Math 1 Quiz: Verifying Properties With Coordinates

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Verifying Properties With CoordinatesQuestion 1 of 18

Triangle $PQR$ has vertices $P(-3, 1)$ , $Q(5, 7)$ , and $R(1, -5)$ . A student wants to verify that this triangle is isosceles by showing that two sides have equal length. After calculating $PQ = \sqrt{100} = 10$ and $PR = \sqrt{52} = 2\sqrt{13}$ , what should be the next step in the verification process?

Calculate

QR

and check if

QR = PQ = 10

or if

QR = PR = 2\sqrt{13}

to confirm equal side lengths

Calculate the slope of each side and verify that two sides have slopes that are negative reciprocals indicating perpendicularity

Calculate the midpoint of side

PQ

and verify that it lies on the perpendicular bisector of the triangle

Calculate the area using the coordinate formula and verify that it matches the area formula for isosceles triangles

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Math 1 Quiz

Math 1 Quiz: Verifying Properties With Coordinates

Practice Verifying Properties With Coordinates 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 Verifying Properties With Coordinates, 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

Calculate $QR$ and check if $QR = PQ = 10$ or if $QR = PR = 2\sqrt{13}$ to confirm equal side lengths (correct answer)
Calculate the slope of each side and verify that two sides have slopes that are negative reciprocals indicating perpendicularity
Calculate the midpoint of side $PQ$ and verify that it lies on the perpendicular bisector of the triangle
Calculate the area using the coordinate formula and verify that it matches the area formula for isosceles triangles

Explanation: To verify an isosceles triangle, you need to show that exactly two sides have equal length. The student has calculated two of the three side lengths, so the logical next step is to calculate the third side length QR and check if it equals either PQ or PR. Choice B describes checking for a right triangle, not isosceles. Choice C is unnecessarily complex and doesn't directly verify the isosceles property. Choice D is incorrect because there's no special area formula that would verify the isosceles property through coordinates.

Question 2

Points $U(1, 3)$ , $V(7, 1)$ , $W(5, -5)$ , and $X(-1, -3)$ form quadrilateral $UVWX$ . A student verifies that $UV \parallel WX$ and $UX \parallel VW$ by calculating equal slopes for opposite sides. The student concludes $UVWX$ is a parallelogram. To determine if it's actually a rectangle, what should be calculated next?

Calculate the lengths of both diagonals $UW$ and $VX$ and verify they are equal, which confirms rectangle properties
Calculate the slopes of both diagonals and verify their product is $-1$ , confirming perpendicular diagonals for rectangles
Calculate the area using the shoelace formula and verify it equals the product of two adjacent side lengths
Calculate the slopes of adjacent sides and verify their product is $-1$ , confirming that all angles are right angles (correct answer)

Explanation: When you need to determine if a parallelogram is actually a rectangle, you're looking for the defining property that distinguishes rectangles from general parallelograms: all angles must be 90°. Since the student has already confirmed this is a parallelogram (opposite sides are parallel), the next step is verifying that adjacent sides meet at right angles. You can do this by calculating the slopes of two adjacent sides and checking if their product equals

-1

. When two lines are perpendicular, their slopes are negative reciprocals, so multiplying them gives

-1

. If one pair of adjacent sides is perpendicular in a parallelogram, then all angles are 90° due to the parallel side relationships. Choice A is incorrect because equal diagonal lengths is a property of rectangles, but it's not the most direct way to verify the rectangle property from the given information. Choice B contains a fundamental error—perpendicular diagonals are actually a property of rhombuses, not rectangles. Rectangle diagonals are equal in length but not necessarily perpendicular. Choice C is wrong because while you could calculate area using the shoelace formula, comparing it to the product of adjacent side lengths doesn't directly verify the rectangle property. Remember: To prove a parallelogram is a rectangle, focus on showing that adjacent sides are perpendicular. The slope multiplication test (product =

-1

) is the most efficient approach when you have coordinates.

Question 3

Triangle $GHI$ has vertices $G(-1, -2)$ , $H(5, 1)$ , and $I(2, 4)$ . A student calculates that the triangle has side lengths $GH = 3\sqrt{5}$ , $HI = \sqrt{18} = 3\sqrt{2}$ , and $GI = 3\sqrt{5}$ . The student concludes this is an isosceles right triangle. Which part of this conclusion requires further verification?

The isosceles property is incorrect because $3\sqrt{5} \neq 3\sqrt{2}$ , so no two sides are actually equal in length
The right triangle property is incorrect because isosceles triangles cannot have right angles due to geometric constraints on angle measures
Both properties need verification because the side length calculations contain errors that invalidate the entire geometric analysis
The right triangle property needs verification by checking if the sides satisfy the Pythagorean theorem relationship $a^2 + b^2 = c^2$ (correct answer)

Explanation: When analyzing triangles using coordinates, you need to verify both the side lengths and any claimed geometric properties independently. The student has calculated the side lengths and noticed that two sides are equal (

GH = GI = 3\sqrt{5}

), which correctly identifies the isosceles property. However, claiming it's also a right triangle requires additional verification. To confirm a right triangle, you must check whether the sides satisfy the Pythagorean theorem:

a^2 + b^2 = c^2

, where

c

is the longest side. Here, the longest side is

3\sqrt{5}

(since

\sqrt{25} > \sqrt{18}

), so you'd need to verify:

(3\sqrt{2})^2 + (3\sqrt{5})^2 = (3\sqrt{5})^2

. This gives

18 + 45 = 45

, or

63 = 45

, which is false. So while the triangle is indeed isosceles, it's not a right triangle. Choice A is wrong because the isosceles property is actually correct—two sides do equal

3\sqrt{5}

. Choice B is incorrect because isosceles triangles can absolutely be right triangles; there's no geometric constraint preventing this. Choice C is wrong because the side length calculations appear to be correct based on the distance formula. Choice D correctly identifies that the right triangle property needs verification through the Pythagorean theorem, which is the standard method for confirming right angles in coordinate geometry. Study tip: Always verify geometric properties independently. Just because side lengths suggest one property doesn't automatically confirm another—use the appropriate theorem or test for each claim.

Question 4

Quadrilateral $JKLM$ has vertices $J(1, 3)$ , $K(7, 1)$ , $L(5, -5)$ , and $M(-1, -3)$ . A student claims this quadrilateral is a rectangle because all angles are right angles. To verify this claim using coordinate methods, which approach would be most efficient?

Calculate slopes of adjacent sides and verify that their products equal $-1$ for all four corner angles of the quadrilateral (correct answer)
Calculate the lengths of all sides and both diagonals, then verify that opposite sides are equal and diagonals are equal
Calculate the area using the shoelace formula and verify that it equals the product of two adjacent side lengths
Calculate the midpoints of all sides and verify that connecting them forms a rectangle with half the original dimensions

Explanation: To verify a rectangle, you need to confirm that all angles are right angles. The most direct coordinate method is to calculate slopes of adjacent sides and verify that their product is -1 (indicating perpendicularity) at each vertex. Choice B would work but is less direct since equal diagonals alone don't guarantee a rectangle (an isosceles trapezoid also has equal diagonals). Choice C only verifies area, not the rectangular property. Choice D describes a property that's true for any parallelogram, not specifically rectangles.

Question 5

A student is verifying that triangle $DEF$ with vertices $D(1, 2)$ , $E(7, 4)$ , and $F(3, 8)$ is scalene. After calculating $DE = 2\sqrt{10}$ and $EF = 2\sqrt{5}$ , the student concludes the triangle is scalene because $DE \neq EF$ . What is wrong with this reasoning?

The student made calculation errors; both $DE$ and $EF$ should equal $2\sqrt{10}$ , making the triangle isosceles rather than scalene
The student needs to calculate the third side $DF$ and verify that all three sides have different lengths before concluding scalene (correct answer)
The student should verify the triangle inequality holds for all three side combinations before determining the triangle type classification
The student confused scalene with isosceles; having two unequal sides actually indicates an isosceles triangle with the third side equal to one of these

Explanation: A scalene triangle has all three sides of different lengths. The student only calculated two sides and found them unequal, but this is insufficient. The third side DF could equal either DE or EF, making the triangle isosceles instead of scalene. All three side lengths must be calculated and shown to be mutually unequal. Choice A incorrectly suggests calculation errors. Choice C mentions triangle inequality, which verifies existence but not the scalene property. Choice D incorrectly defines scalene and isosceles triangles.

Question 6

Points $A(1, 1)$ , $B(4, 5)$ , $C(8, 2)$ , and $D(5, -2)$ form quadrilateral $ABCD$ . A student calculates the midpoint of diagonal $AC$ as $(4.5, 1.5)$ and the midpoint of diagonal $BD$ as $(4.5, 1.5)$ . The student concludes that $ABCD$ is a parallelogram. What additional verification would strengthen this conclusion?

Calculate the slopes of all four sides to verify that opposite sides are parallel, confirming the parallelogram property independently (correct answer)
Calculate the lengths of both diagonals to verify they are equal, which would upgrade the classification from parallelogram to rectangle
Calculate the area using both the shoelace formula and the cross product method to verify consistent results for the parallelogram
Calculate the slopes of both diagonals to verify they are perpendicular, which would upgrade the classification from parallelogram to rhombus

Explanation: The student correctly used the fact that if diagonals bisect each other, then the quadrilateral is a parallelogram. However, verifying that opposite sides are parallel (equal slopes) provides an independent confirmation of the parallelogram property, strengthening the conclusion. Choice B would test for rectangle properties but doesn't strengthen the parallelogram conclusion. Choice C provides computational verification but not geometric verification. Choice D would test for rhombus properties, not strengthen the basic parallelogram verification.

Question 7

Triangle $ABC$ has vertices $A(2, 5)$ , $B(8, 3)$ , and $C(4, -3)$ . A student wants to verify that the triangle is obtuse by showing that one angle is greater than $90°$ . Using coordinate methods, which approach would be most efficient?

Calculate all three side lengths, then use the Law of Cosines to find the largest angle and verify it exceeds $90°$
Calculate the slopes of all three sides, then use the angle formula with slopes to determine which angle is obtuse
Calculate all three side lengths, identify the longest side, then use the converse of the Pythagorean theorem to test for obtuseness (correct answer)
Calculate the centroid and circumcenter, then verify that the circumcenter lies outside the triangle indicating an obtuse triangle

Explanation: The most efficient coordinate method is to calculate the three side lengths, identify the longest side (opposite the largest angle), then check if c² > a² + b² where c is the longest side. If this inequality holds, the triangle is obtuse. Choice A requires more complex trigonometric calculations. Choice B using slope angle formulas is complicated for obtuse angle verification. Choice D requires advanced concepts and more complex calculations than necessary.

Question 8

Triangle $ABC$ has vertices $A(2, 1)$ , $B(8, 3)$ , and $C(4, 7)$ . To verify that this triangle is a right triangle, a student calculates the slopes of all three sides: $m_{AB} = \frac{1}{3}$ , $m_{BC} = -2$ , and $m_{AC} = 3$ . What conclusion should the student draw?

The triangle is a right triangle because slopes $m_{AB}$ and $m_{BC}$ are negative reciprocals, indicating perpendicular sides $AB$ and $BC$
The triangle is a right triangle because slopes $m_{BC}$ and $m_{AC}$ multiply to give $-6$ , which indicates perpendicular sides $BC$ and $AC$
The triangle is not a right triangle because none of the slope pairs multiply to exactly $-1$ as required for perpendicular sides (correct answer)
The triangle is a right triangle because slopes $m_{AB}$ and $m_{AC}$ are both positive while $m_{BC}$ is negative, indicating one obtuse angle

Explanation: For two lines to be perpendicular, their slopes must be negative reciprocals, meaning their product equals -1. Checking the products: (1/3)(-2) = -2/3, (-2)(3) = -6, and (1/3)(3) = 1. None of these products equal -1, so no sides are perpendicular, and the triangle is not a right triangle. Choice A incorrectly identifies 1/3 and -2 as negative reciprocals (they would need to multiply to -1). Choice B incorrectly states that a product of -6 indicates perpendicularity. Choice D confuses the sign of slopes with angle measurement.

Question 9

Quadrilateral $EFGH$ has vertices $E(-3, 2)$ , $F(1, 6)$ , $G(5, 2)$ , and $H(1, -2)$ . A student calculates that the diagonals $EG$ and $FH$ are perpendicular and bisect each other. However, when the student calculates the side lengths, they are not all equal. What type of quadrilateral is $EFGH$ ?

A rectangle, because perpendicular diagonals that bisect each other create right angles at all vertices
A rhombus, because perpendicular bisecting diagonals define rhombuses regardless of side lengths
A parallelogram, because diagonals that bisect each other guarantee opposite sides are parallel and equal (correct answer)
A kite, because perpendicular diagonals indicate two pairs of adjacent equal sides

Explanation: When diagonals bisect each other, the quadrilateral is a parallelogram. The perpendicular diagonals would make it a rhombus IF all sides were equal, but since the sides are unequal, it's a parallelogram with perpendicular diagonals. Choice A is incorrect because perpendicular diagonals don't guarantee right angles at vertices. Choice B requires equal sides for a rhombus. Choice D is incorrect because kites don't necessarily have diagonals that bisect each other.

Question 10

Points $W(-2, 5)$ , $X(4, 3)$ , $Y(2, -3)$ , and $Z(-4, -1)$ form quadrilateral $WXYZ$ . A student wants to verify whether this quadrilateral is a rhombus. Which sequence of calculations would provide the most direct verification?

Calculate all four side lengths and verify they are equal, then calculate diagonal lengths and verify they are perpendicular bisectors
Calculate all four side lengths and verify they are equal, then calculate the slopes of both diagonals and verify their product is $-1$ (correct answer)
Calculate the slopes of opposite sides and verify they are equal, then calculate all side lengths and verify they are equal
Calculate the midpoints of both diagonals and verify they coincide, then calculate the area using two different methods and verify consistency

Explanation: A rhombus is defined as a parallelogram with all sides equal. The most direct verification is to: (1) confirm all four sides have equal length, and (2) verify the diagonals are perpendicular (slopes multiply to -1), which is a characteristic property of rhombuses. Choice A mentions perpendicular bisectors but doesn't specify how to verify this efficiently. Choice C checks for parallelogram properties but doesn't directly verify the perpendicular diagonal property. Choice D is unnecessarily complex and doesn't directly test the defining properties of a rhombus.

Question 11

Quadrilateral $PQRS$ has vertices $P(0, 3)$ , $Q(4, 6)$ , $R(7, 2)$ , and $S(3, -1)$ . To determine if $PQRS$ is a square, a student first verifies that all sides are equal in length and that all angles are right angles. However, the verification is still incomplete. What additional property must be checked?

Verify that the diagonals are equal in length and bisect each other at right angles to confirm the square properties completely
Verify that the vertices are listed in the correct order around the perimeter to ensure the quadrilateral doesn't self-intersect (correct answer)
Verify that the area calculated by the shoelace formula equals the square of one side length to confirm geometric consistency
Verify that the center point is equidistant from all vertices and that this distance equals half the diagonal length

Explanation: If all sides are equal and all angles are right angles, the figure is geometrically a square, but we must verify that the vertices are connected in order around the perimeter (P→Q→R→S→P) rather than forming a self-intersecting quadrilateral. Choice A describes properties that would be automatically satisfied if the figure is truly a square with vertices in order. Choice C provides a consistency check but doesn't address the fundamental issue of vertex ordering. Choice D also describes properties that follow automatically from the square definition.

Question 12

Triangle $DEF$ has vertices $D(-1, 2)$ , $E(3, 5)$ , and $F(0, -1)$ . A student claims the triangle is isosceles and calculates $DE = 5$ , $EF = \sqrt{45}$ , and $DF = \sqrt{18}$ . The student then states that since $\sqrt{45} = 3\sqrt{5}$ and $\sqrt{18} = 3\sqrt{2}$ , the triangle is isosceles because both expressions contain the factor 3. What error did the student make?

The student correctly simplified the radicals but failed to recognize that $3\sqrt{5} \neq 3\sqrt{2}$ , so no sides are equal (correct answer)
The student incorrectly calculated the distance $DE$ ; it should be $\sqrt{25} = 5$ , not just $5$
The student should have compared $\sqrt{45}$ and $2\sqrt{18} = 6\sqrt{2}$ to find the equal sides
The student incorrectly simplified $\sqrt{45}$ ; it should be $9\sqrt{5}$ , not $3\sqrt{5}$

Explanation: When you're determining if a triangle is isosceles, you need to check if any two sides have equal lengths. This requires carefully calculating distances and properly comparing the results. Let's verify the student's distance calculations first. Using the distance formula

d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

$DE = \sqrt{(3-(-1))^2 + (5-2)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ ✓
$EF = \sqrt{(0-3)^2 + (-1-5)^2} = \sqrt{9 + 36} = \sqrt{45}$ ✓
$DF = \sqrt{(0-(-1))^2 + (-1-2)^2} = \sqrt{1 + 9} = \sqrt{18}$ ✓

The student's calculations are correct. Now let's check their radical simplifications:

$\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$ ✓
$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ ✓

These are also correct. The critical error is in the student's reasoning: they concluded the triangle is isosceles because both

3\sqrt{5}

and

3\sqrt{2}

contain the factor 3. However, having a common factor doesn't make expressions equal. Since

\sqrt{5} \neq \sqrt{2}

, we have

3\sqrt{5} \neq 3\sqrt{2}

, so no two sides are equal. Answer A correctly identifies this logical error. Answer B incorrectly suggests the distance calculation was wrong. Answer C proposes an irrelevant comparison with

2\sqrt{18}

. Answer D incorrectly states that

\sqrt{45} = 9\sqrt{5}

, which would give

\sqrt{45} = 9\sqrt{5} = \sqrt{405}

, clearly wrong. Remember: when simplifying radicals, focus on whether the final values are actually equal, not just whether they share common factors.

Question 13

To prove that quadrilateral $WXYZ$ with vertices $W(-2, 1)$ , $X(3, 3)$ , $Y(5, -2)$ , and $Z(0, -4)$ is a rectangle, a student must verify multiple conditions. Which statement correctly identifies what must be proven?

All four sides must be equal in length, and all angles must be right angles
Opposite sides must be parallel and equal, and at least one angle must be a right angle
Both diagonals must be equal in length, and they must bisect each other at right angles
Opposite sides must be parallel and equal, and the diagonals must be equal in length (correct answer)

Explanation: A rectangle is defined as a parallelogram with four right angles. To prove a quadrilateral is a rectangle using coordinates, we can show it's first a parallelogram (opposite sides parallel and equal) and then show the diagonals are equal in length (which implies right angles). Option A describes a square, not just a rectangle. Option B is insufficient because one right angle doesn't guarantee all angles are right angles. Option C describes properties that would make it a rhombus with perpendicular diagonals, but equal diagonal lengths specifically characterize rectangles among parallelograms.

Question 14

Points $J(2, 3)$ , $K(8, 7)$ , $L(6, 13)$ , and $M(0, 9)$ form quadrilateral $JKLM$ . A student calculates that $JK = 2\sqrt{13}$ , $KL = 2\sqrt{10}$ , $LM = 2\sqrt{13}$ , and $MJ = 2\sqrt{10}$ . Additionally, the diagonals $JL = 2\sqrt{26}$ and $KM = 4\sqrt{13}$ . What is the most specific classification for this quadrilateral?

Rectangle, because opposite sides are equal and the quadrilateral has four right angles
Rhombus, because all four sides are equal in length
Parallelogram, because opposite sides are equal but it's neither a rectangle nor rhombus (correct answer)
Kite, because two pairs of adjacent sides are equal but opposite sides are not equal

Explanation: The quadrilateral has opposite sides equal (JK = LM = 2√13 and KL = MJ = 2√10), which makes it a parallelogram. However, not all sides are equal (so it's not a rhombus), and the diagonals are not equal (JL = 2√26 ≠ KM = 4√13, so it's not a rectangle). Option A is wrong because unequal diagonals mean it's not a rectangle. Option B is wrong because the sides have two different lengths. Option D is wrong because opposite sides are equal, not adjacent sides.

Question 15

Triangle $PQR$ has vertices $P(-3, 2)$ , $Q(5, 6)$ , and $R(1, -4)$ . A student claims this triangle is isosceles. After calculating the distances $PQ = \sqrt{80}$ , $PR = \sqrt{52}$ , and $QR = \sqrt{116}$ , what can be concluded about the triangle?

The triangle is isosceles because $\sqrt{80} = 4\sqrt{5}$ and $\sqrt{116} = 2\sqrt{29}$ share a common factor
The triangle is isosceles because $80 + 52 = 132$ , which is close to $116$
The triangle is not isosceles because all three side lengths are different when simplified (correct answer)
The triangle is isosceles because $\sqrt{52} = 2\sqrt{13}$ is exactly half of $\sqrt{208} = 4\sqrt{13}$

Explanation: For a triangle to be isosceles, at least two sides must have equal length. Simplifying: PQ = √80 = 4√5 ≈ 8.94, PR = √52 = 2√13 ≈ 7.21, and QR = √116 = 2√29 ≈ 10.77. Since all three distances are different, the triangle is not isosceles. Option A incorrectly suggests that sharing factors makes distances equal. Option B confuses the Pythagorean theorem with the definition of isosceles triangles. Option D creates a false relationship that doesn't exist between the calculated distances.

Question 16

Triangle $RST$ has vertices $R(0, 0)$ , $S(6, 8)$ , and $T(14, 2)$ . A student calculates $RS = 10$ , $ST = 10$ , and $RT = 2\sqrt{53}$ . The student claims this is an isosceles triangle and attempts to verify that the altitude from vertex $R$ to side $ST$ bisects $ST$ . What error is the student making?

The student should verify that the altitude from vertex $S$ or $T$ bisects the opposite side, not from vertex $R$
The student should calculate the perpendicular bisector of side $RS$ or $RT$ , not side $ST$ , since these are the equal sides
The student should verify that the altitude from the vertex angle $S$ to the base $RT$ bisects the base, not from $R$ to $ST$ (correct answer)
The student should verify that the median from vertex $R$ to side $ST$ is perpendicular to $ST$ , rather than checking if an altitude bisects the side

Explanation: In an isosceles triangle, the altitude from the vertex angle (where the two equal sides meet) to the base bisects the base. Since RS = ST = 10, vertex S is the vertex angle where the equal sides meet, and RT is the base. The student should verify that the altitude from S to RT bisects RT. Choice A incorrectly suggests using S or T arbitrarily. Choice B incorrectly identifies which sides are equal. Choice D confuses the property - the median from the vertex angle equals the altitude in an isosceles triangle.

Question 17

Triangle $ABC$ has vertices $A(0, 4)$ , $B(6, 0)$ , and $C(0, 0)$ . A student wants to verify this is a right triangle by showing that two sides are perpendicular. After calculating the slopes $AB = -\frac{2}{3}$ , $BC = 0$ , and $AC =$ undefined, what conclusion should be drawn?

The triangle is a right triangle because side $AC$ is vertical and side $BC$ is horizontal, making them perpendicular (correct answer)
The triangle is a right triangle because the product of slopes $AB$ and $BC$ equals $-1$
The triangle is not a right triangle because none of the slope products equal $-1$
The calculation is inconclusive because one slope is undefined, making the perpendicular test impossible

Explanation: When you're testing if a triangle is a right triangle using slopes, you're looking for two sides that are perpendicular. Two lines are perpendicular when their slopes multiply to

-1

, but there's a special case you must remember: vertical and horizontal lines are always perpendicular to each other. Let's examine what the slopes tell us. Side

AC

has an undefined slope, which means it's a vertical line (it goes straight up and down from

(0,0)

(0,4)

). Side

BC

has a slope of

0

, making it horizontal (it goes straight across from

(0,0)

(6,0)

). Vertical and horizontal lines are perpendicular by definition, so we have a right angle at vertex

C

. Looking at the wrong answers: Choice B incorrectly applies the slope product rule to

AB

and

BC

. While

(-\frac{2}{3}) \times 0 = 0 \neq -1

, this doesn't matter because we already found perpendicular sides elsewhere. Choice C makes the error of only checking the slope product rule and ignoring the vertical-horizontal relationship. Choice D incorrectly assumes that having an undefined slope makes the perpendicular test impossible, when actually undefined slopes (vertical lines) paired with zero slopes (horizontal lines) guarantee perpendicularity. Study tip: When checking for perpendicular lines, remember two methods: (1) slopes multiply to

-1

, and (2) one line is vertical (undefined slope) and the other is horizontal (slope = 0). The second case is actually easier to spot than calculating products!

Question 18

A quadrilateral has vertices $E(1, 1)$ , $F(5, 2)$ , $G(4, 6)$ , and $H(0, 5)$ . To determine if $EFGH$ is a parallelogram, a student decides to use the midpoint method. What should the student calculate and what result would confirm the parallelogram property?

Calculate midpoints of all four sides; if opposite midpoints are equidistant from the center, it's a parallelogram
Calculate midpoints of both diagonals; if they are the same point, then the diagonals bisect each other and it's a parallelogram (correct answer)
Calculate midpoints of opposite sides; if the midpoint of $EF$ equals the midpoint of $GH$ , it's a parallelogram
Calculate the midpoint of the quadrilateral; if all vertices are equidistant from this point, it's a parallelogram

Explanation: The diagonal midpoint method tests whether the diagonals bisect each other, which is a defining property of parallelograms. The midpoint of diagonal EG is ((1+4)/2, (1+6)/2) = (2.5, 3.5), and the midpoint of diagonal FH is ((5+0)/2, (2+5)/2) = (2.5, 3.5). Since these are equal, the diagonals bisect each other and EFGH is a parallelogram. Option A describes an incorrect method. Option C tests something unrelated to parallelogram properties. Option D describes properties of a circle or regular polygon, not a parallelogram.