Distance Formula - Coordinate Geometry, CBSE, Class 10, Mathematics PDF Download

INTRODUCTION
In the previous class, you have learnt to locate the position of a point in a coordinate plane of two dimensions, in terms of two coordinates. You have learnt that a linear equation in two variables, of the form ax + by + c = 0 (either a ≠ 0 or b ≠ 0) can be represented graphically as a straight line in the coordinate plane of x and y coordinates. In chapter 4, you have learnt that graph of a quadratic equation ax² + bx + c = 0, a ≠ 0 is an upward parabola if a > 0 and a downward parabola if a < 0.
In this chapter, you will learn to find the distance between two given points in terms of their coordinates and also, the coordinates of the point which divides the line segment joining the two given points internally in the given ratio.

HISTORICAL FACTS

Rene Descartes (1596-1650), the 17th century French-Mathematician, was a thinker and a philosopher. He is called the father of Co-ordinate Geometry because he unified Algebra and Geometry which were earlier two distinct branches of Mathematics. Descartes explained that two numbers called co-ordinates are used to locate the position of a point in a plane.
He was the first Mathematician who unified Algebra and Geometry, so Analytical Geometry is also called Algebraic Geometry. Cartesian plane and Cartesian Product of sets have been named after the great Mathematician.
RECALL
 Cartesian Co-ordinate system :

Let X'OX and Y'OY be two perpendicular straight lines intersecting each other at the point O. Then :
1. X'OX is called the x-axis or the axis of x.
2. Y'OY is called the y-axis or the axis of y.
3. The x-co-ordinate along OX is positive and along OX' negative, y-co-ordinate along OY (upward) is positive and along OY' (downward) is negative.
4. Both X'OX and Y'OY taken together in this order are called the rectangular axes because the angle between them is a right angle.
5. O is called the origin i.e., it is point of intersection of the axes of co-ordinates.

Co-ordinates of a point :
  

6. Abscissa of a point in the plane is its perpendicular distance with proper sign from y-axis.
7. Ordinate of a point in the plane is its perpendicular distance with proper sign from x-axis.
8. The y-co-ordinate of any point on x-axis is zero.
9. The x-co-ordinate of any point on y-axis is zero.
10. Any point in the xy-plane, whose y-co-ordinate is zero, lies on x-axis.
11. Any point in the xy-plane, whose x-co-ordinate is zero, lies on y-axis.
12. The origin has coordinates (0, 0).
13. The ordinates of all points on a horizontal line which is parallel to x-axis are equal i.e. y = constant = 2.

14. The abscissa of all points on a vertical line which is a line parallel to y-axis are equal i.e. x = constant = 4.

Four Quadrants of a Coordinate plane :
 The rectangular axes X'OX and Y'OY divide the plane into four quadrants as below :

15. Any point in the I quadrant has (+ve abscissa, +ve ordinate).
16. Any point in the II quadrant has (–ve abscissa, +ve ordinate).
17. Any point in the III quadrant has (–ve abscissa, –ve ordinate).
18. Any point in the IV quadrant has (+ve abscissa, –ve ordinate).

DISTANCE FORMULA

The distance between two points (x₁, y₁) and (x₂, y₂) in a rectangular coordinate system is equal  to

  

Proof : X'OX and Y'OY are the rectangular coordinate axes. P(x₁, y₁) and Q(x₂, y₂) are the given points. We
draw PA and QB perpendiculars on the x-axis ; PC and QD perpendiculars on the y-axis.

Now, CP (produced) meets BQ in R and PR ⊥ BQ.

We find PR = AB = OB – OA = (x₂ – x₁)
and QR = BQ – BR = BQ – AP = OD – OC = (y₂ –y₁)

In right ΔPRQ, using Pythagoras theorem, we have

PQ² = PR² + QR² = (x₂ – x₁)² + (y₂ – y₁)² PQ =

We have taken the positive square root value because distance between two points is a non-negative quantity.

The distance of a point (x, y) from origin is 

Proof : Let us take a point P (x, y) in the given plane of axes X'OX and Y'OY as shown in the fig. Here, the point P (x, y) is in the first quadrant but it can be taken anywhere in all the four quadrants. We have to find the distance OP, i.e., the distance of the point P from the origin O.

From the point P, draw PM ⊥ OX and PL ⊥ OY. Then we have
OM = x
MP = OL = y
OP² = OM² + MP² = x² + y²

 Test For Geometrical Figures :

(a) For an isosceles triangle : Prove that two sides are equal.
 (b) For an equilateral triangle : Prove that three sides are equal.
 (c) For a right-angled triangle : Prove that the sum of the squares of two sides is equal to the square of the third side.
 (d) For a square                      : Prove that all sides are equal and diagonals are equal.
 (e) For a rhombus                   : Prove that all sides are equal and diagonals are not equal.
 (f) For a rectangle                   : Prove that the opposite sides are equal and diagonals are also equal.
 (g) For a parallelogram           : Prove that the opposite sides are equal in length and diagonals are not equal.

Ex.1 Find points on x-axis which are at a distance of 5 units from the point A(–1, 4).
 Sol. Let the point on x-axis be P(x, 0).
Distance = PA = 5 units  PA² = 25
 (x + 1)² + (0 – 4)² = 25  x² + 2x + 1 + 16 = 25

 x² + 2x + 17 = 25 x² +2x – 8 = 0

Required point on x-axis are (2, 0) and (–4, 0)

Ex.2 What point on y-axis is equidistant from the points (3, 1) and (1, 5) ?
Sol. Since the required point P(say) is on the y-axis, its abscissa (x-co-ordinate) will be zero. Let the ordinate (y-coordinate) of the point be y.

Therefore co-ordinates of the point P are : (0, y) i.e., P(0, y)

Let A and B denote the points (3, 1) and (1, 5) respectively.

PA = PB ...(given)

Squaring we get :

PA² = PB²

 (0 – 3)² + (y – 1)² = (0 – 1)² + (y – 5)²

 9 + y² + 1 – 2y = 1 + y² + 25 – 10y
 y² – 2y + 10 = y² – 10y + 26

 –2y + 10y = 26 – 10 8y = 16 y = 2

The required point on y-axis equidistant from A(3, 1) and B(1, 5) is P(0, 2).

Ex.3  If Q(2, 1) and R(–3, 2) and P(x, y) lies on the right bisector of QR then show that 5x – y + 4 = 0.
 Sol. Let P(x, y) be a point on the right bisector of QR :
Q(2, 1) and R(–3, 2) are equidistant from P(x, y), then we must have :

PQ = PR

Ex.4 The vertices of a triangle are (–2, 0), (2, 3) and (1, –3). Is the triangle equilateral : isosceles or scalene ?

Sol. We denote the given point (–2, 0), (2, 3) and (1, –3) by A, B and C respectively then :
A(–2,0), B(2,3), C(1,–3)

Thus we have AB ≠ BC ≠ CA
 ABC is a scalene triangle