Distance Formula in Coordinate Geometry: Class 10 Guide

The four quad­rants#

The axes divide the plane into four quad­rants, num­bered anti­clock­wise start­ing from the top right.

Points on the axes do not belong to any quad­rant.

The dis­tance for­mula#

State­ment#

The dis­tance between the points $A(x_1, y_1)$ and $B(x_2, y_2)$ is

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

In par­tic­u­lar, the dis­tance of a point $P(x, y)$ from the ori­gin is

$$OP = \sqrt{x^2 + y^2}$$

Why the for­mula is true#

Take $A(x_1, y_1)$ and $B(x_2, y_2)$ with $AB$ nei­ther hor­i­zon­tal nor ver­ti­cal. Through $A$ draw a line par­al­lel to the x-axis, and through $B$ a line par­al­lel to the y-axis. They meet at $C(x_2, y_1)$, and the angle at $C$ is a right angle because the axes are per­pen­dic­u­lar.

• $A$ and $C$ have the same y-coor­di­nate, so $AC = |x_2 - x_1|$.
• $B$ and $C$ have the same x-coor­di­nate, so $BC = |y_2 - y_1|$.

By the Pythago­ras the­o­rem in right tri­an­gle $ACB$,

$$\begin{aligned} AB^2 &= AC^2 + BC^2 \\ &= |x_2 - x_1|^2 + |y_2 - y_1|^2 \\ &= (x_2 - x_1)^2 + (y_2 - y_1)^2 \end{aligned}$$

A square of a num­ber equals the square of its absolute value, so the mod­u­lus signs can be dropped. Since a dis­tance is never neg­a­tive, we take the pos­i­tive square root. If $AB$ is hor­i­zon­tal, then $y_2 = y_1$ and the for­mula gives $|x_2 - x_1|$, which is cor­rect. The ver­ti­cal case works in the same way. So the for­mula holds for every pair of points.

Putting $A = O(0, 0)$ gives the dis­tance from the ori­gin, $\sqrt{x^2 + y^2}$.

The order of the points does not mat­ter: $(x_2 - x_1)^2 = (x_1 - x_2)^2$. So $AB = BA$, as it should.

Uses of the dis­tance for­mula#

Check­ing whether three points are collinear#

Three points $A$, $B$, $C$ lie on one straight line exactly when the largest of the three dis­tances equals the sum of the other two, for exam­ple $AB + BC = AC$. If the largest dis­tance is less than the sum of the other two, the points form a tri­an­gle.

Clas­si­fy­ing tri­an­gles#

• Equi­lat­eral: all three sides equal.
• Isosce­les: two sides equal.
• Sca­lene: no two sides equal.
• Right-angled: the square of the longest side equals the sum of the squares of the other two (con­verse of Pythago­ras).

Clas­si­fy­ing quadri­lat­er­als#

Take the ver­tices $A, B, C, D$ in order. Find the four sides $AB, BC, CD, DA$ and the two diag­o­nals $AC, BD$.

These tests rest on stan­dard facts. A quadri­lat­eral with both pairs of oppo­site sides equal is a par­al­lel­o­gram. A par­al­lel­o­gram with equal diag­o­nals is a rec­tan­gle. A rhom­bus with equal diag­o­nals is a square.

Method in steps#

1. Write the two points as $(x_1, y_1)$ and $(x_2, y_2)$.
2. Find $x_2 - x_1$ and $y_2 - y_1$, tak­ing care with signs.
3. Square both, add them and take the pos­i­tive square root.
4. For shape ques­tions, work out every side and both diag­o­nals first, then com­pare them with the tables above.
5. For an unknown coor­di­nate, square both sides of the dis­tance equa­tion before solv­ing, then check every value you get.

Worked exam­ples#

Exam­ple 1: dis­tance between two points#

Find the dis­tance between $A(2, -3)$ and $B(-4, 5)$.

$$\begin{aligned} AB &= \sqrt{(-4 - 2)^2 + (5 - (-3))^2} \\ &= \sqrt{(-6)^2 + 8^2} \\ &= \sqrt{36 + 64} = \sqrt{100} = 10 \end{aligned}$$

So $AB = 10$ units. This is the tri­an­gle shown in the fig­ure above.

Exam­ple 2: dis­tance from the ori­gin#

How far is $P(-5, 12)$ from the ori­gin?

$$OP = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$

So $P$ is 13 units from the ori­gin.

Exam­ple 3: test­ing collinear­ity#

Check whether $A(-2, -1)$, $B(1, 3)$ and $C(7, 11)$ lie on one line.

$$\begin{aligned} AB &= \sqrt{(1 + 2)^2 + (3 + 1)^2} = \sqrt{9 + 16} = 5 \\ BC &= \sqrt{(7 - 1)^2 + (11 - 3)^2} = \sqrt{36 + 64} = 10 \\ AC &= \sqrt{(7 + 2)^2 + (11 + 1)^2} = \sqrt{81 + 144} = 15 \end{aligned}$$

Here $AB + BC = 5 + 10 = 15 = AC$. So the points are collinear, and $B$ lies between $A$ and $C$.

Exam­ple 4: clas­si­fy­ing a tri­an­gle#

Show that $A(2, 1)$, $B(5, 5)$ and $C(9, 2)$ are the ver­tices of an isosce­les right-angled tri­an­gle.

$$\begin{aligned} AB &= \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \\ BC &= \sqrt{4^2 + (-3)^2} = \sqrt{25} = 5 \\ AC &= \sqrt{7^2 + 1^2} = \sqrt{50} \end{aligned}$$

Two sides are equal, so the tri­an­gle is isosce­les. Also $AB^2 + BC^2 = 25 + 25 = 50 = AC^2$, so by the con­verse of Pythago­ras the angle at $B$ is $90^\circ$. The tri­an­gle is an isosce­les right-angled tri­an­gle with the right angle at $B$.

Exam­ple 5: iden­ti­fy­ing a quadri­lat­eral#

Name the quadri­lat­eral with ver­tices $A(1, 1)$, $B(4, 2)$, $C(3, 5)$ and $D(0, 4)$, taken in order.

Sides:

$$\begin{aligned} AB &= \sqrt{3^2 + 1^2} = \sqrt{10} \\ BC &= \sqrt{(-1)^2 + 3^2} = \sqrt{10} \\ CD &= \sqrt{(-3)^2 + (-1)^2} = \sqrt{10} \\ DA &= \sqrt{1^2 + (-3)^2} = \sqrt{10} \end{aligned}$$

Diag­o­nals:

$$AC = \sqrt{2^2 + 4^2} = \sqrt{20}, \qquad BD = \sqrt{(-4)^2 + 2^2} = \sqrt{20}$$

All four sides are equal, so $ABCD$ is a rhom­bus. Its diag­o­nals are also equal, so $ABCD$ is a square.

Exam­ple 6: an unknown coor­di­nate and an equidis­tant point#

(a) The dis­tance between $P(x, 3)$ and $Q(2, -1)$ is 5 units. Find $x$.

$$\begin{aligned} (x - 2)^2 + (3 + 1)^2 &= 5^2 \\ (x - 2)^2 + 16 &= 25 \\ (x - 2)^2 &= 9 \\ x - 2 &= \pm 3 \end{aligned}$$

So $x = 5$ or $x = -1$. Both val­ues work, because there are two points on the line $y = 3$ that are 5 units from $Q$.

(b) Find the point on the x-axis that is equidis­tant from $A(1, 5)$ and $B(7, -1)$.

A point on the x-axis is $P(x, 0)$. Set­ting $PA^2 = PB^2$:

$$\begin{aligned} (x - 1)^2 + 5^2 &= (x - 7)^2 + (-1)^2 \\ x^2 - 2x + 1 + 25 &= x^2 - 14x + 49 + 1 \\ -2x + 26 &= -14x + 50 \\ 12x &= 24 \\ x &= 2 \end{aligned}$$

The point is $P(2, 0)$. Check: $PA = \sqrt{1 + 25} = \sqrt{26}$ and $PB = \sqrt{25 + 1} = \sqrt{26}$.

(c) Find a rela­tion between $x$ and $y$ if $P(x, y)$ is equidis­tant from $A(5, 2)$ and $B(-1, 6)$.

$$\begin{aligned} (x - 5)^2 + (y - 2)^2 &= (x + 1)^2 + (y - 6)^2 \\ x^2 - 10x + 25 + y^2 - 4y + 4 &= x^2 + 2x + 1 + y^2 - 12y + 36 \\ -10x - 4y + 29 &= 2x - 12y + 37 \\ -12x + 8y - 8 &= 0 \\ 3x - 2y + 2 &= 0 \end{aligned}$$

Every point equidis­tant from $A$ and $B$ sat­is­fies $3x - 2y + 2 = 0$. This line is the per­pen­dic­u­lar bisec­tor of $AB$.

Com­mon mis­takes#

• Sign slips with neg­a­tives: $5 - (-3)$ is $8$, not $2$.
• Writ­ing $\sqrt{a^2 + b^2} = a + b$. The square root of a sum is not the sum of the square roots.
• Mix­ing an x-coor­di­nate with a y-coor­di­nate inside one dif­fer­ence. The order of sub­trac­tion does not mat­ter, because each dif­fer­ence is squared, but both num­bers in a dif­fer­ence must be x-coor­di­nates, or both y-coor­di­nates.
• Decid­ing a quadri­lat­eral is a square from four equal sides alone. A rhom­bus also has four equal sides, so you must com­pare the diag­o­nals too.
• For collinear­ity, adding the wrong pair. The largest dis­tance must equal the sum of the other two.
• Throw­ing away one of the two answers in an unknown-coor­di­nate ques­tion with­out check­ing. Often both are valid.
• Leav­ing answers such as $\sqrt{50}$ when the ques­tion asks for the sim­plest form, which is $5\sqrt{2}$.

Try these#

1. Find the dis­tance between $(-3, 7)$ and $(5, 1)$. Answer: 10 units.
2. Find the dis­tance of $(-8, 15)$ from the ori­gin. Answer: 17 units.
3. Are $(1, -1)$, $(3, 3)$ and $(-1, -5)$ collinear? Answer: Yes; the dis­tances are $2\sqrt{5}$, $2\sqrt{5}$ and $4\sqrt{5}$, and $2\sqrt{5} + 2\sqrt{5} = 4\sqrt{5}$.
4. Find $y$ if the dis­tance between $(2, y)$ and $(-4, 3)$ is 10 units. Answer: $y = 11$ or $y = -5$.
5. Find the point on the y-axis equidis­tant from $(5, 2)$ and $(-3, 6)$. Answer: $(0, 2)$.
6. Name the quadri­lat­eral with ver­tices $(-1, -2)$, $(1, 0)$, $(-1, 2)$, $(-3, 0)$ in order. Answer: A square; each side is $2\sqrt{2}$ and each diag­o­nal is 4.

Key terms#

Ori­gin

The point $(0, 0)$ where the x-axis and y-axis meet.

Abscissa

The x-coor­di­nate of a point: its signed dis­tance from the y-axis.

Ordi­nate

The y-coor­di­nate of a point: its signed dis­tance from the x-axis.

Quad­rant

One of the four regions into which the axes divide the plane.

Dis­tance for­mula

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for $A(x_1, y_1)$ and $B(x_2, y_2)$.

Collinear points

Points that lie on one straight line.

Equidis­tant

At equal dis­tances from two or more given points.

Diag­o­nal

A seg­ment join­ing two oppo­site ver­tices of a quadri­lat­eral.

Com­mon ques­tions#

Does it mat­ter which point I call the first point?#

No. Swap­ping the points changes the sign of each dif­fer­ence, but squar­ing removes the sign, so the dis­tance is the same.

Why do we take only the pos­i­tive square root?#

A dis­tance is a length, and a length can­not be neg­a­tive. The neg­a­tive root has no mean­ing here.

Can I prove collinear­ity with the dis­tance for­mula alone?#

Yes. Find all three dis­tances. If the largest equals the sum of the other two, the points are collinear. If not, they form a tri­an­gle.

How do I tell a rhom­bus from a square?#

Both have four equal sides. Com­pare the diag­o­nals: equal diag­o­nals mean a square, unequal diag­o­nals mean a rhom­bus that is not a square.

Why can an unknown-coor­di­nate ques­tion have two answers?#

Squar­ing leads to a qua­dratic equa­tion. Geo­met­ri­cally, a cir­cle around the fixed point usu­ally cuts the given line in two places, so two posi­tions can be at the required dis­tance.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
4. Stew­art, J., Redlin, L. and Wat­son, S. Pre­cal­cu­lus: Math­e­mat­ics for Cal­cu­lus. Cen­gage Learn­ing.