SME
Grade 12 Math · Problem type
Distance and midpoint in space
Use this for “Find the distance between P(1, 2, 3) and Q(3, 5, 9) and the midpoint of PQ.”
Diagram
Going from P(1, 2, 3) to Q(3, 5, 9), the legs are Δx = 2, Δy = 3 and Δz = 6.
PQ
=
√
2
2
+
3
2
+
6
2
=
7.
The midpoint M(2, 3.5, 6) averages each coordinate.
This is an example. Type your own values to start a new problem.
d
2
=
(
x₂
−
x₁
)²
+
(
y₂
−
y₁
)²
+
(
z₂
−
z₁
)²
xM
x of the midpoint M
worked out (not typed)
yM
y of the midpoint M
worked out (not typed)
zM
z of the midpoint M
worked out (not typed)
Formulas
d
=
√(
(
x₂
−
x₁
)
2
+
(
y₂
−
y₁
)
2
+
(
z₂
−
z₁
)
2
)
7
=
√(
(3 − 1)
2
+
(5 − 2)
2
+
(9 − 3)
2
)
xM
=
x₁
+
x₂
2
2
=
1 + 3
2
yM
=
y₁
+
y₂
2
3.5
=
2 + 5
2
zM
=
z₁
+
z₂
2
6
=
3 + 9
2
Assumptions
•
A point in space has three coordinates (x, y, z); the axes meet at right angles.
•
The distance is √(Δx² + Δy² + Δz²): the Pythagorean theorem across the floor, then up.
•
The midpoint averages each coordinate.
Step-by-step
WE KNOW
x₁ = 1, y₁ = 2, z₁ = 3, x₂ = 3, y₂ = 5, z₂ = 9
FIND
distance PQ (d), x of the midpoint M (xM), y of the midpoint M (yM), z of the midpoint M (zM)
Step 1 · Find distance PQ (d)
Use d = √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²)
Subtract matching coordinates, square, add, and take the square root.
d
=
√(
(3 − 1)
2
+
(5 − 2)
2
+
(9 − 3)
2
)
d
=
√
2
2
+
3
2
+
6
2
d
=
√
4 + 9 + 36
d
=
√
49
d
=
7
Step 2 · Find x of the midpoint M (xM)
Use xM = (x₁ + x₂) ÷ 2
The midpoint’s x is halfway between the two xs: their mean.
xM
=
1 + 3
2
xM
=
4
2
xM = 2
Step 3 · Find y of the midpoint M (yM)
Use yM = (y₁ + y₂) ÷ 2
The midpoint’s y is halfway between the two ys: their mean.
yM
=
2 + 5
2
yM
=
7
2
yM = 3.5
Step 4 · Find z of the midpoint M (zM)
Use zM = (z₁ + z₂) ÷ 2
The midpoint’s z is halfway between the two zs: their mean.
zM
=
3 + 9
2
zM
=
12
2
zM = 6
Check
Put the numbers back in. Both sides should match.
7
=
√(
(3 − 1)
2
+
(5 − 2)
2
+
(9 − 3)
2
)
✓
2
=
1 + 3
2
✓
3.5
=
2 + 5
2
✓
6
=
3 + 9
2
✓
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