Distance Formula | Distance Between Two Points

Distance Formula | Distance Between Two Points

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.
Derivation of distance formula
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            Length of PQ

         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),       
Distance from origin

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,               
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        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,
Circle
                        PA = PB = PC = radius of circle
            Taking PA = PB,
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,           
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
Radius PA


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)
                       
Rectangle
                        Now,                    
Example 4: Solution

                        Since, PQ = RS, QR = PS and diagonals PR = QS, PQRS is a rectangle. Proved.


You can comment your questions or problems regarding distance formula here.

No comments:

Powered by Blogger.