Distance Formula - Math 1

In a coordinate plane, point M is the midpoint of segment connecting A $(2, 9)$ and B $(14, 1)$ . 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?
Answer: 2 First, find the midpoint M: $M = \left(\frac{2+14}{2}, \frac{9+1}{2}\right) = (8, 5)$ . Next, calculate AM: $AM = \sqrt{(8-2)^2+(5-9)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$ . Since triangle AMN has a right angle at M and MN = AM, point N must be at distance $2\sqrt{13}$ from M, and MN must be perpendicular to AM. The vector from A to M is $(6, -4)$ . A perpendicular vector is $(4, 6)$ or $(-4, -6)$ . To get the correct length $2\sqrt{13}$ , we need to normalize: the length of $(4, 6)$ is $\sqrt{16+36} = \sqrt{52} = 2\sqrt{13}$ , which is already correct. So the two possible positions for N are: $N_1 = M + (4, 6) = (8, 5) + (4, 6) = (12, 11)$ and $N_2 = M + (-4, -6) = (8, 5) + (-4, -6) = (4, -1)$ . Choice A is incorrect as there are multiple solutions. Choice C and D suggest more solutions than geometrically possible for this configuration.
Three cell phone towers are located at coordinates A $(1, 2)$ , B $(7, 6)$ , and C $(4, 10)$ . 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?
Answer: Exactly two locations exist for point P Points equidistant from A and B lie on the perpendicular bisector of segment AB. Midpoint of AB is $\left(\frac{1+7}{2}, \frac{2+6}{2}\right) = (4, 4)$ . Slope of AB is $\frac{6-2}{7-1} = \frac{2}{3}$ , so perpendicular bisector has slope $-\frac{3}{2}$ and equation $y - 4 = -\frac{3}{2}(x - 4)$ , or $y = -\frac{3}{2}x + 10$ . Points at distance 5 from C $(4, 10)$ satisfy $(x-4)^2 + (y-10)^2 = 25$ . Substituting the line equation: $(x-4)^2 + (-\frac{3}{2}x + 10 - 10)^2 = 25$ , so $(x-4)^2 + \frac{9}{4}x^2 = 25$ . This expands to $\frac{13}{4}x^2 - 8x + 16 = 25$ , giving $13x^2 - 32x - 36 = 0$ . The discriminant is $32^2 + 4(13)(36) = 1024 + 1872 = 2896 > 0$ , so there are exactly two solutions.
A robot moves from point $(0, 0)$ to point $(a, b)$ where $a > 0$ and $b > 0$ . 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 $6\sqrt{2}$ , what are the values of $a$ and $b$ ?
Answer: $a = 6, b = 6$ The robot's shortest path requires $a + b = 12$ steps (total distance). The straight-line distance is $\sqrt{a^2 + b^2} = 6\sqrt{2}$ , so $a^2 + b^2 = 72$ . From $a + b = 12$ , we get $b = 12 - a$ . Substituting: $a^2 + (12-a)^2 = 72$ . Expanding: $a^2 + 144 - 24a + a^2 = 72$ , so $2a^2 - 24a + 72 = 0$ , which gives $a^2 - 12a + 36 = 0$ , so $(a-6)^2 = 0$ , meaning $a = 6$ and $b = 6$ . Other choices don't satisfy both conditions simultaneously.
Points J, K, L, and M form a quadrilateral where J $(-1, 3)$ , K $(5, 1)$ , L $(3, -5)$ , and M $(-3, -3)$ . Which statement about this quadrilateral is true?
Answer: The diagonals JL and KM are equal in length and perpendicular to each other First, I calculate the lengths of the diagonals. For JL: $d_{JL} = \sqrt{(3-(-1))^2 + (-5-3)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$ . For KM: $d_{KM} = \sqrt{(-3-5)^2 + (-3-1)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$ . The diagonals are equal in length. Next, I check if they're perpendicular by finding their direction vectors. Vector JL: $(3-(-1), -5-3) = (4, -8)$ . Vector KM: $(-3-5, -3-1) = (-8, -4)$ . For perpendicularity, their dot product must be zero: $(4)(-8) + (-8)(-4) = -32 + 32 = 0$ . 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.
A quadrilateral has vertices at A $(1, 3)$ , B $(7, 1)$ , C $(5, 7)$ , and D $(−1, 5)$ . 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?
Answer: $2\sqrt{10}$ Length of side BC = $\sqrt{(5-7)^2 + (7-1)^2} = \sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40}$ . We can simplify: $\sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10}$ . Choice B gives $\sqrt{40}$ which equals $2\sqrt{10}$ but isn't simplified. Choice C gives $6\sqrt{2} = \sqrt{36 \cdot 2} = \sqrt{72}$ , which is incorrect. Choice D gives $2\sqrt{11} = \sqrt{4 \cdot 11} = \sqrt{44}$ , also incorrect. The correct answer is $2\sqrt{10}$ .
Two radio stations have transmission ranges represented by circles. Station A at $(0, 0)$ has range 10 units, and Station B at $(15, 0)$ has range $r$ units. If a receiver at point $(6, 8)$ is exactly on the boundary of both transmission ranges, what is the value of $r$ ?
Answer: $r = \sqrt{145}$ The receiver at $(6, 8)$ must be at distance 10 from Station A and distance $r$ from Station B. Distance from A: $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ ✓. Distance from B: $\sqrt{(6-15)^2 + (8-0)^2} = \sqrt{(-9)^2 + 8^2} = \sqrt{81 + 64} = \sqrt{145}$ . Therefore, $r = \sqrt{145}$ . Choice A gives $r = 12$ , but $12^2 = 144 \neq 145$ . Choice C gives $r = 13$ , but $13^2 = 169 \neq 145$ . Choice D gives a different radical that doesn't match the calculation.

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