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Geometry and Trigonometry - Distance and midpoint

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The distance between two points P1(x1,y1)P_1(x_1, y_1) and P2(x2,y2)P_2(x_2, y_2) in a 2D plane is the length of the straight line segment connecting them, derived from the Pythagorean theorem.

A right-angled triangle formed by the distance between two points P1 and P2 showing the horizontal and vertical differences.
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The midpoint MM of a line segment is the point that divides the segment into two equal parts. Its coordinates are the averages of the xx and yy coordinates of the endpoints.

A line segment AB with the midpoint M located exactly in the center.
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In 3D space, the distance formula extends to include the zz-coordinate, calculating the length of a vector in three dimensions using (Δx)2+(Δy)2+(Δz)2\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}.

A 3D box illustrating the diagonal distance between opposite vertices across three dimensions.
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The midpoint in 3D is found by separately averaging the xx, yy, and zz coordinates of the two endpoints: M=(x1+x22,y1+y22,z1+z22)M = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}).

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To find an unknown endpoint when given one endpoint and the midpoint, you can use the midpoint formula algebraically or by considering the 'step' (displacement) from the endpoint to the midpoint and repeating it.

📐Formulae

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

d3D=(x2−x1)2+(y2−y1)2+(z2−z1)2d_{3D} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

M3D=(x1+x22,y1+y22,z1+z22)M_{3D} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)

💡Examples

Problem 1:

Find the distance between the points A(2,−3)A(2, -3) and B(6,0)B(6, 0).

Solution:

d=(6−2)2+(0−(−3))2d = \sqrt{(6 - 2)^2 + (0 - (-3))^2} d=(4)2+(3)2d = \sqrt{(4)^2 + (3)^2} d=16+9=25=5d = \sqrt{16 + 9} = \sqrt{25} = 5

Explanation:

Substitute the coordinates into the 2D distance formula. Note that (0−(−3))(0 - (-3)) becomes +3+3.

Problem 2:

A line segment has endpoints M(10,4,−2)M(10, 4, -2) and N(4,8,6)N(4, 8, 6). Determine the coordinates of the midpoint.

Solution:

xmid=10+42=7x_{mid} = \frac{10 + 4}{2} = 7 ymid=4+82=6y_{mid} = \frac{4 + 8}{2} = 6 zmid=−2+62=2z_{mid} = \frac{-2 + 6}{2} = 2 Midpoint=(7,6,2)Midpoint = (7, 6, 2)

Explanation:

To find the midpoint in 3D, calculate the average of the xx, yy, and zz coordinates separately.

Problem 3:

The midpoint of segment PQPQ is (3,5)(3, 5). If point PP is (1,2)(1, 2), find the coordinates of point QQ.

Solution:

3=1+xQ2  ⟹  6=1+xQ  ⟹  xQ=53 = \frac{1 + x_Q}{2} \implies 6 = 1 + x_Q \implies x_Q = 5 5=2+yQ2  ⟹  10=2+yQ  ⟹  yQ=85 = \frac{2 + y_Q}{2} \implies 10 = 2 + y_Q \implies y_Q = 8 Q=(5,8)Q = (5, 8)

Explanation:

Set up the midpoint formula equations using the known midpoint and one endpoint, then solve for the unknown coordinates xQx_Q and yQy_Q.

Problem 4:

A surveyor is measuring a rectangular plot. Point RR is located at (1,2)(1, 2) and point SS is at (7,10)(7, 10). Calculate the exact length of the diagonal RSRS.

Graph showing points R(1,2) and S(7,10) connected by a line segment.

Solution:

  1. Identify coordinates: (x1,y1)=(1,2)(x_1, y_1) = (1, 2) and (x2,y2)=(7,10)(x_2, y_2) = (7, 10).
  2. Use the distance formula: d=(7−1)2+(10−2)2d = \sqrt{(7 - 1)^2 + (10 - 2)^2}
  3. Simplify: d=62+82d = \sqrt{6^2 + 8^2}
  4. d=36+64=100d = \sqrt{36 + 64} = \sqrt{100}
  5. d=10d = 10 units.

Explanation:

To find the distance, we calculate the difference between the xx and yy coordinates, square them, sum them, and take the square root. This is the magnitude of the vector from RR to SS.

Problem 5:

A transmission tower is anchored at point A(−4,5)A(-4, 5) and its base is at point B(8,−1)B(8, -1). A technician needs to install a sensor at the midpoint MM of the support cable ABAB. Find the coordinates of MM.

Coordinate plane showing a segment from A(-4, 5) to B(8, -1) with the midpoint M at (2, 2).

Solution:

  1. Identify coordinates: (x1,y1)=(−4,5)(x_1, y_1) = (-4, 5) and (x2,y2)=(8,−1)(x_2, y_2) = (8, -1).
  2. Apply the midpoint formula: M=(−4+82,5+(−1)2)M = \left( \frac{-4 + 8}{2}, \frac{5 + (-1)}{2} \right)
  3. Calculate xx-coordinate: x=42=2x = \frac{4}{2} = 2
  4. Calculate yy-coordinate: y=42=2y = \frac{4}{2} = 2
  5. M=(2,2)M = (2, 2).

Explanation:

The midpoint is found by averaging the horizontal and vertical positions of the two endpoints. The resulting coordinate (2,2)(2, 2) lies exactly halfway between the anchor and the base.