Distance Between Two Points

The Distance Formula in Space

Key Point (Distance formula): for $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$,

$$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

Why (the box argument): build the rectangular box with $P$ and $Q$ at opposite corners and edges parallel to the axes. Pythagoras in the base gives the base-diagonal$^2 = (x_2-x_1)^2 + (y_2-y_1)^2$; Pythagoras again in the vertical right triangle adds $(z_2-z_1)^2$. The 2D formula just grows one more square.

Special case: the distance from the origin to $Q(x, y, z)$ is $OQ = \sqrt{x^2 + y^2 + z^2}$.

Worked check: $P(1, -3, 4)$, $Q(-4, 1, 2)$: $PQ = \sqrt{25 + 16 + 4} = \sqrt{45} = 3\sqrt{5}$.

Collinearity test: three points are collinear when the two shorter distances add up to the longest: for $P(-2,3,5)$, $Q(1,2,3)$, $R(7,0,-1)$, the distances are $PQ = \sqrt{14}$, $QR = 2\sqrt{14}$, $PR = 3\sqrt{14}$, and $PQ + QR = PR$ — collinear.

Classifying Figures and Describing Loci

Classification by distances. Compute all side lengths (squared!), then compare:

An isosceles triangle has two equal sides; a right triangle satisfies $a^2 + b^2 = c^2$ (check the largest square against the sum of the other two — the check can fail, so always verify rather than assume); a parallelogram has equal opposite sides, and it is a rectangle only if the diagonals are also equal (a quadrilateral with $AC = \sqrt{3} \ne BD = \sqrt{155}$ is a parallelogram but not a rectangle).

Loci: geometry conditions become equations. Write $P(x, y, z)$, translate the condition through the distance formula, and simplify:

Equidistance from two fixed points gives a plane (the 3D perpendicular bisector — e.g. $10x + 6y - 18z - 29 = 0$ in Example 8 below); a constant sum of distances from $(\pm 4, 0, 0)$ equal to 10 gives $9x^2 + 25y^2 + 25z^2 = 225$; constant distance from a fixed point gives a sphere; and $PA^2 + PB^2 = k^2$ expands to a sphere-type equation.

[Board Tip] Compare distances through their squares — $\sqrt{18} = 3\sqrt{2}$ hides equalities that $18 = 18$ shows instantly, and no marks are lost to surd arithmetic.

Solved Examples

Example 1: A first distance

Find the distance between $P(1, -3, 4)$ and $Q(-4, 1, 2)$.

Solution:

Step 1 — Difference each coordinate. $\Delta x = -4 - 1 = -5$, $\Delta y = 1 - (-3) = 4$, $\Delta z = 2 - 4 = -2$.

Step 2 — Square, add, root. $PQ = \sqrt{25 + 16 + 4} = \sqrt{45}$.

Step 3 — Simplify the surd. $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$ units.

Takeaway: Signs of the differences never matter — squaring erases them, so difference in either order.

Example 2: Four quick distances

Find the distances between: (i) $(2,3,5)$, $(4,3,1)$; (ii) $(-3,7,2)$, $(2,4,-1)$; (iii) $(-1,3,-4)$, $(1,-3,4)$; (iv) $(2,-1,3)$, $(-2,1,3)$.

Solution:

Step 1 — (i). $\sqrt{2^2 + 0^2 + (-4)^2} = \sqrt{20} = 2\sqrt{5}$.

Step 2 — (ii). $\sqrt{5^2 + (-3)^2 + (-3)^2} = \sqrt{25 + 9 + 9} = \sqrt{43}$.

Step 3 — (iii). $\sqrt{2^2 + (-6)^2 + 8^2} = \sqrt{4 + 36 + 64} = \sqrt{104} = 2\sqrt{26}$.

Step 4 — (iv). $\sqrt{(-4)^2 + 2^2 + 0^2} = \sqrt{20} = 2\sqrt{5}$.

Takeaway: A zero difference (equal coordinates) just drops a term — spot those first and the arithmetic halves.

Example 3: Collinearity

Show that $P(-2, 3, 5)$, $Q(1, 2, 3)$ and $R(7, 0, -1)$ are collinear.

Solution:

Step 1 — Compute all three distances. $PQ = \sqrt{9 + 1 + 4} = \sqrt{14}$; $QR = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14}$; $PR = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14}$.

Step 2 — Test the degenerate triangle condition. $PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} = PR$.

Step 3 — Conclude. The "triangle" is flat: $Q$ lies on segment $PR$, so the points are collinear. ∎

Takeaway: Same-surd answers ($\sqrt{14}$, $2\sqrt{14}$, $3\sqrt{14}$) are the signature of collinear points — the common factor is the direction's length scale.

Example 4: Testing for a right angle

Are $A(3, 6, 9)$, $B(10, 20, 30)$, $C(25, -41, 5)$ the vertices of a right-angled triangle?

Solution:

Step 1 — Squared sides only. $AB^2 = 7^2 + 14^2 + 21^2 = 49 + 196 + 441 = 686$; $BC^2 = 15^2 + 61^2 + 25^2 = 225 + 3721 + 625 = 4571$; $CA^2 = 22^2 + 47^2 + 4^2 = 484 + 2209 + 16 = 2709$.

Step 2 — Check Pythagoras against the largest. Largest is $BC^2 = 4571$; the other two give $686 + 2709 = 3395 \ne 4571$.

Step 3 — Conclude. No pairing works (the largest square is the only candidate for the hypotenuse): not right-angled.

Takeaway: Always test — never assume a triangle is right-angled because the numbers look engineered; here the check fails.

Example 5: Isosceles and right triangles

Verify: (i) $(0,7,-10)$, $(1,6,-6)$, $(4,9,-6)$ form an isosceles triangle; (ii) $(0,7,10)$, $(-1,6,6)$, $(-4,9,6)$ form a right triangle.

Solution:

Step 1 — (i) Squared sides. $AB^2 = 1 + 1 + 16 = 18$, $BC^2 = 9 + 9 + 0 = 18$, $CA^2 = 16 + 4 + 16 = 36$.

Step 2 — (i) Compare. $AB = BC = \sqrt{18} = 3\sqrt{2}$: two equal sides — isosceles. ∎

Step 3 — (ii) Same squares, stronger conclusion. $AB^2 = 18$, $BC^2 = 18$, $CA^2 = 36$: now check $AB^2 + BC^2 = 18 + 18 = 36 = CA^2$ — right-angled at $B$. ∎

Takeaway: The same distance data can hold more than one property — compute once, test every classification.

Example 6: A parallelogram

Verify that $A(-1,2,1)$, $B(1,-2,5)$, $C(4,-7,8)$, $D(2,-3,4)$ are the vertices of a parallelogram.

Solution:

Step 1 — Squared sides in order. $AB^2 = 4 + 16 + 16 = 36$; $BC^2 = 9 + 25 + 9 = 43$; $CD^2 = 4 + 16 + 16 = 36$; $DA^2 = 9 + 25 + 9 = 43$.

Step 2 — Compare opposite pairs. $AB = CD = 6$ and $BC = DA = \sqrt{43}$: both pairs of opposite sides equal.

Step 3 — Conclude. $ABCD$ is a parallelogram. ∎

Takeaway: Keep the vertices in cyclic order — comparing $AB$ with $CD$ (not $BC$) is what "opposite sides" means.

Example 7: The equidistant plane

Find the set of points equidistant from $A(1, 2, 3)$ and $B(3, 2, -1)$.

Solution:

Step 1 — Translate the condition. $PA = PB$; square to avoid roots: $(x-1)^2 + (y-2)^2 + (z-3)^2 = (x-3)^2 + (y-2)^2 + (z+1)^2$.

Step 2 — Expand and cancel. The $x^2, y^2, z^2$ terms cancel (and so do the $y$-terms, since both points share $y = 2$): $-2x - 6z + 14 = -6x + 2z + 14$, which collects to $4x - 8z = 0$.

Step 3 — Conclude. $x - 2z = 0$ — a plane (the 3D perpendicular bisector of $AB$).

Takeaway: Squaring an equidistance condition always kills the squares — the locus of $PA = PB$ is forever linear: a plane.

Example 8: Another equidistant plane

Find the set of points equidistant from $A(3, 4, -5)$ and $B(-2, 1, 4)$.

Solution:

Step 1 — Square both distances. $(x-3)^2 + (y-4)^2 + (z+5)^2 = (x+2)^2 + (y-1)^2 + (z-4)^2$.

Step 2 — Expand. Left: $-6x - 8y + 10z + 50$; right: $4x - 2y - 8z + 21$ (after the squares cancel).

Step 3 — Collect. $-6x - 8y + 10z + 50 = 4x - 2y - 8z + 21$ gives $10x + 6y - 18z - 29 = 0$.

Takeaway: Move everything to one side at the END — collecting term-by-term mid-expansion is where sign errors breed.

Example 9: A constant sum of distances

Find the equation of the set of points $P$ whose distances from $A(4, 0, 0)$ and $B(-4, 0, 0)$ sum to 10.

Solution:

Step 1 — Set up. $\sqrt{(x-4)^2 + y^2 + z^2} + \sqrt{(x+4)^2 + y^2 + z^2} = 10$.

Step 2 — Isolate one root and square. After the first squaring and simplification: $5\sqrt{(x+4)^2 + y^2 + z^2} = 25 + 4x$.

Step 3 — Square again and collect. $25(x^2 + 8x + 16 + y^2 + z^2) = 625 + 200x + 16x^2$ collapses to $9x^2 + 25y^2 + 25z^2 = 225$.

Takeaway: This is the ellipse's string definition lifted to 3D — with $a = 5$, $c = 4$, $b^2 = 9$, the equation is its solid-of-revolution analogue.

Example 10: A sum of squared distances

Find the set of points $P$ with $PA^2 + PB^2 = 2k^2$, where $A = (3, 4, 5)$ and $B = (-1, 3, -7)$.

Solution:

Step 1 — Write both squared distances. $(x-3)^2 + (y-4)^2 + (z-5)^2 + (x+1)^2 + (y-3)^2 + (z+7)^2 = 2k^2$.

Step 2 — Expand and merge. $2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z + (9+16+25+1+9+49) = 2k^2$.

Step 3 — Collect the constant. The constants sum to 109: $2x^2 + 2y^2 + 2z^2 - 4x - 14y + 4z = 2k^2 - 109$ — a sphere-type equation.

Takeaway: Squared-distance conditions keep their squares, so they produce spheres; plain equidistance cancels them and produces planes.

Example 11: Distance from the origin — a drill

Which is farther from the origin: $A(2, 3, 6)$ or $B(4, 4, 4)$?

Solution:

Step 1 — Compare squares. $OA^2 = 4 + 9 + 36 = 49$; $OB^2 = 16 + 16 + 16 = 48$.

Step 2 — Read off. $49 > 48$: $A$ is farther ($OA = 7$, $OB = 4\sqrt{3} \approx 6.93$).

Takeaway: Never take roots to compare distances — $49$ vs $48$ settles instantly what $7$ vs $6.928\ldots$ obscures.

Example 12: An equilateral check

Show that $A(1, 1, 0)$, $B(0, 1, 1)$, $C(1, 0, 1)$ form an equilateral triangle.

Solution:

Step 1 — Squared sides. $AB^2 = 1 + 0 + 1 = 2$; $BC^2 = 1 + 1 + 0 = 2$; $CA^2 = 0 + 1 + 1 = 2$.

Step 2 — Conclude. All sides $\sqrt{2}$: equilateral. ∎

Step 3 — See the symmetry. The three points permute the coordinates of $(1, 1, 0)$ — permutation symmetry forces equal distances before any arithmetic.

Takeaway: Points that are coordinate-permutations of each other are automatically "equally spread" — a symmetry worth spotting in exams.