IB Mathematics: Analysis and Approaches Quiz: Coordinate Geometry
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Coordinate GeometryQuestion 1 of 20

Point M(3, 1) is the midpoint of the line segment AB. Point A has coordinates (-1, 4). Find the length of the segment MB.

5
10
525\sqrt{2}
252\sqrt{5}
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Coordinate Geometry

Practice Coordinate Geometry in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Coordinate Geometry, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

Point M(3, 1) is the midpoint of the line segment AB. Point A has coordinates (-1, 4). Find the length of the segment MB.

  1. 5 (correct answer)
  2. 10
  3. 525\sqrt{2}
  4. 252\sqrt{5}
Explanation: The length of the segment MB is the distance from M to B. However, since M is the midpoint of AB, the length of MB is equal to the length of AM. We can find the length of AM directly from the given coordinates of A and M. Using the distance formula: Length of AM=(3−(−1))2+(1−4)2\text{Length of AM} = \sqrt{(3 - (-1))^2 + (1 - 4)^2} =(4)2+(−3)2= \sqrt{(4)^2 + (-3)^2} =16+9=25=5= \sqrt{16 + 9} = \sqrt{25} = 5 Since AM = MB, the length of MB is 5. An alternative, longer method is to first find the coordinates of B and then calculate the distance MB.

Question 2

Point M is the midpoint of AB and point N is the midpoint of AM. If A is (-5, 12) and B is (7, -4), find the coordinates of N.

  1. (-2, 8) (correct answer)
  2. (-0.5, 5)
  3. (1, 4)
  4. (4, 0)
Explanation: This is a two-step problem. First, find the coordinates of M, the midpoint of AB. M=(−5+72,12+(−4)2)=(22,82)=(1,4)M = \left(\frac{-5+7}{2}, \frac{12+(-4)}{2}\right) = \left(\frac{2}{2}, \frac{8}{2}\right) = (1, 4) Next, find the coordinates of N, the midpoint of AM. A is (-5, 12) and M is (1, 4). N=(−5+12,12+42)=(−42,162)=(−2,8)N = \left(\frac{-5+1}{2}, \frac{12+4}{2}\right) = \left(\frac{-4}{2}, \frac{16}{2}\right) = (-2, 8)

Question 3

The distance between point A(t, 2) and point B(1, 5) is 5. Find the sum of the possible values of t.

  1. -2
  2. 0
  3. 2 (correct answer)
  4. 8
Explanation: The distance formula is d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}. We are given d=5. Squaring both sides, d²=25. (t−1)2+(2−5)2=52(t-1)^2 + (2-5)^2 = 5^2 (t−1)2+(−3)2=25(t-1)^2 + (-3)^2 = 25 (t−1)2+9=25(t-1)^2 + 9 = 25 (t−1)2=16(t-1)^2 = 16 Taking the square root of both sides gives two possibilities: t−1=4ort−1=−4t-1 = 4 \quad \text{or} \quad t-1 = -4 t=5ort=−3t = 5 \quad \text{or} \quad t = -3 The possible values of t are 5 and -3. The sum of these values is 5 + (-3) = 2. Alternatively, expanding (t−1)2=16(t-1)^2=16 gives t2−2t+1=16t^2-2t+1=16, which simplifies to the quadratic equation t2−2t−15=0t^2-2t-15=0. The sum of the roots of a quadratic equation ax2+bx+c=0ax^2+bx+c=0 is −b/a-b/a. Here, the sum is −(−2)/1=2-(-2)/1 = 2.

Question 4

Three points in space are P(1, 2, 3), Q(3, 5, 7), and R(-2, k, -3). The points P, Q, and R are collinear. Find the value of k.

  1. -5/2 (correct answer)
  2. -1
  3. 3/2
  4. 13/2
Explanation: For three points to be collinear, the vectors between them must be parallel. Let's find the vectors PQ⃗\vec{PQ} and PR⃗\vec{PR}. PQ⃗=Q−P=(3−1,5−2,7−3)=(2,3,4)\vec{PQ} = Q - P = (3-1, 5-2, 7-3) = (2, 3, 4) PR⃗=R−P=(−2−1,k−2,−3−3)=(−3,k−2,−6)\vec{PR} = R - P = (-2-1, k-2, -3-3) = (-3, k-2, -6) If the vectors are parallel, then PR⃗=c⋅PQ⃗\vec{PR} = c \cdot \vec{PQ} for some scalar c. We can find c by comparing the components. From the x-components: −3=c⋅2⇒c=−3/2-3 = c \cdot 2 \Rightarrow c = -3/2. From the z-components: −6=c⋅4⇒c=−6/4=−3/2-6 = c \cdot 4 \Rightarrow c = -6/4 = -3/2. The values of c are consistent. Now use this value of c for the y-components: k−2=c⋅3k-2 = c \cdot 3 k−2=(−3/2)⋅3=−9/2k-2 = (-3/2) \cdot 3 = -9/2 k=2−9/2=4/2−9/2=−5/2k = 2 - 9/2 = 4/2 - 9/2 = -5/2

Question 5

A line segment has one endpoint at A(-3, 5) and its midpoint at M(2, 1). What are the coordinates of the other endpoint, B?

  1. (-0.5, 3)
  2. (-1, 6)
  3. (7, -3) (correct answer)
  4. (5, -4)
Explanation: Let the coordinates of endpoint B be (x, y). The midpoint formula states that M=(xA+xB2,yA+yB2)M = \left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}\right). We are given A(-3, 5) and M(2, 1). We can set up two equations: −3+x2=2and5+y2=1\frac{-3+x}{2} = 2 \quad \text{and} \quad \frac{5+y}{2} = 1 Solving for x: −3+x=4⇒x=7-3+x = 4 \Rightarrow x = 7 Solving for y: 5+y=2⇒y=−35+y = 2 \Rightarrow y = -3 Therefore, the coordinates of endpoint B are (7, -3).

Question 6

Point M(3, 1) is the midpoint of the line segment AB. Point A has coordinates (-1, 4). Find the length of the segment MB.

  1. 5 (correct answer)
  2. 10
  3. 525\sqrt{2}
  4. 252\sqrt{5}
Explanation: The length of the segment MB is the distance from M to B. However, since M is the midpoint of AB, the length of MB is equal to the length of AM. We can find the length of AM directly from the given coordinates of A and M. Using the distance formula: Length of AM=(3−(−1))2+(1−4)2\text{Length of AM} = \sqrt{(3 - (-1))^2 + (1 - 4)^2} =(4)2+(−3)2= \sqrt{(4)^2 + (-3)^2} =16+9=25=5= \sqrt{16 + 9} = \sqrt{25} = 5 Since AM = MB, the length of MB is 5. An alternative, longer method is to first find the coordinates of B and then calculate the distance MB.

Question 7

The distance between the midpoints of the segments AB and CD is 17\sqrt{17}. The coordinates are A(1, 2), B(3, 8), C(-2, 5), and D(k, 3). Find a possible value for k.

  1. -4
  2. -2 (correct answer)
  3. 0
  4. 2
Explanation: First, find the midpoint of segment AB: MAB=(1+32,2+82)=(2,5)M_{AB} = \left(\frac{1+3}{2}, \frac{2+8}{2}\right) = (2, 5) Next, find the midpoint of segment CD in terms of k: MCD=(−2+k2,5+32)=(k−22,4)M_{CD} = \left(\frac{-2+k}{2}, \frac{5+3}{2}\right) = \left(\frac{k-2}{2}, 4\right) The distance between these midpoints is 17\sqrt{17}, so: (k−22−2)2+(4−5)2=17\left( \frac{k-2}{2} - 2 \right)^2 + (4-5)^2 = 17 (k−62)2+1=17\left( \frac{k-6}{2} \right)^2 + 1 = 17 (k−6)24=16\frac{(k-6)^2}{4} = 16 (k−6)2=64(k-6)^2 = 64 k−6=±8k-6 = \pm 8 This gives k = 14 or k = -2. Among the given choices, k = -2 is a possible value.

Question 8

A line segment has one endpoint at A(-3, 5) and its midpoint at M(2, 1). What are the coordinates of the other endpoint, B?

  1. (-0.5, 3)
  2. (-1, 6)
  3. (7, -3) (correct answer)
  4. (5, -4)
Explanation: Let the coordinates of endpoint B be (x, y). The midpoint formula states that M=(xA+xB2,yA+yB2)M = \left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}\right). We are given A(-3, 5) and M(2, 1). We can set up two equations: −3+x2=2and5+y2=1\frac{-3+x}{2} = 2 \quad \text{and} \quad \frac{5+y}{2} = 1 Solving for x: −3+x=4⇒x=7-3+x = 4 \Rightarrow x = 7 Solving for y: 5+y=2⇒y=−35+y = 2 \Rightarrow y = -3 Therefore, the coordinates of endpoint B are (7, -3).

Question 9

Find the length of the median from vertex A for a triangle with vertices A(5, -1), B(0, 4), and C(-2, -2).

  1. 252\sqrt{5}
  2. 525\sqrt{2}
  3. 58\sqrt{58}
  4. 2102\sqrt{10} (correct answer)
Explanation: When you encounter a question about finding the length of a median, remember that a median connects a vertex to the midpoint of the opposite side. Your approach should be: find the midpoint, then calculate the distance. First, find the midpoint of side BC (opposite vertex A). Using the midpoint formula (x1+x22,y1+y22)\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) with B(0, 4) and C(-2, -2): Midpoint M = (0+(−2)2,4+(−2)2)=(−1,1)\left(\frac{0+(-2)}{2}, \frac{4+(-2)}{2}\right) = (-1, 1) Now find the distance from A(5, -1) to M(-1, 1) using the distance formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} d=(−1−5)2+(1−(−1))2=(−6)2+(2)2=36+4=40d = \sqrt{(-1-5)^2 + (1-(-1))^2} = \sqrt{(-6)^2 + (2)^2} = \sqrt{36 + 4} = \sqrt{40} Simplifying: 40=4×10=210\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10} This confirms answer D is correct. The wrong answers likely come from calculation errors: A) 252\sqrt{5} equals 20\sqrt{20}, suggesting someone found the wrong midpoint or made an arithmetic mistake. B) 525\sqrt{2} equals 50\sqrt{50}, possibly from incorrectly calculating the distance components. C) 58\sqrt{58} might result from using the wrong vertices or forgetting to find the midpoint first. Study tip: Always break median problems into two clear steps: find the midpoint of the opposite side, then calculate the distance. Double-check your midpoint calculation before moving to the distance formula, as errors compound quickly in coordinate geometry.

Question 10

A sphere has a diameter with endpoints P(-2, 4, 1) and Q(4, -2, 7). Find the square of the radius of the sphere.

  1. 18
  2. 27 (correct answer)
  3. 36
  4. 108
Explanation: The length of the diameter is the distance between points P and Q. The radius is half the length of the diameter. First, find the square of the diameter's length (d²). d2=(4−(−2))2+(−2−4)2+(7−1)2d^2 = (4 - (-2))^2 + (-2 - 4)^2 + (7 - 1)^2 d2=(6)2+(−6)2+(6)2d^2 = (6)^2 + (-6)^2 + (6)^2 d2=36+36+36=108d^2 = 36 + 36 + 36 = 108 The radius r is half the diameter, so r=d/2r = d/2. The square of the radius is r2=(d/2)2=d2/4r^2 = (d/2)^2 = d^2/4. r2=1084=27r^2 = \frac{108}{4} = 27 Alternatively, one could find the center of the sphere (the midpoint of PQ) and then find the distance from the center to one of the endpoints. Center C = ((-2+4)/2, (4-2)/2, (1+7)/2) = (1, 1, 4). Then r² = distance from C to P squared = (1−(−2))2+(1−4)2+(4−1)2=32+(−3)2+32=9+9+9=27(1 - (-2))^2 + (1 - 4)^2 + (4 - 1)^2 = 3^2 + (-3)^2 + 3^2 = 9 + 9 + 9 = 27.

Question 11

Find the perimeter of a triangle with vertices A(0, 0, 3), B(0, 4, 0), and C(5, 0, 0).

  1. 15
  2. 10 + 41\sqrt{41}
  3. 5 + 34\sqrt{34} + 41\sqrt{41} (correct answer)
  4. 5 + 41\sqrt{41} + 50\sqrt{50}
Explanation: The perimeter is the sum of the lengths of the three sides AB, BC, and AC. We use the 3D distance formula to find the length of each side. Length of AB=(0−0)2+(4−0)2+(0−3)2=02+42+(−3)2=16+9=25=5\text{Length of AB} = \sqrt{(0-0)^2 + (4-0)^2 + (0-3)^2} = \sqrt{0^2+4^2+(-3)^2} = \sqrt{16+9} = \sqrt{25} = 5 Length of BC=(5−0)2+(0−4)2+(0−0)2=52+(−4)2+02=25+16=41\text{Length of BC} = \sqrt{(5-0)^2 + (0-4)^2 + (0-0)^2} = \sqrt{5^2+(-4)^2+0^2} = \sqrt{25+16} = \sqrt{41} Length of AC=(5−0)2+(0−0)2+(0−3)2=52+02+(−3)2=25+9=34\text{Length of AC} = \sqrt{(5-0)^2 + (0-0)^2 + (0-3)^2} = \sqrt{5^2+0^2+(-3)^2} = \sqrt{25+9} = \sqrt{34} The perimeter is the sum of these lengths: Perimeter = 5 + 41\sqrt{41} + 34\sqrt{34}.

Question 12

The midpoint of the line segment joining A(2, -3, 1) and B is M(4, 1, -2). Find the distance from the origin to point B.

  1. 14\sqrt{14}
  2. 21\sqrt{21}
  3. 86\sqrt{86} (correct answer)
  4. 116\sqrt{116}
Explanation: First, find the coordinates of point B(x, y, z). The midpoint M is given by M=(xA+xB2,yA+yB2,zA+zB2)M = \left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}, \frac{z_A+z_B}{2}\right). We have: (2+x2,−3+y2,1+z2)=(4,1,−2)\left(\frac{2+x}{2}, \frac{-3+y}{2}, \frac{1+z}{2}\right) = (4, 1, -2) This gives three equations: 2+x2=4⇒2+x=8⇒x=6\frac{2+x}{2} = 4 \Rightarrow 2+x=8 \Rightarrow x=6 −3+y2=1⇒−3+y=2⇒y=5\frac{-3+y}{2} = 1 \Rightarrow -3+y=2 \Rightarrow y=5 1+z2=−2⇒1+z=−4⇒z=−5\frac{1+z}{2} = -2 \Rightarrow 1+z=-4 \Rightarrow z=-5 So, point B is (6, 5, -5). The distance from the origin O(0, 0, 0) to B is: d=(6−0)2+(5−0)2+(−5−0)2=36+25+25=86d = \sqrt{(6-0)^2 + (5-0)^2 + (-5-0)^2} = \sqrt{36+25+25} = \sqrt{86}

Question 13

A line segment has endpoints A(a, -2, 5) and B(3, b, 1). Its midpoint is M(-1, 4, c). Find the value of a + b + c.

  1. 5
  2. 8 (correct answer)
  3. 10
  4. 12
Explanation: The coordinates of the midpoint M are the average of the coordinates of the endpoints A and B. We set up equations for each coordinate: a+32=−1⇒a+3=−2⇒a=−5\frac{a+3}{2} = -1 \Rightarrow a+3 = -2 \Rightarrow a = -5 −2+b2=4⇒−2+b=8⇒b=10\frac{-2+b}{2} = 4 \Rightarrow -2+b = 8 \Rightarrow b = 10 5+12=c⇒62=c⇒c=3\frac{5+1}{2} = c \Rightarrow \frac{6}{2} = c \Rightarrow c = 3 Now, we find the sum a + b + c: a+b+c=−5+10+3=8a + b + c = -5 + 10 + 3 = 8

Question 14

Point A is (2, 3) and point B is (10, k). The midpoint M of the segment AB lies on the line y = 2x - 7. Find the value of k.

  1. -1
  2. 1
  3. 3
  4. 7 (correct answer)
Explanation: First, find the coordinates of the midpoint M in terms of k. M=(2+102,3+k2)=(6,3+k2)M = \left(\frac{2+10}{2}, \frac{3+k}{2}\right) = \left(6, \frac{3+k}{2}\right) Since M lies on the line y = 2x - 7, substitute the coordinates of M into the equation: 3+k2=2(6)−7\frac{3+k}{2} = 2(6) - 7 3+k2=12−7=5\frac{3+k}{2} = 12 - 7 = 5 3+k=103+k = 10 k=7k = 7

Question 15

A triangle PQR has vertices P(-3, 2), Q(5, 2), and R(k, 7). Given that the triangle is isosceles with base PQ, find the value of k.

  1. -3
  2. 1 (correct answer)
  3. 2
  4. 5
Explanation: If PQ is the base of the isosceles triangle, then the lengths of the other two sides, PR and QR, must be equal. The vertex R must lie on the perpendicular bisector of the base PQ. The segment PQ is horizontal, so its perpendicular bisector is a vertical line. The x-coordinate of this line is the midpoint of the x-coordinates of P and Q. x=−3+52=22=1x = \frac{-3+5}{2} = \frac{2}{2} = 1 Since R(k, 7) must lie on this line, its x-coordinate k must be 1. Alternatively, set the distances PR and QR equal: PR² = QR². (k−(−3))2+(7−2)2=(k−5)2+(7−2)2(k - (-3))^2 + (7-2)^2 = (k-5)^2 + (7-2)^2 (k+3)2+52=(k−5)2+52(k+3)^2 + 5^2 = (k-5)^2 + 5^2 (k+3)2=(k−5)2(k+3)^2 = (k-5)^2 k2+6k+9=k2−10k+25k^2 + 6k + 9 = k^2 - 10k + 25 16k=1616k = 16 k=1k=1

Question 16

The points A(1, 2), B(5, 4), and C(3, 8) are three vertices of a parallelogram ABCD, in order. Find the sum of the coordinates of the fourth vertex D.

  1. 1
  2. 5 (correct answer)
  3. 7
  4. 17
Explanation: In a parallelogram ABCD, the diagonals AC and BD bisect each other. This means they share the same midpoint. Let the coordinates of D be (x, y). First, find the midpoint of diagonal AC: MAC=(1+32,2+82)=(2,5)M_{AC} = \left(\frac{1+3}{2}, \frac{2+8}{2}\right) = (2, 5) This must also be the midpoint of diagonal BD: MBD=(5+x2,4+y2)M_{BD} = \left(\frac{5+x}{2}, \frac{4+y}{2}\right) Setting the coordinates equal: 5+x2=2⇒5+x=4⇒x=−1\frac{5+x}{2} = 2 \Rightarrow 5+x=4 \Rightarrow x=-1 4+y2=5⇒4+y=10⇒y=6\frac{4+y}{2} = 5 \Rightarrow 4+y=10 \Rightarrow y=6 So, the coordinates of D are (-1, 6). The sum of the coordinates is -1 + 6 = 5.

Question 17

A sphere has a diameter with endpoints P(-2, 4, 1) and Q(4, -2, 7). Find the square of the radius of the sphere.

  1. 18
  2. 27 (correct answer)
  3. 36
  4. 108
Explanation: The length of the diameter is the distance between points P and Q. The radius is half the length of the diameter. First, find the square of the diameter's length (d²). d2=(4−(−2))2+(−2−4)2+(7−1)2d^2 = (4 - (-2))^2 + (-2 - 4)^2 + (7 - 1)^2 d2=(6)2+(−6)2+(6)2d^2 = (6)^2 + (-6)^2 + (6)^2 d2=36+36+36=108d^2 = 36 + 36 + 36 = 108 The radius r is half the diameter, so r=d/2r = d/2. The square of the radius is r2=(d/2)2=d2/4r^2 = (d/2)^2 = d^2/4. r2=1084=27r^2 = \frac{108}{4} = 27 Alternatively, one could find the center of the sphere (the midpoint of PQ) and then find the distance from the center to one of the endpoints. Center C = ((-2+4)/2, (4-2)/2, (1+7)/2) = (1, 1, 4). Then r² = distance from C to P squared = (1−(−2))2+(1−4)2+(4−1)2=32+(−3)2+32=9+9+9=27(1 - (-2))^2 + (1 - 4)^2 + (4 - 1)^2 = 3^2 + (-3)^2 + 3^2 = 9 + 9 + 9 = 27.

Question 18

Find the perimeter of a triangle with vertices A(0, 0, 3), B(0, 4, 0), and C(5, 0, 0).

  1. 15
  2. 10 + 41\sqrt{41}
  3. 5 + 34\sqrt{34} + 41\sqrt{41} (correct answer)
  4. 5 + 41\sqrt{41} + 50\sqrt{50}
Explanation: The perimeter is the sum of the lengths of the three sides AB, BC, and AC. We use the 3D distance formula to find the length of each side. Length of AB=(0−0)2+(4−0)2+(0−3)2=02+42+(−3)2=16+9=25=5\text{Length of AB} = \sqrt{(0-0)^2 + (4-0)^2 + (0-3)^2} = \sqrt{0^2+4^2+(-3)^2} = \sqrt{16+9} = \sqrt{25} = 5 Length of BC=(5−0)2+(0−4)2+(0−0)2=52+(−4)2+02=25+16=41\text{Length of BC} = \sqrt{(5-0)^2 + (0-4)^2 + (0-0)^2} = \sqrt{5^2+(-4)^2+0^2} = \sqrt{25+16} = \sqrt{41} Length of AC=(5−0)2+(0−0)2+(0−3)2=52+02+(−3)2=25+9=34\text{Length of AC} = \sqrt{(5-0)^2 + (0-0)^2 + (0-3)^2} = \sqrt{5^2+0^2+(-3)^2} = \sqrt{25+9} = \sqrt{34} The perimeter is the sum of these lengths: Perimeter = 5 + 41\sqrt{41} + 34\sqrt{34}.

Question 19

The distance between the midpoints of the segments AB and CD is 17\sqrt{17}. The coordinates are A(1, 2), B(3, 8), C(-2, 5), and D(k, 3). Find a possible value for k.

  1. -4
  2. -2 (correct answer)
  3. 0
  4. 2
Explanation: First, find the midpoint of segment AB: MAB=(1+32,2+82)=(2,5)M_{AB} = \left(\frac{1+3}{2}, \frac{2+8}{2}\right) = (2, 5) Next, find the midpoint of segment CD in terms of k: MCD=(−2+k2,5+32)=(k−22,4)M_{CD} = \left(\frac{-2+k}{2}, \frac{5+3}{2}\right) = \left(\frac{k-2}{2}, 4\right) The distance between these midpoints is 17\sqrt{17}, so: (k−22−2)2+(4−5)2=17\left( \frac{k-2}{2} - 2 \right)^2 + (4-5)^2 = 17 (k−62)2+1=17\left( \frac{k-6}{2} \right)^2 + 1 = 17 (k−6)24=16\frac{(k-6)^2}{4} = 16 (k−6)2=64(k-6)^2 = 64 k−6=±8k-6 = \pm 8 This gives k = 14 or k = -2. Among the given choices, k = -2 is a possible value.

Question 20

Point P lies on the y-axis and is equidistant from A(-3, 4) and B(5, -2). Find the coordinates of P.

  1. (0, -1)
  2. (0, 1)
  3. (0, 1/3)
  4. (0, -1/3) (correct answer)
Explanation: When you encounter a problem about a point being equidistant from two given points, you're dealing with the perpendicular bisector concept. Since point P lies on the y-axis, it has coordinates (0, y) for some value y that you need to find. To find P, use the fact that it's equidistant from A(-3, 4) and B(5, -2). This means the distance from P to A equals the distance from P to B. Using the distance formula: (0−(−3))2+(y−4)2=(0−5)2+(y−(−2))2\sqrt{(0-(-3))^2 + (y-4)^2} = \sqrt{(0-5)^2 + (y-(-2))^2} Simplifying: 9+(y−4)2=25+(y+2)2\sqrt{9 + (y-4)^2} = \sqrt{25 + (y+2)^2} Square both sides to eliminate the square roots: 9+(y−4)2=25+(y+2)29 + (y-4)^2 = 25 + (y+2)^2 Expand the squared terms: 9+y2−8y+16=25+y2+4y+49 + y^2 - 8y + 16 = 25 + y^2 + 4y + 4 25−8y=29+4y25 - 8y = 29 + 4y −4=12y-4 = 12y y=−13y = -\frac{1}{3} Therefore, P is at (0, -1/3), which is answer D. Answer A (0, -1) results from making calculation errors in the algebra. Answer B (0, 1) comes from a sign error when solving for y. Answer C (0, 1/3) occurs if you get the correct fraction but miss the negative sign. Remember: equidistant problems often involve setting up equal distance expressions and squaring both sides to eliminate square roots. Always double-check your algebra, especially with fractions and negative signs.