We can find the distance between two points when the co-ordinates of the two points
are given. We calculate that distance by using a formula called distance formula which is derived as
follows.
Let P(x1, y1) and Q(x2,
y2) be any two given points in the coordinate plane. Draw perpendicular
PL from P to the x-axis. Then, OL = x1 and PL = y1. Draw
perpendicular QM from Q to the x-axis. Then, OM = x2 and QM = y2.
Again, draw perpendicular PN from P to
the line segment QM. Then,
PN = LM = OM
– OL = x2 – x2
QN = QM–NM
= QM–PL = y2–y1
From right-angled DPNQ, by Pythagoras theorem,
PQ2
= PN2 + QN2
= (x2–x1)2
+ (y2–y1)2 
This
is the distance formula between two
points P(x1, y1) and Q(x2, y2). Only positive square root is
to be taken because PQ being the distance between two points is positive.
Distance from origin
The
distance of any the point (x, y) from the origin (0, 0),
Remarks:
1. The formula remains the same if the points P(x1, y1)
and Q(x2, y2) are taken in different quadrants.
2. If a point lies on x-axis, its ordinate is zero. Therefore, any
point on x-axis is (x, 0).
3. If a point lies on y-axis, its abscissa is zero. Therefore, any
point on y-axis is (0, y).
4. To prove that a quadrilateral is a,
(i) rhombus, show that all sides are equal.
(ii) square, show that all sides are equal and diagonals are also
equal.
(iii) parallelogram, show that the opposite sides are equal.
(iv) rectangle, show that the opposite sides are equal and diagonals
are also equal.
5. To prove that a triangle is a,
(i) scalene, show that none of the sides are equal.
(ii) isosceles, show that two sides are equal.
(iii) equilateral, show that all sides are equal.
(iv) right angled triangle, show that square of one side is equal to
the sum of the squares of other two sides.
Application of distance formula:
We
can apply the distance formula to find the distance between any two points on
the co-ordinate plane. And this can be applied for various geometrical and
mathematical purposes such as to find the co-ordinates of any point under the
given conditions, to find the centre of circum-circle, to verify a square,
rectangle or parallelogram etc. Look at the following workout examples:
Workout Examples
Example 1: Find the distance between the
points A(-8, 2) and B(4, 7).
Solution: Here,
A(-8, 2)
B(4, 7)
∴ x1 = -8 x2 = 4
y1 = 2 y2 = 7
By
using distance formula,
Example 2: If the distance between two
points (-3, 4) and (a, 1) is 5 units, find the possible values of a.
Solution: Here,
Given points are (-3, 4)
and (a, 1)
Distance = 5 units 
or, a2
+ 6a + 18 = 52 [squaring]
or, a2
+ 6a + 18 = 25
or, a2
+ 6a + 18 – 25 = 0
or, a2
+ 6a – 7 = 0
or, a2
+ (7 – 1)a – 7 = 0
or, a2
+ 7a – a – 7 = 0
or, a(a + 7) –
1(a + 7) = 0
or, (a + 7)(a
– 1) = 0
Either, a + 7 = 0
or, a = -7
∴ Or, a – 1 = 0
or, a = 1
∴ a = 1 or -7
Example 3: Find the centre and radius of
the circum-circle of a trianle having vertices (-4, 1), (-2, 5) and (1, -4).
Solution: Here,
A(-4, 1), B(-2, 5) and
C(1 -4)
Let, P(x, y) be the
centre of the circum-circle,
∴ PA = PB = PC = radius of circle
Taking PA =
PB,
or, x2
+2.x.4+42+y2-2.y.1+12 = x2+2.x.2+22+y2-2.y.5+52
or, x2
+8x+16+y2-2y+1 = x2+4x+4+y2-10y+25
or, 8x–2y-4x+10y
= 4+25-16-1
or, 4x+8y = 12
or, 4(x+2y) =
12
or, x+2y = 3
------------> (i)
Again, taking PB = PC,
or, x2+2.x.2+22+y2-2.y.5+52
= x2-2.x.1+12+y2+2.y.4+42
or, x2+4x+4+y2-10y+25
= x2-2x+1+y2+8y+16
or, 4x–10y+2x-8y
= 1+16-4-25
or, 6x-18y =
-12
or, 6(x-3y) =
-12
or, x-3y = -2
------------> (ii)
Subtracting equatin (ii) from equatin (i), we get,
x+2y = 3
x-3y = -2
5y = 5
or, y = 1
Putting y = 1 in equatin (i), we get,
x + 2×1 =
3
or, x + 2 = 3
or, x = 3 – 2
or, x = 1
∴ Centre of circle is P(x, y) = P(1, 1), and
Example 4: Prove that the points (1, 3),
(2, 5), (6, 3) and (5, 1) are the vertices of a rectangle.
Solution: Let,
P(1, 3), Q(2, 5), R(6,
3) and S(5, 1)
Since, PQ = RS, QR = PS
and diagonals PR = QS, PQRS is a rectangle. Proved.
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formula here.
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