Math 1 Quiz: Coordinate Proofs

18 questions · exam conditions

0:00

Coordinate ProofsQuestion 1 of 18

In triangle $LMN$ with vertices $L(0, 4)$ , $M(6, 0)$ , and $N(-2, -2)$ , a coordinate proof attempts to show the triangle is acute by verifying that all three angles are less than $90°$ . Which geometric insight explains why checking perpendicular relationships between sides is insufficient for this proof?

Perpendicular side checks only examine adjacent angle relationships, missing the comprehensive angle analysis needed for acute triangle classification

Perpendicular relationships between sides would indicate right or obtuse triangles, contradicting the acute triangle hypothesis being tested

The absence of perpendicular side relationships is necessary but not sufficient to prove all angles are acute in the triangle

Perpendicular side relationships only identify right angles, but acute triangle verification requires confirming all angles are strictly less than

90°

← Back to quizzes

Math 1 Quiz

Math 1 Quiz: Coordinate Proofs

Practice Coordinate Proofs in Math 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Coordinate Proofs, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Perpendicular side checks only examine adjacent angle relationships, missing the comprehensive angle analysis needed for acute triangle classification
Perpendicular relationships between sides would indicate right or obtuse triangles, contradicting the acute triangle hypothesis being tested
The absence of perpendicular side relationships is necessary but not sufficient to prove all angles are acute in the triangle
Perpendicular side relationships only identify right angles, but acute triangle verification requires confirming all angles are strictly less than $90°$ (correct answer)

Explanation: When proving a triangle is acute, you need to verify that all three angles measure strictly less than

90°

. The key insight here is understanding what perpendicular relationships between sides can and cannot tell you about angle measures. Checking for perpendicular sides helps identify right angles (exactly

90°

), but this approach has a critical limitation. If two sides are perpendicular, you've found a

90°

angle, which means the triangle is right, not acute. If no sides are perpendicular, you've only confirmed there are no right angles—but this doesn't guarantee all angles are acute. Some angles could still be obtuse (greater than

90°

). Choice D correctly identifies this limitation: perpendicular checks only detect right angles, but proving a triangle is acute requires the stronger condition that all angles are strictly less than

90°

. Choice A incorrectly suggests the issue is about "adjacent angle relationships"—perpendicular checks examine the relationship between any two sides, not just adjacent angles. Choice B is backwards; finding perpendicular sides would indeed contradict the acute hypothesis, but the question asks why the absence of perpendicular relationships is insufficient. Choice C uses confusing language about "necessary but not sufficient"—while technically the absence of perpendicular sides is necessary for an acute triangle, this phrasing doesn't clearly explain the core geometric limitation. For coordinate geometry proofs involving triangle classification, remember that ruling out right angles (through perpendicular checks) is just the first step. You must also rule out obtuse angles using methods like the dot product or distance-based angle calculations.

Question 2

A coordinate proof shows that quadrilateral $PQRS$ with vertices $P(1, 2)$ , $Q(4, 6)$ , $R(7, 2)$ , and $S(4, -2)$ is a rhombus by verifying all sides are equal. However, the proof fails to address why the quadrilateral is not a square. What additional calculation would distinguish between these two classifications?

Calculate the slopes of adjacent sides to verify that not all angles are $90°$ , since squares require four right angles (correct answer)
Calculate the diagonal lengths to verify they are not equal, since squares require congruent diagonals unlike rhombuses
Calculate the area using both the diagonal method and side-length method to verify the results differ for non-square rhombuses
Calculate the perpendicular distances between opposite sides to verify the figure lacks the uniform width property of squares

Explanation: To distinguish a rhombus from a square, we need to verify that not all angles are 90°. Calculate slopes: slope of PQ = (6-2)/(4-1) = 4/3, slope of QR = (2-6)/(7-4) = -4/3. Since (4/3)(-4/3) = -16/9 ≠ -1, adjacent sides are not perpendicular, so not all angles are 90°. Therefore, it's a rhombus but not a square. Option B is incorrect because rhombuses don't require unequal diagonals. Option C is unnecessarily complex. Option D doesn't directly address the square vs. rhombus distinction.

Question 3

In a coordinate proof that quadrilateral $EFGH$ is a trapezoid, a student shows that $EF \parallel GH$ by calculating equal slopes. What additional condition must be verified to ensure the quadrilateral is a trapezoid and not a parallelogram?

Verify that sides $EH$ and $FG$ are not parallel by showing their slopes are not equal (correct answer)
Verify that the parallel sides $EF$ and $GH$ are not equal in length by calculating their distances
Verify that angles $E$ and $H$ are not equal by comparing slopes of adjacent sides at these vertices
Verify that the non-parallel sides $EH$ and $FG$ are not equal in length to avoid creating an isosceles trapezoid

Explanation: A trapezoid is defined as a quadrilateral with exactly one pair of parallel sides. Since the student proved EF ∥ GH, they must now verify that EH and FG are NOT parallel to ensure there's only one pair of parallel sides. If EH ∥ FG as well, then EFGH would be a parallelogram, not a trapezoid. This is verified by calculating slopes of EH and FG and confirming they're not equal. Option B is irrelevant to the trapezoid definition. Options C and D address properties that don't distinguish trapezoids from parallelograms.

Question 4

In coordinate plane, quadrilateral $ABCD$ has vertices $A(2, 1)$ , $B(6, 3)$ , $C(8, 7)$ , and $D(4, 5)$ . To prove that $ABCD$ is a parallelogram using coordinate geometry, which combination of calculations provides the most efficient proof?

Calculate all four side lengths and verify opposite sides are equal in length
Calculate slopes of all four sides and verify opposite sides are parallel
Calculate diagonals' lengths and verify they bisect each other at the same point
Calculate midpoints of both diagonals and verify they are the same point (correct answer)

Explanation: The most efficient proof is to show that the diagonals bisect each other by calculating their midpoints. Midpoint of diagonal AC: ((2+8)/2, (1+7)/2) = (5, 4). Midpoint of diagonal BD: ((6+4)/2, (3+5)/2) = (5, 4). Since the midpoints are identical, the diagonals bisect each other, proving ABCD is a parallelogram. This requires only two midpoint calculations. Option A requires four distance calculations. Option B requires four slope calculations. Option C requires calculating diagonal lengths AND midpoints, making it less efficient than D.

Question 5

Points $A(3, 1)$ , $B(7, 4)$ , $C(4, 8)$ , and $D(0, 5)$ form a quadrilateral. A student claims this is a parallelogram because slopes of $AB$ and $CD$ are both $\frac{3}{4}$ , and slopes of $BC$ and $AD$ are both $-\frac{4}{3}$ . What makes this reasoning incomplete for a rigorous coordinate proof?

The student should verify that opposite sides are equal in length, not just parallel, to confirm the parallelogram property
The student should verify that the quadrilateral is convex by checking that vertices are ordered properly around the perimeter (correct answer)
The student should verify that adjacent sides are not perpendicular to ensure the figure is a parallelogram rather than a rectangle
The student should verify that the diagonals bisect each other by calculating midpoints to complete the parallelogram proof

Explanation: The student's slope calculations prove that opposite sides are parallel, which is sufficient for the parallelogram property IF the quadrilateral is convex and vertices are ordered correctly. However, the student didn't verify that points A, B, C, D are actually consecutive vertices of a convex quadrilateral. They could form a self-intersecting quadrilateral or be ordered incorrectly. Option A is wrong because parallel opposite sides are sufficient for parallelograms. Option C confuses parallelogram vs. rectangle distinction. Option D suggests an alternative proof method but isn't necessary when slope method is used correctly.

Question 6

Triangle $ABC$ has vertices $A(-2, 3)$ , $B(4, 1)$ , and $C(2, 7)$ . In proving that this triangle is isosceles using coordinate geometry, which approach provides the most complete mathematical justification?

Calculate all three side lengths and verify that exactly two distances are equal, confirming two equal sides (correct answer)
Calculate two side lengths and verify they are equal, then conclude the triangle is isosceles without checking the third
Calculate the slopes of all three sides and verify that exactly one perpendicular relationship exists between sides
Calculate one side length and compare it to coordinate differences to verify the triangle has symmetrical properties

Explanation: An isosceles triangle is defined as having exactly two equal sides, so the most complete justification requires calculating all three side lengths and verifying that exactly two are equal. |AB| = √[(4-(-2))² + (1-3)²] = √[36+4] = √40 = 2√10. |BC| = √[(2-4)² + (7-1)²] = √[4+36] = √40 = 2√10. |AC| = √[(2-(-2))² + (7-3)²] = √[16+16] = √32 = 4√2. Since |AB| = |BC| ≠ |AC|, the triangle is isosceles with two equal sides. Option A provides complete verification. Option B might be sufficient but doesn't confirm the third side is different. Options C and D don't directly address the isosceles property.

Question 7

Triangle $DEF$ has vertices $D(-4, 1)$ , $E(2, 5)$ , and $F(6, -1)$ . To prove that the triangle is isosceles using coordinates, which approach requires the least computational work while providing complete justification?

Calculate all three side lengths using the distance formula, then identify which two sides are equal
Calculate slopes of all three sides, then use perpendicular bisector properties to identify equal sides
Calculate two side lengths strategically chosen based on visual symmetry, then verify the third if needed (correct answer)
Calculate the perpendicular distance from each vertex to the opposite side to identify the altitude relationships

Explanation: Strategic calculation is most efficient. By examining the coordinates, we can identify potential equal sides and calculate only those distances first. |DE| = √[(2-(-4))² + (5-1)²] = √[36+16] = √52. |EF| = √[(6-2)² + (-1-5)²] = √[16+36] = √52. Since |DE| = |EF|, the triangle is isosceles, requiring only two calculations. Option A requires all three calculations unnecessarily. Option B is overly complex for this problem. Option D involves perpendicular distance formulas that are more computational work than needed.

Question 8

In a coordinate proof that triangle $ABC$ with vertices $A(0, 0)$ , $B(6, 0)$ , and $C(3, 3\sqrt{3})$ is equilateral, a student calculated: $|AB| = 6$ , $|BC| = \sqrt{(6-3)^2 + (0-3\sqrt{3})^2} = \sqrt{9 + 27} = 6$ , and $|AC| = \sqrt{9 + 27} = 6$ . What additional geometric insight makes this coordinate proof more complete?

The coordinates show that vertex $C$ is positioned at the apex of an equilateral triangle with $AB$ as base
The height $3\sqrt{3}$ equals $\frac{\sqrt{3}}{2}$ times the base length, confirming the equilateral triangle altitude formula (correct answer)
The coordinates demonstrate that angle $ACB$ has measure $60°$ because $C$ lies on the perpendicular bisector of $AB$
The coordinates verify that the centroid, circumcenter, and orthocenter coincide at point $(3, \sqrt{3})$

Explanation: The geometric insight is that the height 3√3 confirms the equilateral triangle relationship. For an equilateral triangle with side length 6, the altitude is (√3/2) × 6 = 3√3. This validates that point C is correctly positioned to create an equilateral triangle, not just any triangle with three equal sides calculated arbitrarily. Option A is vague and doesn't add mathematical insight. Option C is incorrect about the angle measure reasoning. Option D has the wrong coordinates for the triangle centers and doesn't relate to the proof strategy.

Question 9

Points $M(a, b)$ , $N(c, d)$ , $O(e, f)$ , and $P(g, h)$ form a rectangle in the coordinate plane. A student proves this by showing: (1) $MN \perp NO$ , (2) $NO \perp OP$ , (3) $OP \perp PM$ , and (4) $|MN| = |OP|$ and $|NO| = |PM|$ . Which step in this proof strategy is redundant?

Step (3) is redundant because if three consecutive angles are right angles, the fourth must also be a right angle (correct answer)
Step (4) is redundant because if all four angles are right angles, opposite sides are automatically equal in a quadrilateral
Step (2) is redundant because perpendicular relationships are transitive, so $MN \perp NO$ and $NO \perp OP$ imply $MN \parallel OP$
Step (1) is redundant because the rectangle property can be established through the parallel side and equal side conditions alone

Explanation: In any quadrilateral, if three consecutive angles are right angles, the fourth angle must also be right angle since the sum of interior angles is 360°. Therefore, proving OP ⊥ PM in step (3) is redundant once we've established MN ⊥ NO and NO ⊥ OP. Step (4) is still necessary because a quadrilateral with four right angles could be a rectangle or a square, and we need opposite sides equal to confirm it's a rectangle. The other options misstate geometric principles or proof requirements.

Question 10

Points $P(2, 1)$ , $Q(6, 3)$ , and $R(4, 7)$ form a triangle. A student uses the coordinate method to prove the triangle is obtuse by showing that $PQ^2 + QR^2 < PR^2$ . Which error in geometric reasoning led to this incorrect approach?

The student applied the Pythagorean theorem incorrectly by comparing the wrong combination of sides for the obtuse angle test
The student confused the obtuse triangle test with the acute triangle test, using the wrong inequality direction for the comparison
The student assumed that the largest side is opposite the obtuse angle without first identifying which side is actually longest (correct answer)
The student used squared distances instead of actual distances, which invalidates the Pythagorean theorem relationships for obtuse triangles

Explanation: In an obtuse triangle, the obtuse angle is opposite the longest side. To test if triangle PQR is obtuse, one must first identify the longest side, then check if the square of that longest side is greater than the sum of squares of the other two sides. The student assumed PR was the longest side without verification. Calculate: |PQ|² = (6-2)² + (3-1)² = 20, |QR|² = (4-6)² + (7-3)² = 20, |PR|² = (4-2)² + (7-1)² = 40. Since |PR| is longest, we test: 40 > 20 + 20? Since 40 = 40, the triangle is actually right, not obtuse. The student's fundamental error was not identifying which side was longest before applying the obtuse triangle test.

Question 11

In a coordinate proof, Marcus wants to show that the diagonals of rectangle $PQRS$ are congruent. He calculates the diagonal lengths as $|PR| = \sqrt{(x_3-x_1)^2 + (y_3-y_1)^2}$ and $|QS| = \sqrt{(x_4-x_2)^2 + (y_4-y_2)^2}$ and finds they are equal. What does this algebraic result confirm about the geometric properties of rectangles?

It confirms that rectangles have diagonals that bisect each other at right angles, making them special parallelograms
It confirms that rectangles have congruent diagonals, which distinguishes them from general parallelograms (correct answer)
It confirms that rectangles have perpendicular diagonals, which is necessary for the rectangle classification
It confirms that rectangles have diagonals of equal length that also serve as lines of symmetry

Explanation: Marcus's calculation showing

|PR| = |QS|

proves the diagonals are congruent (equal in length). This is a key property that distinguishes rectangles from general parallelograms, which don't necessarily have congruent diagonals. Choice A incorrectly states diagonals are perpendicular (that's true for rhombuses, not rectangles). Choice C also incorrectly claims perpendicular diagonals. Choice D adds the incorrect claim about lines of symmetry, which isn't what the equal lengths establish.

Question 12

To prove that quadrilateral $ABCD$ with vertices $A(0, 0)$ , $B(3, 4)$ , $C(8, 4)$ , and $D(5, 0)$ is a trapezoid, Carmen calculates the slopes of all four sides and finds $m_{AB} = \frac{4}{3}$ , $m_{BC} = 0$ , $m_{CD} = -\frac{4}{3}$ , and $m_{DA} = 0$ . What do these slope calculations reveal about the quadrilateral's classification?

Sides $AB$ and $CD$ are parallel since $m_{AB} = -m_{CD}$ , making $ABCD$ a trapezoid
No sides are parallel since all slopes are different, so $ABCD$ is not a trapezoid
Sides $BC$ and $DA$ are parallel since both have slope 0, making $ABCD$ a trapezoid with one pair of parallel sides (correct answer)
Both pairs of opposite sides are parallel since $m_{BC} = m_{DA} = 0$ and $|m_{AB}| = |m_{CD}|$ , making $ABCD$ a parallelogram

Explanation: When classifying quadrilaterals using coordinate geometry, you need to examine slope relationships to determine which sides, if any, are parallel. Two lines are parallel when they have identical slopes, and a trapezoid is defined as a quadrilateral with exactly one pair of parallel sides. Looking at Carmen's slope calculations, you can see that sides

BC

and

DA

both have slope 0, meaning they're both horizontal lines and therefore parallel to each other. Since

AB

has slope

\frac{4}{3}

and

CD

has slope

-\frac{4}{3}

, these sides are not parallel (parallel lines must have identical slopes, not opposite slopes). This gives us exactly one pair of parallel sides, which confirms

ABCD

is a trapezoid. Choice A incorrectly assumes that slopes of

\frac{4}{3}

and

-\frac{4}{3}

indicate parallel lines. These are actually negative reciprocals, which would indicate perpendicular lines if we were looking at adjacent sides. Choice B misses that

BC

and

DA

have the same slope (0). Choice D incorrectly concludes the quadrilateral is a parallelogram—while

BC

and

DA

are parallel,

AB

and

CD

are not parallel since

\frac{4}{3} \neq -\frac{4}{3}

. Remember: parallel lines have identical slopes, not just equal absolute values. When classifying quadrilaterals, check each pair of opposite sides separately to determine how many pairs are parallel—this tells you whether you have a trapezoid (one pair) or parallelogram (both pairs).

Question 13

To prove that quadrilateral $ABCD$ with vertices $A(2, 1)$ , $B(6, 3)$ , $C(4, 7)$ , and $D(0, 5)$ is a rectangle, Maria calculates the slopes of all four sides and finds that opposite sides are parallel. What additional coordinate calculation must she perform to complete her proof?

Show that the diagonals are congruent by calculating $|AC| = |BD|$
Show that adjacent sides are perpendicular by verifying that the product of their slopes equals $-1$ (correct answer)
Show that the diagonals bisect each other by finding their common midpoint
Show that all four sides are congruent by calculating $|AB| = |BC| = |CD| = |DA|$

Explanation: To prove a quadrilateral is a rectangle using coordinates, one must show it's a parallelogram (opposite sides parallel, which Maria did) AND that it has right angles. The most efficient way is to show adjacent sides are perpendicular using slopes: if slopes multiply to -1, the sides are perpendicular. Choice A would prove it's a rectangle but is less direct. Choice C proves it's a parallelogram but doesn't add new information. Choice D would prove it's a rhombus, not necessarily a rectangle.

Question 14

In proving that triangle $ABC$ with vertices $A(-2, 3)$ , $B(4, 1)$ , and $C(2, -5)$ is a right triangle, Kai calculates the slopes: $m_{AB} = -\frac{1}{3}$ and $m_{BC} = 3$ . He concludes the triangle is a right triangle because $m_{AB} \cdot m_{BC} = -1$ . What geometric relationship has Kai established, and where is the right angle located?

Sides $AB$ and $BC$ are perpendicular, so the right angle is at vertex $B$ where these sides meet (correct answer)
Sides $AB$ and $BC$ are parallel, so the right angle is formed by the altitude from $A$ to side $BC$
The triangle has a right angle, but its location requires checking the third side $AC$ as well
Sides $AB$ and $BC$ are perpendicular, creating two right angles at vertices $A$ and $C$ respectively

Explanation: When the product of two slopes equals -1, the lines are perpendicular. Since sides

AB

and

BC

are perpendicular and they share vertex

B

, the right angle is at

B

. Choice B incorrectly states the sides are parallel. Choice C is wrong because finding one right angle is sufficient to prove a triangle is a right triangle. Choice D is impossible since a triangle can have only one right angle.

Question 15

In a coordinate proof involving triangle $STU$ , David calculates the midpoints of all three sides: $M_{ST} = (2, 3)$ , $M_{SU} = (4, 1)$ , and $M_{TU} = (6, 3)$ . He then finds the lengths $|M_{ST}M_{SU}| = 2\sqrt{2}$ , $|M_{SU}M_{TU}| = 2\sqrt{2}$ , and $|M_{ST}M_{TU}| = 4$ . What geometric theorem is David likely investigating with these calculations?

The Triangle Inequality Theorem, by comparing the sum of two sides to the third side
The Triangle Midpoint Theorem, by proving that the three midpoints are collinear
The Pythagorean Theorem, by checking if the triangle formed by the midpoints is a right triangle
The Midsegment Theorem, by showing that segments connecting midpoints are parallel to and half the length of the triangle's sides (correct answer)

Explanation: When you encounter a coordinate geometry problem involving midpoints of triangle sides and distance calculations, you're likely dealing with properties of the medial triangle - the triangle formed by connecting the three midpoints. Looking at David's calculations, he found that the triangle formed by the three midpoints has sides of length

2\sqrt{2}

2\sqrt{2}

, and

4

. The Midsegment Theorem states that when you connect two midpoints of a triangle's sides, the resulting segment is parallel to the third side and exactly half its length. David is investigating this relationship by analyzing the medial triangle's properties. Choice A is incorrect because the Triangle Inequality Theorem simply states that the sum of any two sides must be greater than the third side - David isn't comparing sums to individual sides. Choice B misses the mark because the Triangle Midpoint Theorem involves proving collinearity, but David calculated distances between separate points, not whether they lie on the same line. Choice C might seem tempting since

(\sqrt{8})^2 + (\sqrt{8})^2 = 16 = 4^2

, making this a right triangle, but that's just a byproduct of his investigation, not the main theorem being explored. Choice D correctly identifies that David is exploring the Midsegment Theorem. His distance calculations of the medial triangle's sides will help him determine the original triangle's side lengths (which should be

4\sqrt{2}

4\sqrt{2}

, and

8

) and verify the parallel relationships. Study tip: When you see midpoint calculations in coordinate geometry, immediately think midsegments and the 1:2 ratio relationship with the original triangle's sides.

Question 16

In a coordinate proof, Aiden shows that quadrilateral $JKLM$ is a parallelogram by proving $\overrightarrow{JK} = \overrightarrow{ML}$ and $\overrightarrow{KL} = \overrightarrow{JM}$ using vector methods. Specifically, he calculates $\overrightarrow{JK} = (4, 2)$ and $\overrightarrow{ML} = (4, 2)$ , then $\overrightarrow{KL} = (-1, 3)$ and $\overrightarrow{JM} = (-1, 3)$ . What fundamental geometric property do these equal vectors establish?

The diagonals bisect each other, which is a necessary condition for parallelograms
Adjacent sides are perpendicular, making the quadrilateral a rectangle rather than just a parallelogram
Opposite sides are parallel and congruent, which is the definition of a parallelogram (correct answer)
All four sides are congruent, making the quadrilateral a rhombus rather than just a parallelogram

Explanation: When you encounter vector proofs involving quadrilaterals, focus on what the equal vectors actually represent geometrically. Vectors capture both direction and magnitude, so when two vectors are equal, they represent segments that are both parallel and congruent. In this problem, Aiden found that

\overrightarrow{JK} = \overrightarrow{ML} = (4, 2)

and

\overrightarrow{KL} = \overrightarrow{JM} = (-1, 3)

. Since these vectors are equal, sides

JK

and

ML

are parallel and congruent, as are sides

KL

and

JM

. Notice that

JK

and

ML

are opposite sides of the quadrilateral, and so are

KL

and

JM

. This directly establishes that opposite sides are parallel and congruent, which is precisely the definition of a parallelogram. Answer C is correct. Answer A is incorrect because diagonal bisection is a different property entirely—you'd need to examine the midpoints of the diagonals, not the side vectors. Answer B misinterprets the relationship between the vectors. To check for perpendicularity, you'd calculate the dot product:

\overrightarrow{JK} \cdot \overrightarrow{KL} = (4)(-1) + (2)(3) = -4 + 6 = 2 \neq 0

, so the sides aren't perpendicular. Answer D is wrong because the vectors have different magnitudes:

|\overrightarrow{JK}| = \sqrt{16 + 4} = 2\sqrt{5}

while

|\overrightarrow{KL}| = \sqrt{1 + 9} = \sqrt{10}

, so the sides aren't all congruent. Remember: equal vectors between opposite sides of a quadrilateral immediately prove you have a parallelogram—no additional calculations needed.

Question 17

To prove that point $M(3, 4)$ is the midpoint of segment $\overline{PQ}$ where $P(1, 2)$ and $Q(5, 6)$ , Sophia uses the midpoint formula: $M = \left(\frac{1+5}{2}, \frac{2+6}{2}\right) = (3, 4)$ . Her classmate argues this doesn't constitute a complete proof. What additional verification should Sophia include to strengthen her coordinate proof?

Calculate $|PM| + |MQ|$ and verify it equals $|PQ|$ to confirm $M$ lies on segment $\overline{PQ}$
Calculate $|PM|$ and $|MQ|$ separately and verify they are equal to confirm $M$ bisects the segment
Calculate the slope of $\overline{PM}$ and $\overline{MQ}$ to verify they are equal, confirming collinearity
No additional verification is needed; the midpoint formula application is sufficient proof that $M$ is the midpoint (correct answer)

Explanation: The midpoint formula directly gives the coordinates of the point that is equidistant from both endpoints and lies on the segment connecting them. When Sophia applies the formula correctly and gets the coordinates of point

M

, this constitutes a complete proof. The other choices suggest redundant verifications: if

M

is the midpoint by the formula, then

|PM| = |MQ|

and

|PM| + |MQ| = |PQ|

automatically follow. Choice C suggests checking collinearity, which is also automatically satisfied by the midpoint formula.

Question 18

In a coordinate proof showing that triangle $PQR$ is isosceles, Jake calculates $|PQ| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = 5$ and $|PR| = \sqrt{(x_3-x_1)^2 + (y_3-y_1)^2} = 5$ . His teacher asks him to explain the geometric significance of this algebraic result. Which response best connects the algebra to the geometry?

The equal distances prove that points $Q$ and $R$ are equidistant from point $P$ , making $P$ the vertex angle of the isosceles triangle (correct answer)
The equal distances prove that the triangle has two congruent base angles at vertices $Q$ and $R$ , satisfying the isosceles triangle theorem
The equal distances prove that sides $PQ$ and $PR$ are parallel, which is a necessary condition for an isosceles triangle
The equal distances prove that the perpendicular bisector of side $QR$ passes through vertex $P$ , confirming the triangle's symmetry

Explanation: Jake's calculation shows

|PQ| = |PR|

, meaning two sides from vertex

P

are congruent. This makes

P

the vertex where the two equal sides meet (the vertex angle), and

Q

and

R

are equidistant from

P

. Choice B describes a consequence but not the direct geometric meaning of equal side lengths. Choice C is incorrect (parallel sides don't make triangles isosceles). Choice D describes a property that follows but isn't the direct geometric interpretation of the distance calculation.