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AI SL · 12.3

Coordinate Geometry

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12.3.1

3D coordinates: distance and midpoint

12.3.1

3D coordinates: distance and midpoint

Target: I can find the midpoint of two points in 2D or 3D, and use a midpoint to find unknown coordinates. (+5 more)

By the end of this section

Key points

Distance and midpoint

  • The distance between \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\).
  • The midpoint is the average of each coordinate: \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)\).
  • The distance from the origin to a point is the square root of the sum of the squares of its coordinates.
  • Squaring removes the sign, so an equation for an unknown coordinate usually has two solutions. Check which one the question wants.

Cuboids and surfaces

  • The diagonal of a cuboid \(l \times w \times h\) is \(\sqrt{l^2+w^2+h^2}\): Pythagoras twice, first across the base, then up.
  • To read coordinates from a diagram, put O at a corner on the ground. The x, y and z-coordinates are the distances along the three edges.
  • The shortest path over the surface of a cuboid is a straight line on an unfolded net. Try the three unfoldings: \(\sqrt{(a+b)^2+c^2}\), \(\sqrt{(a+c)^2+b^2}\), \(\sqrt{(b+c)^2+a^2}\). Take the smallest.

Lines and shapes in 2D

  • The gradient through two points is \(\frac{y_2-y_1}{x_2-x_1}\). A line \(ax+by+d=0\) has gradient \(-\frac{a}{b}\).
  • Perpendicular lines have gradients that multiply to −1. The shortest distance from a point to a line is along the perpendicular.
  • The diagonals of a square and of a rhombus bisect each other at right angles. In a square they are also equal in length.
  • To turn a step \((p, q)\) through 90°, use \((-q, p)\).

On the calculator

Solving two equations together

  • Write each equation in the form ax + by = c.
  • Use the calculator to solve the pair for x and y.
  • Check the point in both equations.

Screenshot and video to come

Solving for an unknown coordinate

  • Write the distance equation with the unknown, for example \((k-4)^2 + 61 = 81\).
  • Use the solver on this equation.
  • Give both values of k: they are the two points at that distance.

Screenshot and video to come

Explore: distance and midpoint in 3D

Drag A and B across, and use the y slider to move a point in and out. AB is the diagonal of the box between them: Pythagoras across the base, then up. Then try "Find the missing coordinate".

Coordinates of point A

Distance
 The same in two steps Midpoint: the average of each coordinate

Drawn in oblique projection: y goes in at an angle and is drawn shorter, as on paper. With a point selected, the arrow keys move it in x and z.

Drag A or B to move it.

  1. Why is the angle at P a right angle? Which two lines meet there?
  2. Put A at O. How does the distance formula become √(x² + y² + z²), the diagonal of a cuboid?
  3. Swap A and B. Does the distance change? Does the midpoint? Why?
  4. In "Find the missing coordinate", why are there two values of k when you are given a distance, but only one point B when you are given a midpoint?

Explore: two lines, one solution

Each equation is a straight line. Type whole numbers for a, b and c in each line and watch where the lines meet. Try the three buttons first.

Line 1 x + y =
Line 2 x + y =

Solution of the pair

Change a number to move a line.

  1. Where the lines cross, the point is on both lines. Why does that make its coordinates the solution of both equations?
  2. Make the lines parallel. What is the same about the numbers a and b in the two lines?
  3. Keep them parallel and change c in one line. When do they become the same line? How many solutions then?
  4. Could a pair of straight lines meet exactly twice? Why not?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Midpoint, then distance from the origin

Points A and B have coordinates \(A(5, -12, 7)\) and \(B(-3, 4, -1)\). M is the midpoint of AB. Find the distance from the origin O to M.

Working
  1. Midpoint

    \(M = \left(\frac{5+(-3)}{2}, \frac{-12+4}{2}, \frac{7+(-1)}{2}\right) = (1, -4, 3)\)

    Add each pair of coordinates and halve.

  2. Distance from O

    \(OM = \sqrt{1+16+9} = \sqrt{26} = 5.10\)

    O is (0, 0, 0), so the differences are the coordinates of M.

Video walkthrough coming
An unknown coordinate

The point \(P(3a, a, 5)\) is twice as far from the origin as the point \(Q(1, -2, 4)\). Find the positive value of a.

Working
  1. Distance OQ

    \(OQ = \sqrt{1^2+(-2)^2+4^2} = \sqrt{21}\)

    Keep it as a surd. Square later.

  2. Equation

    \(OP = 2\sqrt{21}\), so \((3a)^2+a^2+5^2 = 4 \times 21\)

    Square both sides: \((2\sqrt{21})^2 = 84\).

  3. Solve

    \(10a^2 + 25 = 84\), so \(a^2 = 5.90\), \(a = \pm 2.43\)

    Squaring gives two values. The question asks for the positive one: a = 2.43.

Video walkthrough coming
The diagonal of a room

A room is a cuboid 5.5 m long, 3.2 m wide and 2.6 m high. A cable runs in a straight line from corner A on the floor to the opposite corner G on the ceiling. Find the length of the cable.

5.5 m3.2 m2.6 mABCDEFGH
Working
  1. Across the floor

    \(AC^2 = 5.5^2 + 3.2^2 = 40.49\)

    Pythagoras in the floor rectangle.

  2. Up to the ceiling

    \(AG^2 = AC^2 + 2.6^2 = 47.25\)

    Triangle ACG has a right angle at C.

  3. Length

    \(AG = 6.87\) m

    Same as \(\sqrt{l^2+w^2+h^2}\). With coordinates: A(0, 0, 0) and G(5.5, 3.2, 2.6).

Video walkthrough coming
Unknowns from a midpoint

The midpoint of \(A(a, 3, -2)\) and \(B(5, b, c)\) is \(M(4, 1, 3)\). Find a, b and c.

Working
  1. One equation each

    \(\frac{a+5}{2}=4,\quad \frac{3+b}{2}=1,\quad \frac{-2+c}{2}=3\)

    The midpoint is the average of each coordinate, so each coordinate gives its own equation.

  2. Solve

    \(a = 3,\quad b = -1,\quad c = 8\)

    Multiply by 2, then subtract the known number.

Video walkthrough coming
A square from its diagonal

ABCD is a square. A is \((1, -2)\) and C is \((7, 2)\). Find the coordinates of B and D, and the side length.

ABCD
Working
  1. Midpoint of AC

    \(M = \left(\frac{1+7}{2}, \frac{-2+2}{2}\right) = (4, 0)\)

    The diagonals of a square bisect each other. M is also the midpoint of BD.

  2. Half a diagonal

    From M to C: \((3, 2)\). Turn 90°: \((-2, 3)\)

    The diagonals are perpendicular and equal. Swap the numbers and change one sign to turn a step through 90°.

  3. B and D

    \(B = (4-2, 0+3) = (2, 3)\), \(D = (4+2, 0-3) = (6, -3)\)

    Go from M one way for B and the other way for D. Check that MB is as long as MA.

  4. Side

    \(AB = \sqrt{(2-1)^2+(3-(-2))^2} = \sqrt{26} = 5.10\)

    Use any two neighbouring vertices.

Video walkthrough coming
A line, its perpendicular and a distance

The line L passes through \(P(-2, 1)\) and \(Q(4, 5)\). Find the shortest distance from \(R(5, -1)\) to L.

−4−22468−4−22468xyPQRF
Working
  1. Equation of L

    Gradient \(= \frac{5-1}{4-(-2)} = \frac23\). \(y - 1 = \frac23(x+2)\), so \(2x - 3y + 7 = 0\)

    The gradient of a line through two points is the change in y over the change in x.

  2. Perpendicular through R

    Gradient \(-\frac32\). \(y + 1 = -\frac32(x-5)\), so \(3x + 2y - 13 = 0\)

    Perpendicular gradients multiply to −1. The shortest distance is along the perpendicular.

  3. Where they meet

    Solve the pair: \(F = \left(\frac{25}{13}, \frac{47}{13}\right)\)

    Use the calculator to solve the two equations together.

  4. Distance RF

    \(RF = \sqrt{(5-\frac{25}{13})^2+(-1-\frac{47}{13})^2} = 5.55\)

    Distance formula in two dimensions.

Video walkthrough coming
The shortest path over a box

A beetle crawls over the surface of a closed box 9 cm long, 4 cm wide and 3 cm high, from corner A to the opposite corner G. It stays on the surface. Find the shortest distance it can crawl.

9 cm4 cm3 cmABCDEFGH
Working
  1. Unfold

    Open out two faces into one rectangle. A and G are then opposite corners of the rectangle.

    A straight line on the flat net is the shortest path on the surface. There are three ways to unfold.

  2. The three rectangles

    \(\sqrt{(4+3)^2+9^2} = 11.4\); \(\sqrt{(9+3)^2+4^2} = 12.6\); \(\sqrt{(9+4)^2+3^2} = 13.3\)

    Each rectangle has one edge equal to the sum of two dimensions, the other equal to the third.

  3. Shortest

    Shortest \(= 11.4\) cm

    Compare all three and take the smallest. It is the one with the longest edge kept alone.

Video walkthrough coming