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Coordinate Geometry

    Distance Formula

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10th class > Coordinate Geometry > Distance Formula

Distance Formula

Let us consider the following situation: A town B is located 36 km east and 15 km north of the town A. How would you find the distance from town A to town B without actually measuring it. Let us see. This situation can be represented graphically as shown. You may use the Pythagoras Theorem to calculate this distance.

Consider the figure above. What is the distance from b to a?

Since both the points lie on the x-axis, we just need to walk on the x axis. So the distance is nothing but ob-oa = units.

Now, what is the distance from d to c? These points are on the y-axis and their distance is units.

Can we find the distance between a and c? The distance between a and c is units. (hint: apply pythogaras theorem).

Next, can you find the distance of A from C? Since OA = 4 units and OC = 3 units, the distance of A from C, i.e., AC = 32+42 = units.

Similarly, you can find the distance of B from D = BD = units.

Now, if we consider two points not lying on coordinate axis, can we find the distance between them? !

We shall use Pythagoras theorem to do so. Let us see an example.

This is also easy enough. We just need to visualize a scenario which we know. Draw a line from P to Q.

The line is just hanging in the air. We still don't have enough information. Why don't you draw a perpendicular from P to x-axis namely, PR?

We are getting somewhere, but still we dont have any formation that we can recognize. Lets continue. Draw a perpendicular from Q to x-axis namely, QS.

Now we are getting somewhere. We just need one more final line segment. Can you guess and draw?

! Now you have a right angle triangle and once we have the right angle triangle we can use our ever dependable Pythogoras theorem. Using the theorem we get the distance PQ is 22.

Consider the points P(6, 4) and Q(–5, –3). Draw QS perpendicular to the x-axis. Also draw a perpendicular PT from the point P on QS (extended) to meet y-axis at the point R. A perpendicular from the point P on QS is drawn to meet it at the point T.

Then PT = units and QT = units.

Using the Pythagoras Theorem to the right triangle PTQ, we get PQ = 112+72 = units.

Let us now find the distance between any two points P(x1,y1) and Q(x2,y2).

Then, OR = x1, OS = x2.

So, RS = x2−x1 =

Also, SQ = y2, ST = = y1

So, = y2−y1

Now, applying the Pythagoras theorem in triangle PTQ, we get:

PQ2 = PT2 + QT2 = x2−x12+ y2−y12

Therefore, PQ = x2−x12+y2−y12

Note that since distance is always non-negative, we take only the positive square root.

So, the distance between the points P(x1,y1) and Q(x2,y2) is:

PQ = x2−x12+y2−y12

which is called the distance formula.

Remarks:

  1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by OP = x2+y2

  2. We can also write, PQ = x1−x22+y1−y22 as the points are interchangeable.

Let's move on to doing some related problems.

1. Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed.

Solution : Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. We have:

Check for triangle

  • Let's find the length of the sides using distance formula.
  • We have: PQ = 3+22+2+32 = 2+2= = (upto two decimal places)
  • QR = −2−22+−3−32 = 2+2 = = (upto two decimal places)
  • PR = 3−22+2−32 = 2+2 = = (upto two decimal places)
  • Check if the sum of any two of these sides is greater than the third side.
  • Therefore, the points P, Q and R form a triangle.
  • Also, PQ2+PR2=QR2 , by the converse of Pythagoras theorem, we have angle P = 90°.
  • Therefore, PQR is a right triangle.

2. Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.

Solution : Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points. One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal. Now,

AB = 1−42+7−22 = 9+25=

BC = 4+12+2+12 = = 34

CD = −1+42+−1−42 = = 34

DA = 1+42+7−42 = = 34

AC = 1+12+7+12 = =

BD = 4+42+2−42 = 64+4= 68

Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal.

Thereore, ABCD is a square.

Alternative Solution : We find the four sides and one diagonal, say, AC as above. Here AD2+DC2 = + = = AC2.

Therefore, by the converse of Pythagoras theorem, angle D = °.

A quadrilateral with all four sides equal and one angle 90° is a square. So, ABCD is a square.

3. Above figure shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively. Do you think they are seated in a line? Give reasons for your answer.

Check for line

  • Let's find the distance between the students using distance formula.
  • We have: AB = 6−32+4−12 = 9+9= =
  • BC = 8−62+6−42 = 4+4= =
  • AC = 8−32+6−12 = 25+25= =
  • We see that: AB + BC = 32+22 =
  • This is further equal to
  • Thus, the points A, B and C are collinear. Therefore, they are seated in a line.

4. Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).

Solution : Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).

We are given that AP = BP. So, AP2=BP2

Therefore, x−72+y−12=x−32+y−52

On expanding, we get:

x2 + + y2−2y+1 = x2 – + 9+y2−10y +

x−y=2 which is the required relation.

Remark: Note that the graph of the equation x – y = 2 is a line. From earlier studies, we know that a point which is equidistant from A and B lies on the perpendicular bisector of AB.

Therefore, the graph of x – y = 2 is the perpendicular bisector of AB (see below).

5. Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).

Solution : We know that a point on the y-axis is of the form .

So, let the point P(0, y) be equidistant from A and B. Then:

6−02+5−y2=−4−02+3−y2

On expanding,

+ 25 + y2−10y = + 9 + y2 – 6y

4y = 36

y =

So, the required point is (0,9).

Let us check our solution : AP = 6−02+5−92 = + =

BP = −4−02+3−92 = + =

Note : Using the remark above, we see that (0, 9) is the intersection of the y-axis and the perpendicular bisector of AB.

Additional Concept

Area of a Triangle

We all know the area of a triangle = 12·base·height. We also know about Heron's formula(s·s−a·s−b·s−c) to find the area of a triangle.

But here we are in the coordinate system. In the coordinate system we have points. If we know the points of the vertices of a triangle can we find it's area?

Consider the triangle below:

We can find the distances between AB,BC and AC. If we have the distances we can apply Heron's formula to find the area. But let us see if there is any other way to calculate the area.

As usual, lets draw perpendiculars to the x axis and see if it throws up some known structures.

Nice. Now we have triangles and .

Area of triangle ABC=Area of trapezium ABQP+Area of trapezium APRC- Area of trapezium BQRC

area of a trapezium = 12 (sum of parallel sides)(distance between them)

Substituting for the trapeziums in the above equation we get

Area of triangle ABC= 12x1y2−y3+x2y3−y1+x3y1−y2