Distance Between Two Points

The idea in one breath#

If two points lie on the same hor­i­zon­tal or ver­ti­cal line, find­ing the dis­tance between them is easy. You just sub­tract. For any other pair, there is a neat trick. The seg­ment join­ing them is the hypotenuse of a right tri­an­gle whose legs are hor­i­zon­tal and ver­ti­cal, and the Baud­hāyana–Pythago­ras the­o­rem does the rest. In this les­son you will build the for­mula, see why the signs of the dif­fer­ences never mat­ter, and then use it to test shapes and to find a mid­point.

Hor­i­zon­tal and ver­ti­cal seg­ments#

If two points share a $y$-coor­di­nate, the seg­ment between them is hor­i­zon­tal and its length is the dif­fer­ence of the $x$-coor­di­nates. From $(2, 5)$ to $(9, 5)$, for exam­ple, the length is $9 - 2 = 7$. It works the same way when the $x$-coor­di­nate is shared. From $(4, -3)$ to $(4, 6)$ the length is $6 - (-3) = 9$.

Take the larger minus the smaller, or find the dif­fer­ence and drop the sign. Either way, remem­ber that a length is never neg­a­tive.

A slanted seg­ment#

Let us try $A(2, 3)$ and $B(8, 11)$. Walk from $A$ hor­i­zon­tally to the point $C(8, 3)$ directly below $B$, then ver­ti­cally up to $B$. Tri­an­gle $ACB$ is right-angled at $C$, with

$$AC = 8 - 2 = 6, \qquad CB = 11 - 3 = 8$$

and so, by the Baud­hāyana–Pythago­ras the­o­rem,

$$AB = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

The dis­tance for­mula#

For $P(x_1, y_1)$ and $Q(x_2, y_2)$, the hor­i­zon­tal leg is $x_2 - x_1$ and the ver­ti­cal leg is $y_2 - y_1$, so

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

Here is why you need not worry about signs. Squar­ing removes any minus sign, so it does not mat­ter which point you call $P$ and which you call $Q$, or whether the coor­di­nates are neg­a­tive. If you reflect the whole pic­ture in an axis, the signs change but the dis­tances do not.

Test­ing a shape#

Once you can find dis­tances, you can answer quite a few ques­tions about a shape with­out draw­ing it.

Isosce­les or equi­lat­eral: com­pare the three sides and see how many match.

Right-angled: check whether the squares of two sides add up to the square of the third.

Collinear: three points lie on one line exactly when the two shorter dis­tances add up to the longest one.

The mid­point#

Find­ing the point halfway along a seg­ment is even eas­ier. For the seg­ment from $(x_1, y_1)$ to $(x_2, y_2)$, the coor­di­nates of the halfway point are just the aver­ages.

$$\displaystyle M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

Why does this work? Going from the first point to the sec­ond, $x$ changes by $x_2 - x_1$. At the halfway mark it has changed by half of that, giv­ing $\displaystyle x_1 + \frac{x_2 - x_1}{2} = \frac{x_1 + x_2}{2}$. The same rea­son­ing works for $y$.

Prac­tice#

1. Find the dis­tance between (a) $(1, 1)$ and $(4, 5)$; (b) $(-6, 2)$ and $(6, -3)$; (c) $(0, 0)$ and $(-9, -12)$.
2. Show that the tri­an­gle with ver­tices $(0, 3)$, $(4, 0)$ and $(4, 6)$ is isosce­les.
3. Are $(2, 1)$, $(5, 3)$ and $(11, 7)$ collinear?
4. Show that $(1, 1)$, $(4, 1)$ and $(1, 5)$ are the ver­tices of a right tri­an­gle and find its hypotenuse.
5. Which of $(-8, 6)$, $(9, 4)$ and $(0, -10)$ lie on the cir­cle of radius $10$ cen­tred at the ori­gin? Which is inside it?
6. Find the mid­point of $(7, -2)$ and $(-3, 8)$.
7. $M(0, 5)$ is the mid­point of $P(-6, 2)$ and $Q$. Find $Q$.
8. The ver­tices of a square are $(2, 1)$, $(-1, 2)$, $(-2, -1)$ and $(1, -2)$ in order. Ver­ify that all four sides are equal and that the two diag­o­nals are equal, and find the area.

Answers#