
Answer: A
The midpoints are (1, 3) and (5, 3), so the segment is horizontal with length 5 - 1 = 4, half of AB = 8, as the midsegment theorem predicts. Choice B is the length of AB itself. Choice C is the y-coordinate of the midpoints, not a length.
Question 15 of 20 · Short Answer
Prove or disprove that R(-2, -2), S(4, 1), T(3, 3) and U(-3, 0) form a rectangle.
Slopes: RS = 3/6 = 1/2, ST = 2/(-1) = -2, TU = (-3)/(-6) = 1/2, UR = (-2)/1 = -2. Opposite sides have equal slopes, so RSTU is a parallelogram, and (1/2)(-2) = -1, so its angles are right angles. RSTU is a rectangle. Check: the diagonals RT and SU both have length √50.
Question 16 of 20 · Short Answer
A circle is centered at the origin and contains the point (3, 0). Prove or disprove that (-2, √5) lies on the circle.
The radius is 3. The distance from the origin to (-2, √5) is √(4 + 5) = √9 = 3, which equals the radius. The point lies on the circle.
Question 17 of 20 · Short Answer
For any a > 0 and b > 0, prove that the triangle with vertices O(0, 0), P(2a, 0) and Q(a, b) is isosceles.
OQ = √(a² + b²) and PQ = √((a - 2a)² + b²) = √(a² + b²). OQ = PQ for every a and b, so the triangle is isosceles. Using letters makes this a proof for every such triangle, not one example.
Question 18 of 20 · Short Answer
A circle has center (-1, 2) and passes through (2, 6). Prove or disprove that (3, -2) lies on the circle. If it does not, is it inside or outside?
The radius is √(3² + 4²) = 5. The distance from (-1, 2) to (3, -2) is √(16 + 16) = √32 ≈ 5.66. The point is not on the circle; it is outside, because √32 > 5.
Question 19 of 20 · Short Answer
Prove or disprove that the triangle with vertices (-1, 0), (5, 3) and (1, 11) is a right triangle.
The slope from (-1, 0) to (5, 3) is 3/6 = 1/2, and the slope from (5, 3) to (1, 11) is 8/(-4) = -2. Since (1/2)(-2) = -1, the two sides are perpendicular. The triangle is a right triangle with the right angle at (5, 3).
Question 20 of 20 · Short Answer
Prove that A(-3, -2), B(3, 0), C(5, 4) and D(-1, 2) form a parallelogram by showing that its diagonals bisect each other.
Midpoint of AC = ((-3 + 5)/2, (-2 + 4)/2) = (1, 1). Midpoint of BD = ((3 - 1)/2, (0 + 2)/2) = (1, 1). The diagonals share the midpoint (1, 1), so they bisect each other and ABCD is a parallelogram.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.GPE.B.4 mean?
It means students use coordinates and algebra to prove or disprove simple geometric claims. Instead of measuring a sketch, students compute slopes, distances and midpoints and use them to show, for example, that four points form a rectangle or that a point lies on a circle. The standard's own examples are exactly these two tasks.
Is HSG.GPE.B.4 taught in Geometry or Algebra?
HSG.GPE.B.4 is usually taught in Geometry. It uses algebra skills from Algebra I (slope, simplifying square roots, working with variables), which is why it sits in the domain "Expressing Geometric Properties with Equations." In integrated courses it often appears in Math II.
What formulas do students need for coordinate proofs?
Three formulas cover almost every proof at this level:
Slope (y₂ - y₁)/(x₂ - x₁), for parallel (equal slopes) and perpendicular (product -1) segments
Distance √((x₂ - x₁)² + (y₂ - y₁)²), for congruent segments and points on a circle
Midpoint ((x₁ + x₂)/2, (y₁ + y₂)/2), for segments that bisect each other
How do you prove that four points form a rectangle?
Show that the quadrilateral is a parallelogram and that it has a right angle. With slopes: opposite sides have equal slopes, and two adjacent sides have slopes whose product is -1. Another route is to show that the diagonals bisect each other (same midpoint) and are congruent (same length). Be careful with the order of the vertices: A, B, C, D must go around the figure.
How do you prove that a point lies on a circle?
Find the radius, then show that the point is exactly that far from the center. In the official example, the circle centered at the origin through (0, 2) has radius 2, and the distance from the origin to (1, √3) is √(1 + 3) = 2, so the point lies on the circle. If the distance is smaller than the radius the point is inside, and if it is larger the point is outside.
What does "prove or disprove" mean in HSG.GPE.B.4?
It means students must decide whether the claim is true and then justify their answer either way. A proof checks every condition in the definition. A disproof needs only one condition that fails, but the student must name it, for example "the slopes of JK and KL multiply to 2/3, not -1, so angle K is not a right angle."
Why do some coordinate proofs use letters like (a, 0) and (0, b) instead of numbers?
A proof with numbers shows that the claim is true for one figure. To prove a theorem about every parallelogram or every right triangle, the coordinates must describe all of them, so they use letters. Placing a vertex at the origin and a side along the x-axis does not lose any cases, because any figure can be moved there without changing its lengths or angles, and it makes the algebra much shorter.
What are common mistakes in coordinate proofs?
A common mistake is concluding "rectangle" or "square" from side lengths alone, since equal sides do not guarantee right angles. Others are:
subtracting the coordinates in a different order in the numerator and the denominator of a slope
forgetting the square root in the distance formula
calling slopes like 2/3 and 3/2 perpendicular (they must be negative reciprocals)
treating a sketch or one numerical example as a general proof
How is a coordinate proof different from a two-column proof?
A two-column proof (HSG.CO.C.9-11) reasons from definitions, postulates and theorems without coordinates. A coordinate proof places the figure on the coordinate plane and turns each property into a calculation. Both need a clear chain of reasoning and a conclusion. Many theorems, such as "the diagonals of a parallelogram bisect each other," can be proved both ways, and comparing the two proofs is a useful discussion.
How does HSG.GPE.B.4 connect to later math?
It prepares students for writing the equation of a circle from its center and radius (HSG.GPE.A.1), which is the distance formula written as an equation, and for computing perimeters and areas with coordinates (HSG.GPE.B.7). The habit of turning a geometric statement into an algebraic one is also the basis of analytic geometry in Precalculus.
07
Related Standards
5 standards
These standards connect to HSG.GPE.B.4: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.G.B.8Prerequisite
Apply the Pythagorean Theorem to find the distance between two points in a coordinate system
