ICSE Class 9 Maths Chapter 28 Distance Formula

Distance Formula

Introduction

For any two known (given) points in a co-ordinate (Cartesian) plane, the knowledge of co-ordinate geometry may be used to find:

(i) the distance between the given points,

(ii) the co-ordinates of a point which divides the line joining the given points in a given ratio,

(iii) the co-ordinates of the mid-point of the line segment joining the two given points,

(iv) equation of the straight line through the given points,

(v) equation of the perpendicular bisector of a line segment, etc.

The Distance Formula

To Find The Distance Between Two Given Points

Let the two given points be A (\(x_1\), \(y_1\)) and B (\(x_2\), \(y_2\)).

It is clear from the adjoining figure that in right-angled triangle ABC,

AC = \(x_2\) - \(x_1\) = difference between abscissae of A and B

BC = \(y_2\) - \(y_1\) = difference between ordinates of A and B.

Using Pythagoras' Theorem, we get:

AB\(^2\) = AC\(^2\) + BC\(^2\)

= (\(x_2\) - \(x_1\))\(^2\) + (\(y_2\) - \(y_1\))\(^2\)

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

Therefore, the distance between two given points (\(x_1\), \(y_1\)) and (\(x_2\), \(y_2\))

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

= \(\sqrt{\text{(difference of abscissae)}^2 + \text{(difference of ordinates)}^2}\)

Note: In co-ordinate geometry, all the formulae are obtained by taking the points in the first quadrant (as in the case of the distance formula, obtained above), but every formula will always remain true in whatever quadrant the points may lie. Only the proper signs must be used with the co-ordinates.

Teacher's Note

The distance formula is the foundation of GPS technology and mapping applications used daily in navigation and ride-sharing services.

Example 1: Find The Distance Between The Points (3, 6) And (0, 2)

Let (3, 6) = (\(x_1\), \(y_1\)) and (0, 2) = (\(x_2\), \(y_2\))

Distance between the given points = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

= \(\sqrt{(0 - 3)^2 + (2 - 6)^2}\)

= \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5

Answer: 5

The distance of any point (x, y) from the origin (0, 0)

= \(\sqrt{(x - 0)^2 + (y - 0)^2}\) = \(\sqrt{x^2 + y^2}\)

Example 2: Find The Distance Between The Origin And The Point

(i) (-12, 5) (ii) (15, -8).

Since, distance between origin and (x, y) = \(\sqrt{x^2 + y^2}\)

Distance between origin and the point (-12, 5)

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

Answer: 13

Distance between origin and the point (15, -8)

= \(\sqrt{(15)^2 + (-8)^2}\) = \(\sqrt{225 + 64}\) = 17

Answer: 17

Remember: If a point lies on the x-axis; its ordinate is zero, therefore a point on the x-axis is taken as (x, 0), such as (3, 0), (-5, 0), etc. In the same way, if a point lies on the y-axis, its abscissa is zero, therefore a point on the y-axis is taken as (0, y), such as (0, 4), (0, -7), etc.

Example 3: Find The Co-ordinates Of Points On The X-axis Which Are At A Distance Of 5 Units From The Point (6, -3)

Let the co-ordinates of the point on the x-axis be (x, 0)

Since, distance = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

5 = \(\sqrt{(x - 6)^2 + (0 + 3)^2}\)

Taking (6, -3) = (\(x_1\), \(y_1\)) and (x, 0) = (\(x_2\), \(y_2\))

25 = \(x^2\) - 12x + 36 + 9

On squaring

\(x^2\) - 12x + 20 = 0

x = 2, or x = 10

On solving the quadratic equation

Required points on the x-axis are (2, 0) and (10, 0)

Answer:

Example 4: KM Is A Straight Line Of 13 Units. If K Has The Co-ordinates (2, 5) And M Has The Co-ordinates (x, -7), Find The Values Of x

Let K (2, 5) = (\(x_1\), \(y_1\)) and M (x, -7) = (\(x_2\), \(y_2\))

KM = 13 units \(\Rightarrow\) \(\sqrt{(x - 2)^2 + (-7 - 5)^2}\) = 13

\(\Rightarrow\) \(x^2\) - 4x + 4 + 144 = 169

\(\Rightarrow\) \(x^2\) - 4x - 21 = 0

\(\Rightarrow\) (x - 7) (x + 3) = 0 \(\Rightarrow\) x = 7 or x = -3

Answer:

Example 5: Which Point On The Y-axis Is Equidistant From The Points (12, 3) And (-5, 10)

Let the required point on the y-axis be (0, y).

Given (0, y) is equidistant from (12, 3) and (-5, 10)

i.e. distance between (0, y) and (12, 3)

= distance between (0, y) and (-5, 10)

\(\Rightarrow\) \(\sqrt{(12 - 0)^2 + (3 - y)^2}\) = \(\sqrt{(-5 - 0)^2 + (10 - y)^2}\)

\(\Rightarrow\) 144 + 9 + \(y^2\) - 6y = 25 + 100 + \(y^2\) - 20y

\(\Rightarrow\) 14y = -28 and y = -2

Required point on the y-axis = (0, -2)

Answer:

Example 6: Use The Distance Formula To Show That The Points A(1, -1), B(6, 4) And C(4, 2) Are Collinear

Three points A, B and C are collinear

if and only if: (i) AB + BC = AC i.e. A - B - C

or, (ii) AB + AC = BC i.e. B - A - C

or, (iii) AC + BC = AB i.e. A - C - B

Since, AB = \(\sqrt{(6 - 1)^2 + (4 + 1)^2}\) = \(\sqrt{25 + 25}\) = \(\sqrt{50}\) = \(5\sqrt{2}\)

BC = \(\sqrt{(4 - 6)^2 + (2 - 4)^2}\) = \(\sqrt{4 + 4}\) = \(\sqrt{8}\) = \(2\sqrt{2}\)

and, AC = \(\sqrt{(4 - 1)^2 + (2 + 1)^2}\) = \(\sqrt{9 + 9}\) = \(\sqrt{18}\) = \(3\sqrt{2}\)

\(\Rightarrow\) BC + AC = \(2\sqrt{2}\) + \(3\sqrt{2}\) = \(5\sqrt{2}\) = AB

Given points A, B and C are collinear.

Answer:

Example 7: Show That The Points A (8, 3), B (0, 9) And C (14, 11) Are The Vertices Of An Isosceles Right-angled Triangle

AB = \(\sqrt{(0 - 8)^2 + (9 - 3)^2}\) = \(\sqrt{64 + 36}\) = \(\sqrt{100}\) = 10

BC = \(\sqrt{(14 - 0)^2 + (11 - 9)^2}\) = \(\sqrt{196 + 4}\) = \(\sqrt{200}\) = \(10\sqrt{2}\)

CA = \(\sqrt{(8 - 14)^2 + (3 - 11)^2}\) = \(\sqrt{36 + 64}\) = \(\sqrt{100}\) = 10

AB\(^2\) + CA\(^2\) = 100 + 100 = 200 = BC\(^2\)

BC\(^2\) = AB\(^2\) + CA\(^2\) \(\Rightarrow\) the triangle is right-angled triangle.

and, AB = CA \(\Rightarrow\) the triangle is isosceles.

Hence, the triangle ABC is an isosceles right-angled triangle.

Remember: A given quadrilateral will be a:

(i) parallelogram; if both the pairs of opposite sides are equal.

(ii) rectangle; if both the pairs of opposite sides are equal and diagonals are also equal.

(iii) rhombus; if all the sides are equal.

(iv) square; if all the sides are equal and diagonals are also equal.

Example 8: Find The Area Of A Circle, Whose Centre Is (5, -3) And Which Passes Through The Point (-7, 2)

(Take \(\pi\) = 3.14)

The radius (r) of the circle = distance between the points (5, -3) and (-7, 2)

= \(\sqrt{(-7 - 5)^2 + (2 + 3)^2}\)

= \(\sqrt{144 + 25}\) = \(\sqrt{169}\) = 13

Area of the circle = \(\pi r^2\)

= 3.14 \(\times\) 13\(^2\) = 530.66 sq. units

Answer:

Example 9: Find The Points On The X-axis Whose Distances From The Points A(7, 6) And B(-3, 4) Are In The Ratio 1 : 2

Let the required point P on x-axis = (x, 0)

A = (7, 6) and B = (-3, 4)

Given: \(\frac{PA}{PB}\) = \(\frac{1}{2}\) \(\Rightarrow\) 2 PA = PB

i.e. \(2\sqrt{(x - 7)^2 + (0 - 6)^2}\) = \(\sqrt{(x + 3)^2 + (0 - 4)^2}\)

\(\Rightarrow\) \(4(x^2 - 14x + 49 + 36)\) = \(x^2 + 6x + 9 + 16\)

\(\Rightarrow\) \(4x^2 - 56x + 196 + 144\) = \(x^2 + 6x + 25\)

\(\Rightarrow\) \(3x^2 - 62x + 315\) = 0

\(\Rightarrow\) \(3x^2 - 27x - 35x + 315\) = 0

\(\Rightarrow\) (x - 9) (3x - 35) = 0

\(\Rightarrow\) x = 9 or x = \(\frac{35}{3}\)

Required points on x-axis are: (9, 0) and (\(\frac{35}{3}\), 0)

Answer:

Teacher's Note

Coordinate geometry helps surveyors and engineers determine distances and positions on land, which is essential for infrastructure development and construction projects.

Circumcentre Of A Triangle

It is the point which is equidistant from the vertices of the triangle i.e. if P is the circumcentre of the triangle ABC, then:

PA = PB = PC = Circumradius.

(If a circle is drawn with P as centre and PA or PB or PC as radius, the circle will pass through all the three vertices of the triangle).

Example 10: Find The Co-ordinates Of The Circumcentre Of The Triangle ABC; Whose Vertices A, B And C Are (4, 6), (0, 4) And (6, 2) Respectively

Let the circumcentre be P (x, y).

then, PA = PB

\(\Rightarrow\) \(\sqrt{(x - 4)^2 + (y - 6)^2}\) = \(\sqrt{(x - 0)^2 + (y - 4)^2}\)

\(\Rightarrow\) \(x^2 - 8x + 16 + y^2 - 12y + 36\) = \(x^2 + y^2 - 8y + 16\)

\(\Rightarrow\) -8x - 4y = -36

\(\Rightarrow\) 2x + y = 9 .....I

and PA = PC

\(\Rightarrow\) \(\sqrt{(x - 4)^2 + (y - 6)^2}\) = \(\sqrt{(x - 6)^2 + (y - 2)^2}\)

\(\Rightarrow\) \(x^2 - 8x + 16 + y^2 - 12y + 36\) = \(x^2 - 12x + 36 + y^2 - 4y + 4\)

\(\Rightarrow\) 4x - 8y = -12

\(\Rightarrow\) x - 2y = -3 .....II

On solving I and II, we get: x = 3 and y = 3

The circumcentre of the given triangle = (3, 3)

Answer:

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