Math 1 Quiz: Distance Formula
6 questions · exam conditions
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Distance FormulaQuestion 1 of 6

A quadrilateral has vertices at A(1,3)(1, 3), B(7,1)(7, 1), C(5,7)(5, 7), and D(−1,5)(−1, 5). To verify if this quadrilateral is a rhombus, a student calculates the lengths of all four sides. Which of the following represents the length of side BC?

2102\sqrt{10}
40\sqrt{40}
626\sqrt{2}
2112\sqrt{11}
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Math 1 Quiz

Math 1 Quiz: Distance Formula

Practice Distance Formula 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 Distance Formula, 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.

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Question 1

A quadrilateral has vertices at A(1,3)(1, 3), B(7,1)(7, 1), C(5,7)(5, 7), and D(−1,5)(−1, 5). To verify if this quadrilateral is a rhombus, a student calculates the lengths of all four sides. Which of the following represents the length of side BC?

  1. 2102\sqrt{10} (correct answer)
  2. 40\sqrt{40}
  3. 626\sqrt{2}
  4. 2112\sqrt{11}
Explanation: Length of side BC = (5−7)2+(7−1)2=(−2)2+62=4+36=40\sqrt{(5-7)^2 + (7-1)^2} = \sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40}. We can simplify: 40=4⋅10=210\sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10}. Choice B gives 40\sqrt{40} which equals 2102\sqrt{10} but isn't simplified. Choice C gives 62=36⋅2=726\sqrt{2} = \sqrt{36 \cdot 2} = \sqrt{72}, which is incorrect. Choice D gives 211=4⋅11=442\sqrt{11} = \sqrt{4 \cdot 11} = \sqrt{44}, also incorrect. The correct answer is 2102\sqrt{10}.

Question 2

A robot moves from point (0,0)(0, 0) to point (a,b)(a, b) where a>0a > 0 and b>0b > 0. The robot can only move right (increasing x) or up (increasing y) one unit at a time. If the robot takes the shortest possible path and the total distance traveled is 12 units, while the straight-line distance from start to finish is 626\sqrt{2}, what are the values of aa and bb?

  1. a=6,b=6a = 6, b = 6 (correct answer)
  2. a=8,b=4a = 8, b = 4 or a=4,b=8a = 4, b = 8
  3. a=9,b=3a = 9, b = 3 or a=3,b=9a = 3, b = 9
  4. a=7,b=5a = 7, b = 5 or a=5,b=7a = 5, b = 7
Explanation: The robot's shortest path requires a+b=12a + b = 12 steps (total distance). The straight-line distance is a2+b2=62\sqrt{a^2 + b^2} = 6\sqrt{2}, so a2+b2=72a^2 + b^2 = 72. From a+b=12a + b = 12, we get b=12−ab = 12 - a. Substituting: a2+(12−a)2=72a^2 + (12-a)^2 = 72. Expanding: a2+144−24a+a2=72a^2 + 144 - 24a + a^2 = 72, so 2a2−24a+72=02a^2 - 24a + 72 = 0, which gives a2−12a+36=0a^2 - 12a + 36 = 0, so (a−6)2=0(a-6)^2 = 0, meaning a=6a = 6 and b=6b = 6. Other choices don't satisfy both conditions simultaneously.

Question 3

Three cell phone towers are located at coordinates A(1,2)(1, 2), B(7,6)(7, 6), and C(4,10)(4, 10). A mobile phone at point P receives signals from all three towers. If the phone is equidistant from towers A and B, and the distance from P to tower C is 5 units, how many possible locations are there for point P?

  1. No valid locations exist for point P
  2. Exactly one location exists for point P
  3. Exactly two locations exist for point P (correct answer)
  4. Infinitely many locations exist for point P
Explanation: Points equidistant from A and B lie on the perpendicular bisector of segment AB. Midpoint of AB is (1+72,2+62)=(4,4)\left(\frac{1+7}{2}, \frac{2+6}{2}\right) = (4, 4). Slope of AB is 6−27−1=23\frac{6-2}{7-1} = \frac{2}{3}, so perpendicular bisector has slope −32-\frac{3}{2} and equation y−4=−32(x−4)y - 4 = -\frac{3}{2}(x - 4), or y=−32x+10y = -\frac{3}{2}x + 10. Points at distance 5 from C(4,10)(4, 10) satisfy (x−4)2+(y−10)2=25(x-4)^2 + (y-10)^2 = 25. Substituting the line equation: (x−4)2+(−32x+10−10)2=25(x-4)^2 + (-\frac{3}{2}x + 10 - 10)^2 = 25, so (x−4)2+94x2=25(x-4)^2 + \frac{9}{4}x^2 = 25. This expands to 134x2−8x+16=25\frac{13}{4}x^2 - 8x + 16 = 25, giving 13x2−32x−36=013x^2 - 32x - 36 = 0. The discriminant is 322+4(13)(36)=1024+1872=2896>032^2 + 4(13)(36) = 1024 + 1872 = 2896 > 0, so there are exactly two solutions.

Question 4

Two radio stations have transmission ranges represented by circles. Station A at (0,0)(0, 0) has range 10 units, and Station B at (15,0)(15, 0) has range rr units. If a receiver at point (6,8)(6, 8) is exactly on the boundary of both transmission ranges, what is the value of rr?

  1. r=12r = 12
  2. r=145r = \sqrt{145} (correct answer)
  3. r=13r = 13
  4. r=153r = \sqrt{153}
Explanation: The receiver at (6,8)(6, 8) must be at distance 10 from Station A and distance rr from Station B. Distance from A: 62+82=36+64=100=10\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 ✓. Distance from B: (6−15)2+(8−0)2=(−9)2+82=81+64=145\sqrt{(6-15)^2 + (8-0)^2} = \sqrt{(-9)^2 + 8^2} = \sqrt{81 + 64} = \sqrt{145}. Therefore, r=145r = \sqrt{145}. Choice A gives r=12r = 12, but 122=144≠14512^2 = 144 \neq 145. Choice C gives r=13r = 13, but 132=169≠14513^2 = 169 \neq 145. Choice D gives a different radical that doesn't match the calculation.

Question 5

In a coordinate plane, point M is the midpoint of segment connecting A(2,9)(2, 9) and B(14,1)(14, 1). Point N is located such that triangle AMN has a right angle at M, and MN has the same length as AM. How many possible locations are there for point N?

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 4
Explanation: First, find the midpoint M: M=(2+142,9+12)=(8,5)M = \left(\frac{2+14}{2}, \frac{9+1}{2}\right) = (8, 5). Next, calculate AM: AM=(8−2)2+(5−9)2=36+16=52=213AM = \sqrt{(8-2)^2+(5-9)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}. Since triangle AMN has a right angle at M and MN = AM, point N must be at distance 2132\sqrt{13} from M, and MN must be perpendicular to AM. The vector from A to M is (6,−4)(6, -4). A perpendicular vector is (4,6)(4, 6) or (−4,−6)(-4, -6). To get the correct length 2132\sqrt{13}, we need to normalize: the length of (4,6)(4, 6) is 16+36=52=213\sqrt{16+36} = \sqrt{52} = 2\sqrt{13}, which is already correct. So the two possible positions for N are: N1=M+(4,6)=(8,5)+(4,6)=(12,11)N_1 = M + (4, 6) = (8, 5) + (4, 6) = (12, 11) and N2=M+(−4,−6)=(8,5)+(−4,−6)=(4,−1)N_2 = M + (-4, -6) = (8, 5) + (-4, -6) = (4, -1). Choice A is incorrect as there are multiple solutions. Choice C and D suggest more solutions than geometrically possible for this configuration.

Question 6

Points J, K, L, and M form a quadrilateral where J(−1,3)(-1, 3), K(5,1)(5, 1), L(3,−5)(3, -5), and M(−3,−3)(-3, -3). Which statement about this quadrilateral is true?

  1. The diagonals JL and KM are equal in length and perpendicular to each other (correct answer)
  2. The diagonals JL and KM are equal in length but not perpendicular to each other
  3. The diagonals JL and KM are perpendicular but not equal in length
  4. The diagonals JL and KM are neither equal in length nor perpendicular to each other
Explanation: First, I calculate the lengths of the diagonals. For JL: dJL=(3−(−1))2+(−5−3)2=16+64=80=45d_{JL} = \sqrt{(3-(-1))^2 + (-5-3)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}. For KM: dKM=(−3−5)2+(−3−1)2=64+16=80=45d_{KM} = \sqrt{(-3-5)^2 + (-3-1)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}. The diagonals are equal in length. Next, I check if they're perpendicular by finding their direction vectors. Vector JL: (3−(−1),−5−3)=(4,−8)(3-(-1), -5-3) = (4, -8). Vector KM: (−3−5,−3−1)=(−8,−4)(-3-5, -3-1) = (-8, -4). For perpendicularity, their dot product must be zero: (4)(−8)+(−8)(−4)=−32+32=0(4)(-8) + (-8)(-4) = -32 + 32 = 0. Since the dot product is zero, the diagonals are perpendicular. Therefore, the diagonals are both equal in length and perpendicular. Choice B is wrong because they are perpendicular. Choice C is wrong because they are equal in length. Choice D is wrong because they satisfy both conditions.