Distance Between Two Points in the Plane, Class 9

Points on a ver­ti­cal line

In the same way, $(x, y_1)$ and $(x, y_2)$ lie on a line par­al­lel to the y-axis, and the dis­tance between them is $|y_2 - y_1|$. In the floor plan of Reiaan's room, this is how we mea­sure the bath­room door B₁B₂, which runs along the y-axis; the room door and the wardrobe's front edge W₁W₂ lie along the x-axis, so they use the rule for hor­i­zon­tal lines.

Worked exam­ple 1: straight along a grid line

Find the dis­tance between (a) $(-4, 3)$ and $(5, 3)$, and (b) $(2, -6)$ and $(2, 1)$.

(a) The y-coor­di­nates are equal, so the dis­tance is $|5 - (-4)| = |9| = 9$ units.

(b) The x-coor­di­nates are equal, so the dis­tance is $|1 - (-6)| = |7| = 7$ units. Sub­tract­ing the other way gives $|-6 - 1| = |-7| = 7$ as well, so the order does not mat­ter.

Dis­tances along a slant

Split­ting the jour­ney into two moves

The book looks at tri­an­gle ADM with A$(3, 4)$, D$(7, 1)$ and M$(9, 6)$, an acute-angled tri­an­gle in the first quad­rant. None of its sides is par­al­lel to an axis. How long is AD?

Its Think and Reflect prompts point the way. Going from A$(3, 4)$ to D$(7, 1)$:

• along the x-axis we move from 3 to 7, a dis­tance of $7 - 3 = 4$ units;
• along the y-axis we move from 4 down to 1, a dis­tance of $4 - 1 = 3$ units.

Yes, these two moves give us AD. Mark the point C$(3, 1)$, straight below A and level with D. AC is ver­ti­cal and CD is hor­i­zon­tal, so the angle at C is a right angle, and AD is the hypotenuse of the right-angled tri­an­gle ACD.

By the Baud­hāyana–Pythago­ras the­o­rem,

$$AD = \sqrt{CD^2 + AC^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ units}.$$

The other two sides

From D$(7, 1)$ to M$(9, 6)$ the shifts are $9 - 7 = 2$ across and $6 - 1 = 5$ up:

$$DM = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \text{ units}.$$

From M$(9, 6)$ to A$(3, 4)$ the shifts are $9 - 3 = 6$ across and $6 - 4 = 2$ in height:

$$MA = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \text{ units}.$$

The dis­tance for­mula

The same steps work for any two points A$(x_1, y_1)$ and D$(x_2, y_2)$. Take the helper point F$(x_1, y_2)$. Then AF is ver­ti­cal with length $|y_2 - y_1|$, FD is hor­i­zon­tal with length $|x_2 - x_1|$, and the angle at F is a right angle. The the­o­rem gives

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

Why the signs do not mat­ter. The dif­fer­ences $x_2 - x_1$ and $y_2 - y_1$ might be neg­a­tive, depend­ing on which way we travel. But a num­ber and its neg­a­tive have the same square; for exam­ple $(-3)^2 = 3^2 = 9$. So the for­mula gives the same answer whichever point we call the first, and whether the shifts are left or right, up or down. The squares sim­ply mea­sure the size of each shift.

If the two points hap­pen to be on a hor­i­zon­tal line, then $y_2 - y_1 = 0$ and the for­mula becomes $\sqrt{(x_2 - x_1)^2} = |x_2 - x_1|$, the rule we started with. So the for­mula cov­ers every case.

Worked exam­ple 2: a whole-num­ber dis­tance

Find the dis­tance between P$(1, 2)$ and Q$(7, 10)$.

$$PQ = \sqrt{(7 - 1)^2 + (10 - 2)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ units}.$$

Worked exam­ple 3: points in dif­fer­ent quad­rants

Find the dis­tance between G$(-4, 5)$ in Quad­rant II and H$(2, -3)$ in Quad­rant IV.

The shift across is $2 - (-4) = 6$ and the shift in height is $-3 - 5 = -8$. Take care with the dou­ble minus in the first one.

$$GH = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ units}.$$

The neg­a­tive shift of $-8$ gives the same square as a shift of $8$, just as we expected.

Worked exam­ple 4: what kind of tri­an­gle?

Show that K$(1, 1)$, L$(4, 5)$ and N$(8, 2)$ form an isosce­les right-angled tri­an­gle.

$$\begin{aligned} KL &= \sqrt{(4 - 1)^2 + (5 - 1)^2} = \sqrt{9 + 16} = 5 \\ LN &= \sqrt{(8 - 4)^2 + (2 - 5)^2} = \sqrt{16 + 9} = 5 \\ KN &= \sqrt{(8 - 1)^2 + (2 - 1)^2} = \sqrt{49 + 1} = \sqrt{50} \end{aligned}$$

Two sides are equal ($KL = LN = 5$), so the tri­an­gle is isosce­les. Also $KL^2 + LN^2 = 25 + 25 = 50 = KN^2$. By the con­verse of the Baud­hāyana–Pythago­ras the­o­rem, the angle at L, oppo­site the longest side KN, is a right angle.

Worked exam­ple 5: find­ing an unknown point

Find the point on the x-axis that is equally far from U$(2, 5)$ and V$(-2, 3)$.

A point on the x-axis has the form T$(x, 0)$. We need $TU = TV$. Both are lengths, so it is the same to ask for $TU^2 = TV^2$:

$$\begin{aligned} (x - 2)^2 + (0 - 5)^2 &= (x + 2)^2 + (0 - 3)^2 \\ x^2 - 4x + 4 + 25 &= x^2 + 4x + 4 + 9 \\ -4x + 29 &= 4x + 13 \\ 16 &= 8x \\ x &= 2 \end{aligned}$$

So T $= (2, 0)$. Check: $TU = \sqrt{0 + 25} = 5$ and $TV = \sqrt{16 + 9} = 5$. Equal, as required.

Dis­tances and reflec­tion

Reflect­ing tri­an­gle ADM in the y-axis

Reflect­ing a point in the y-axis keeps its height and moves it to the same dis­tance on the other side of the axis, so $(x, y)$ becomes $(-x, y)$. The tri­an­gle ADM becomes A′D′M′ with

$$A'(-3, 4), \qquad D'(-7, 1), \qquad M'(-9, 6).$$

The helper point C$(3, 1)$ becomes C′$(-3, 1)$. Now all x-coor­di­nates are neg­a­tive, which is a good test of the for­mula.

For A′D′: the shift across is $C'D' = -3 - (-7) = 4$ and the shift in height is $A'C' = 4 - 1 = 3$, so

$$A'D' = \sqrt{4^2 + 3^2} = 5 \text{ units}.$$

Sim­i­larly,

$$\begin{aligned} D'M' &= \sqrt{(-9 - (-7))^2 + (6 - 1)^2} = \sqrt{(-2)^2 + 5^2} = \sqrt{29} \\ M'A' &= \sqrt{(-3 - (-9))^2 + (4 - 6)^2} = \sqrt{6^2 + (-2)^2} = \sqrt{40} \end{aligned}$$

These are exactly the side lengths of ADM. Reflec­tion has kept every length the same.

Why the lengths can­not change

Take any two points $(x_1, y_1)$ and $(x_2, y_2)$. Their images are $(-x_1, y_1)$ and $(-x_2, y_2)$. The new shift across is $-x_2 - (-x_1) = -(x_2 - x_1)$, which is the old shift with its sign changed, and the shift in height is unchanged. Squar­ing removes the sign change, so the dis­tance for­mula gives exactly the same num­ber. This holds for every pair of points, so every length is pre­served.

What stays the same, and what changes

The book asks what the reflec­tion keeps and what it alters.

• Same: all three side lengths, so the tri­an­gle's size and shape are unchanged (its angles and area are unchanged too). The y-coor­di­nates of the ver­tices are unchanged.
• Changed: each x-coor­di­nate has switched sign, the tri­an­gle has moved from Quad­rant I to Quad­rant II, and it is now a mir­ror image, so going A → D → M turns the oppo­site way round (clock­wise instead of anti­clock­wise, or the reverse).

Reflect­ing in the x-axis instead? The same obser­va­tions hold, with the roles swapped: $(x, y)$ becomes $(x, -y)$, the x-coor­di­nates stay, the y-coor­di­nates switch sign, and the tri­an­gle moves from Quad­rant I to Quad­rant IV. The shift in height changes sign, the shift across does not, and squar­ing again keeps every length.

Worked exam­ple 6: reflect­ing in the x-axis

Tri­an­gle PQR has P$(2, 1)$, Q$(6, 1)$ and R$(6, 4)$. Reflect it in the x-axis and com­pare the side lengths.

The images are P′$(2, -1)$, Q′$(6, -1)$ and R′$(6, -4)$.

Orig­i­nal: $PQ = 6 - 2 = 4$, $QR = 4 - 1 = 3$, $PR = \sqrt{4^2 + 3^2} = 5$.

Image: $P'Q' = 6 - 2 = 4$, $Q'R' = |-4 - (-1)| = 3$, $P'R' = \sqrt{4^2 + (-3)^2} = 5$.

The sides are 4, 3 and 5 in both tri­an­gles, so the reflec­tion has pre­served all the lengths.

Com­mon mis­takes

• Adding the coor­di­nates in place of sub­tract­ing them. The for­mula uses dif­fer­ences: $x_2 - x_1$, not $x_2 + x_1$.
• Los­ing a sign when sub­tract­ing a neg­a­tive, for exam­ple writ­ing $2 - (-4) = -2$ in place of 6.
• Writ­ing $\sqrt{a^2 + b^2} = a + b$. For exam­ple, $\sqrt{4^2 + 3^2} = 5$, not 7.
• Mix­ing coor­di­nates, such as sub­tract­ing an x-coor­di­nate from a y-coor­di­nate.
• Round­ing a square root too early. Keep $\sqrt{29}$ exact until the very end.
• Think­ing reflec­tion changes lengths because the coor­di­nates change. Only the signs change, and squar­ing removes them.

Try these

1. Find the dis­tance between $(0, 0)$ and $(5, 12)$. Answer: 13 units.
2. Find the dis­tance between $(-3, 2)$ and $(3, -6)$. Answer: 10 units.
3. Find the dis­tance between $(1, -1)$ and $(1, 7)$. Answer: 8 units.
4. Find the dis­tance between $(-2, -3)$ and $(4, 5)$. Answer: 10 units.
5. Is the tri­an­gle with ver­tices $(0, 0)$, $(3, 4)$ and $(-4, 3)$ isosce­les? Answer: Yes; two sides are 5 and the third is $\sqrt{50}$ (it is also right-angled at the ori­gin).
6. Reflect the tri­an­gle $(-5, 2)$, $(-1, 2)$, $(-1, 5)$ in the y-axis and give its side lengths. Answer: $(5, 2)$, $(1, 2)$, $(1, 5)$; sides 4, 3 and 5, the same as before.

Key terms

Dis­tance for­mula

The length of the seg­ment join­ing $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Absolute value

The size of a num­ber with­out its sign, writ­ten $|a|$; for exam­ple $|-7| = 7$.

Baud­hāyana–Pythago­ras the­o­rem

In a right-angled tri­an­gle, the square on the hypotenuse equals the sum of the squares on the other two sides.

Hypotenuse

The side oppo­site the right angle; the longest side of a right-angled tri­an­gle.

Shift

The change in a coor­di­nate when mov­ing from one point to another, such as $x_2 - x_1$ across.

Reflec­tion in the y-axis

The mir­ror move that sends $(x, y)$ to $(-x, y)$.

Reflec­tion in the x-axis

The mir­ror move that sends $(x, y)$ to $(x, -y)$.

Com­mon ques­tions

Does it mat­ter which point I call $(x_1, y_1)$?

No. Swap­ping the points changes the sign of both dif­fer­ences, and squar­ing removes the signs, so the dis­tance is the same.

Why do we need a right-angled tri­an­gle?

The Baud­hāyana–Pythago­ras the­o­rem only works for right-angled tri­an­gles. The hor­i­zon­tal and ver­ti­cal shifts are always at right angles to each other, so they give us the right-angled tri­an­gle we need.

What if the answer is not a whole num­ber?

Leave it as a square root, such as $\sqrt{29}$, or sim­plify it, such as $\sqrt{40} = 2\sqrt{10}$. Give a dec­i­mal only if the ques­tion asks for one.

Does reflec­tion change the area of a tri­an­gle?

No. All three sides keep their lengths, so the reflected tri­an­gle is an exact copy of the orig­i­nal, flipped over, and has the same area.

How do I find the dis­tance of a point from the ori­gin?

Use $(0, 0)$ as one of the points. The dis­tance of $(x, y)$ from O is $\sqrt{x^2 + y^2}$.

Ref­er­ences

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Ganita Man­jari: Text­book of Math­e­mat­ics, Grade 9 (Part I). NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 9. Dhan­pat Rai Pub­li­ca­tions.
3. Heath, T. L. (trans.) The Thir­teen Books of Euclid's Ele­ments. Dover Pub­li­ca­tions.