ISEE Middle Level Quiz: Coordinate Shape Analysis
16 questions · exam conditions
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Coordinate Shape AnalysisQuestion 1 of 16

A rectangle has vertices at A(2, 1), B(6, 1), and C(6, 6). What is the length of the diagonal AC?

4 units
5 units
41\sqrt{41} units
52\sqrt{52} units
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ISEE Middle Level Quiz

ISEE Middle Level Quiz: Coordinate Shape Analysis

Practice Coordinate Shape Analysis in ISEE Middle Level 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 Shape Analysis, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Middle Level.

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

A rectangle has vertices at A(2, 1), B(6, 1), and C(6, 6). What is the length of the diagonal AC?

  1. 4 units
  2. 5 units
  3. 41\sqrt{41} units (correct answer)
  4. 52\sqrt{52} units
Explanation: The length of the diagonal AC can be found using the distance formula, (x2−x1)2+(y2−y1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}. For points A(2, 1) and C(6, 6), the length is (6−2)2+(6−1)2=42+52=16+25=41\sqrt{(6-2)^2 + (6-1)^2} = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}.

Question 2

A line segment AB on the coordinate plane has a midpoint M(3, 4). If point A is at (-1, 6), what are the coordinates of point B?

  1. (1, 5)
  2. (2, 2)
  3. (7, 2) (correct answer)
  4. (4, -2)
Explanation: Let the coordinates of B be (x,y)(x, y). The midpoint M is the average of the coordinates of A and B. So, 3=−1+x23 = \frac{-1+x}{2} and 4=6+y24 = \frac{6+y}{2}. Solving for x: 6=−1+x6 = -1+x, so x=7x=7. Solving for y: 8=6+y8 = 6+y, so y=2y=2. The coordinates of point B are (7, 2).

Question 3

A quadrilateral has diagonals that are congruent, perpendicular, and bisect each other. What is the most specific name for this shape?

  1. Parallelogram
  2. Rectangle
  3. Rhombus
  4. Square (correct answer)
Explanation: Let's analyze the properties. Diagonals that bisect each other define a parallelogram. If the diagonals are also congruent, the parallelogram is a rectangle. If the diagonals are also perpendicular, the parallelogram is a rhombus. A shape that is both a rectangle and a rhombus must be a square. Therefore, a quadrilateral with diagonals that are congruent, perpendicular, and bisect each other is a square.

Question 4

Point M(5, 1) is equidistant from point P(2, 5) and point Q(x, 5). Given that x is not equal to 2, what is the value of x?

  1. 3
  2. 5
  3. 8 (correct answer)
  4. 9
Explanation: Equidistant means the distance from M to P is equal to the distance from M to Q. We can compare the squares of the distances to avoid square roots. Distance MP squared is (5−2)2+(1−5)2=32+(−4)2=9+16=25(5-2)^2 + (1-5)^2 = 3^2 + (-4)^2 = 9 + 16 = 25. Distance MQ squared is (5−x)2+(1−5)2=(5−x)2+(−4)2=(5−x)2+16(5-x)^2 + (1-5)^2 = (5-x)^2 + (-4)^2 = (5-x)^2 + 16. Set the squared distances equal: 25=(5−x)2+1625 = (5-x)^2 + 16. Subtract 16 from both sides: 9=(5−x)29 = (5-x)^2. Take the square root of both sides: ±3=5−x\pm 3 = 5-x. This gives two possibilities: 3=5−x3 = 5-x or −3=5−x-3 = 5-x. The first equation gives x=2x=2, which is excluded by the problem. The second equation gives x=8x=8.

Question 5

A right triangle has vertices at (0, 0), (8, 0), and (0, 6). What is the perimeter of the triangle?

  1. 14 units
  2. 24 units (correct answer)
  3. 28 units
  4. 48 units
Explanation: The triangle has a horizontal side of length 8 (from (0,0) to (8,0)) and a vertical side of length 6 (from (0,0) to (0,6)). These are the legs of the right triangle. The third side is the hypotenuse. Its length can be found using the Pythagorean theorem: c=a2+b2=82+62=64+36=100=10c = \sqrt{a^2+b^2} = \sqrt{8^2+6^2} = \sqrt{64+36} = \sqrt{100} = 10. The perimeter is the sum of the side lengths: 8+6+10=248 + 6 + 10 = 24 units.

Question 6

What is the area of a triangle with vertices at the points (2, 2), (8, 2), and (4, 6)?

  1. 10 square units
  2. 12 square units (correct answer)
  3. 16 square units
  4. 24 square units
Explanation: The base of the triangle can be the horizontal segment connecting (2, 2) and (8, 2). The length of this base is 8−2=68 - 2 = 6 units. The height of the triangle is the perpendicular distance from the third vertex (4, 6) to the line containing the base (the line y=2). The height is 6−2=46 - 2 = 4 units. The area of a triangle is 12×base×height\frac{1}{2} \times \text{base} \times \text{height}. So, the area is 12×6×4=12\frac{1}{2} \times 6 \times 4 = 12 square units.

Question 7

The vertices of a square are located at (1,1), (1,5), (5,5), and (5,1). What is the sum of the slopes of the two diagonals?

  1. 0 (correct answer)
  2. 1
  3. 2
  4. The sum is undefined.
Explanation: The diagonals connect opposite vertices. The first diagonal connects (1,1) and (5,5). Its slope is 5−15−1=44=1\frac{5-1}{5-1} = \frac{4}{4} = 1. The second diagonal connects (1,5) and (5,1). Its slope is 1−55−1=−44=−1\frac{1-5}{5-1} = \frac{-4}{4} = -1. The sum of the slopes is 1+(−1)=01 + (-1) = 0. This is a property of any rhombus, including a square, whose diagonals are not horizontal and vertical.

Question 8

A square has vertices at (1,1), (4,1), (4,4), and (1,4). If the square is translated 3 units to the left and 5 units up, what are the coordinates of the new center of the square?

  1. (-0.5, 7.5) (correct answer)
  2. (5.5, -2.5)
  3. (-1.5, 6.5)
  4. (2.5, 2.5)
Explanation: First, find the center of the original square. The center is the midpoint of a diagonal. Using the diagonal from (1,1) to (4,4), the midpoint is (1+42,1+42)=(2.5,2.5)(\frac{1+4}{2}, \frac{1+4}{2}) = (2.5, 2.5). The translation moves every point (x,y)(x, y) to (x−3,y+5)(x-3, y+5). Applying this translation to the center's coordinates: (2.5−3,2.5+5)=(−0.5,7.5)(2.5 - 3, 2.5 + 5) = (-0.5, 7.5).

Question 9

A triangle is defined by the vertices A(-1, -1), B(5, -1), and C(2, 3). Which of the following best describes the triangle?

  1. Equilateral and right
  2. Isosceles and right
  3. Scalene and right
  4. Isosceles and not right (correct answer)
Explanation: When you encounter a triangle classification problem with given vertices, you need to determine two things: the side lengths (to classify by sides) and whether any angles are 90° (to determine if it's a right triangle). First, calculate the distances between each pair of vertices using the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}:
  • Side AB: (5−(−1))2+(−1−(−1))2=36+0=6\sqrt{(5-(-1))^2 + (-1-(-1))^2} = \sqrt{36 + 0} = 6
  • Side BC: (2−5)2+(3−(−1))2=9+16=5\sqrt{(2-5)^2 + (3-(-1))^2} = \sqrt{9 + 16} = 5
  • Side AC: (2−(−1))2+(3−(−1))2=9+16=5\sqrt{(2-(-1))^2 + (3-(-1))^2} = \sqrt{9 + 16} = 5
Since two sides have equal length (BC = AC = 5), this is an isosceles triangle. To check if it's a right triangle, use the Pythagorean theorem. For a right triangle, a2+b2=c2a^2 + b^2 = c^2. Testing with the longest side (6) as the hypotenuse: 52+52=25+25=505^2 + 5^2 = 25 + 25 = 50, but 62=366^2 = 36. Since 50≠3650 \neq 36, this is not a right triangle. Now examining the answer choices: Choice A is wrong because the triangle isn't equilateral (all sides aren't equal) or right. Choice B is incorrect because while it's isosceles, it's not right. Choice C is wrong because it's not scalene (all different sides) or right. Choice D correctly identifies the triangle as isosceles and not right. Remember: Always calculate all three side lengths first to classify by sides, then use the Pythagorean theorem to test for right angles.

Question 10

The vertices of a rectangle are (0, 0), (a, 0), (a, b), and (0, b). If the area of the rectangle is 24 and its perimeter is 20, what are the values of a and b?

  1. a = 8, b = 3
  2. a = 12, b = 2
  3. a = 6, b = 4 (correct answer)
  4. a = 5, b = 5
Explanation: The side lengths of the rectangle are 'a' and 'b'. The area is ab=24ab = 24. The perimeter is 2(a+b)=202(a+b) = 20, which simplifies to a+b=10a+b = 10. We need to find two numbers that multiply to 24 and add up to 10. Let's test the options. A) 83=24, 8+3=11 (Incorrect). B) 122=24, 12+2=14 (Incorrect). C) 64=24, 6+4=10 (Correct). D) 55=25 (Incorrect).

Question 11

A circle has a diameter with endpoints at (-5, 8) and (3, -2). What are the coordinates of the center of the circle?

  1. (-2, 6)
  2. (-1, 3) (correct answer)
  3. (-4, 5)
  4. (8, 10)
Explanation: The center of the circle is the midpoint of its diameter. The midpoint formula is (x1+x22,y1+y22)(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}). Using the endpoints (-5, 8) and (3, -2), the center is (−5+32,8+(−2)2)=(−22,62)=(−1,3)(\frac{-5+3}{2}, \frac{8+(-2)}{2}) = (\frac{-2}{2}, \frac{6}{2}) = (-1, 3).

Question 12

A trapezoid has vertices at A(1, 2), B(9, 2), C(7, 6), and D(3, 6). What is the area of the trapezoid?

  1. 22 square units
  2. 48 square units
  3. 32 square units
  4. 24 square units (correct answer)
Explanation: When you encounter a trapezoid with given coordinates, you need to identify the parallel sides and calculate the area using the formula: Area = 12(b1+b2)×h\frac{1}{2}(b_1 + b_2) \times h, where b1b_1 and b2b_2 are the lengths of the parallel bases and hh is the height. First, plot or visualize the points: A(1, 2), B(9, 2), C(7, 6), and D(3, 6). Notice that points A and B both have y-coordinate 2, making AB a horizontal line. Similarly, points C and D both have y-coordinate 6, making CD horizontal. These are your parallel bases. Calculate the base lengths:
  • Base AB: distance from (1, 2) to (9, 2) = 9−1=89 - 1 = 8 units
  • Base CD: distance from (3, 6) to (7, 6) = 7−3=47 - 3 = 4 units
The height is the perpendicular distance between the parallel lines: 6−2=46 - 2 = 4 units. Now apply the area formula: Area = 12(8+4)×4=12(12)(4)=24\frac{1}{2}(8 + 4) \times 4 = \frac{1}{2}(12)(4) = 24 square units. Answer choice A (22) likely results from calculation errors in adding the bases. Answer choice B (48) comes from forgetting the 12\frac{1}{2} in the formula and calculating (8+4)×4=48(8 + 4) \times 4 = 48. Answer choice C (32) might result from using 8×4=328 \times 4 = 32, treating it like a rectangle with only one base. Remember: Always identify which sides are parallel first, then carefully apply the trapezoid area formula with the 12\frac{1}{2} factor.

Question 13

A triangle has vertices P(2, 3), Q(5, 3), and R(2, 7). If the triangle is reflected across the y-axis, what are the new coordinates of vertex R?

  1. (-2, 7) (correct answer)
  2. (2, -7)
  3. (-2, -7)
  4. (7, 2)
Explanation: A reflection across the y-axis transforms a point (x,y)(x, y) to (−x,y)(-x, y). The x-coordinate changes sign, and the y-coordinate remains the same. Applying this rule to vertex R(2, 7), the new coordinates will be (-2, 7).

Question 14

Three points, A(1, 2), B(3, 8), and C(5, 14), are plotted on a coordinate plane. Which statement best describes these points?

  1. They form a right triangle.
  2. They form an isosceles triangle.
  3. They are collinear. (correct answer)
  4. They form an equilateral triangle.
Explanation: To determine the relationship between the points, we can check the slopes between them. The slope between A and B is 8−23−1=62=3\frac{8-2}{3-1} = \frac{6}{2} = 3. The slope between B and C is 14−85−3=62=3\frac{14-8}{5-3} = \frac{6}{2} = 3. Since the slope between A and B is the same as the slope between B and C, the three points lie on the same straight line. Therefore, they are collinear and do not form a triangle.

Question 15

A parallelogram has vertices at A(1, 5), B(6, 5), and C(8, 1). What are the coordinates of the fourth vertex, D?

  1. (2, 1)
  2. (3, 1) (correct answer)
  3. (4, 1)
  4. (13, 1)
Explanation: In a parallelogram ABCD, the vector from A to B is equal to the vector from D to C. The vector AB is (6−1),(5−5)(6-1), (5-5) = (5,0)(5, 0). Let the coordinates of D be (x,y)(x, y). The vector DC is (8−x),(1−y)(8-x), (1-y). Setting the vectors equal: 8−x=58-x = 5 and 1−y=01-y = 0. This gives x=3x = 3 and y=1y = 1. So, vertex D is at (3, 1). Alternatively, the midpoint of the diagonal AC must be the same as the midpoint of the diagonal BD. Midpoint of AC is (1+82,5+12)=(4.5,3)(\frac{1+8}{2}, \frac{5+1}{2}) = (4.5, 3). Midpoint of BD is (6+x2,5+y2)(\frac{6+x}{2}, \frac{5+y}{2}). Setting them equal: 6+x2=4.5⇒6+x=9⇒x=3\frac{6+x}{2} = 4.5 \Rightarrow 6+x=9 \Rightarrow x=3, and 5+y2=3⇒5+y=6⇒y=1\frac{5+y}{2} = 3 \Rightarrow 5+y=6 \Rightarrow y=1.

Question 16

A square has opposite vertices at (0, 5) and (4, 1). What is the area of the square?

  1. 16 square units (correct answer)
  2. 20 square units
  3. 32 square units
  4. 36 square units
Explanation: First, find the square of the length of the diagonal connecting the opposite vertices. Using the distance formula, d2=(4−0)2+(1−5)2=42+(−4)2=16+16=32d^2 = (4-0)^2 + (1-5)^2 = 4^2 + (-4)^2 = 16 + 16 = 32. The area of a square can be calculated as d22\frac{d^2}{2}, where dd is the length of the diagonal. So, the area is 322=16\frac{32}{2} = 16 square units.