Distance Formula

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Example – 01:

• Find the distance between the following pairs of points :
• (2, 3), (4, 1)

Let A(2, 3) ≡ (x1, y1) and B(4, 1) ≡ (x2, y2) be the points

By distance formula

AB2 = (x2 – x1)2 + (y2 – y1)2

∴  AB2 = (4 – 2)2 + (1 – 3) = (2)2 + (-2)= 4 + 4 = 8

∴  AB =8  = 22 unit

Ans: The distance between given points is 22 unit

• (– 5, 7), (– 1, 3)

Let A(-5, 7) ≡ (x1, y1) and B(-1, 3) ≡ (x2, y2) be the points

By distance formula

AB2 = (x2 – x1)2 + (y2 – y1)2

∴  AB2 = (-1 + 5)2 + (3 – 7) = (4)2 + (-4)= 16 + 16 = 32

∴  AB =32  = 42 unit

Ans: The distance between given points is 42 unit

• (6, 8), (– 9, – 12)

Let A(6, 8) ≡ (x1, y1) and B(-9, -12) ≡ (x2, y2) be the points

By distance formula

AB2 = (x2 – x1)2 + (y2 – y1)2

∴  AB2 = (- 9 – 6)2 + (-12 – 8) = (-15)2 + (-20)= 225 + 400 = 625

∴  AB =625  = 25 unit

Ans: The distance between given points is 25 unit

• (-6, -1), (– 6, 11)

Let A(-6, -1) ≡ (x1, y1) and B(-6, 11) ≡ (x2, y2) be the points

By distance formula

AB2 = (x2 – x1)2 + (y2 – y1)2

∴  AB2 = (- 6 + 6)2 + (11 + 1) = (0)2 + (12)= 0 + 144 = 144

∴  AB =144  = 12 unit

Ans: The distance between given points is 12 unit

• (a, b), (– a, – b)

Let A(a, b) ≡ (x1, y1) and B(-a, -b) ≡ (x2, y2) be the points

By distance formula

AB2 = (x2 – x1)2 + (y2 – y1)2

∴  AB2 = (- a – a)2 + (-b – b) = (- 2a)2 + (- 2b)= 4a2 + 4b2 = 4(a2 + 4b2)

• (0, 0), (36, 15)

Let A(0, 0) ≡ (x1, y1) and B(36, 15) ≡ (x2, y2) be the points

By distance formula

AB2 = (x2 – x1)2 + (y2 – y1)2

∴  AB2 = (36 – 0)2 + (15 – 0) = (36)2 + (15)= 1296 + 225 = 1521

∴  AB =1521  = 39 unit

Alternate Direct Method

One of the given point is origin O. Let A(36, 15) ≡ (x, y)

∴  OA2 = x2 + y2 = (36)2 + (15) = 1296 + 225 = 1521

∴  OA =1521  = 39 unit

Ans: The distance between given points is 39 unit

Example – 02:

• Using distance formula show that following sets of points are collinear
• A(0,4), B(2,10) and C(3,13)

Given points are A(0,4), B(2,10) and C(3,13)

Using distance formula

AB2 = (2 – 0)2 + (10 – 4)= (2)2 + (6)2 = 4 + 36 = 40

∴  AB  = 40  =  210  unit …………….. (1)

BC2 = (3 – 2)2 + (13 – 10)= (1)2 + (3)2 = 1 + 9 = 10

∴  BC  = 10   unit  …………….. (2)

AC2 = (3 – 0)2 + (13 – 4)= (3)2 + (9)2 = 9 + 81 = 90

∴  AC  = 90  =  310  unit …………….. (3)

From equations (1), (2) and (3) we have

AC = AB + BC

∴ A – B – C

Hence points A, B and C are collinear

• P(5, 0), Q(10, -3) and R(-5, 6)

Given points are P(5, 0), Q(10, -3) and R(-5, 6)

Using distance formula

PQ2 = (10 – 5)2 + (-3 – 0)= (5)2 + (-3)2 = 25 + 9 = 34

∴  PQ  = 34   unit   …………….. (1)

QR2 = (-5 – 10)2 + (6 + 3)= (-15)2 + (9)2 = 225 + 81 = 306

∴  QR  = 306  =  334   unit …………….. (2)

PR2 = (-5 – 5)2 + (6 – 0)= (-10)2 + (6)2 = 100+ 36 = 136

∴  PR  = 136  =  234   unit …………….. (3)

From equations (1), (2) and (3) we have

QR = PQ + PR

∴ Q – P – R

Hence points P, Q and R are collinear

• L(2, 5), M(5, 7) and N(8, 9)

Given points are L(2, 5), M(5, 7) and N(8, 9)

Using distance formula

LM2 = (5 – 2)2 + (7 – 5)= (3)2 + (2)2 = 9 + 4 = 13

∴  LM  = 13   unit   …………….. (1)

MN2 = (8 – 5)2 + (9 – 7)= (3)2 + (2)2 = 9 + 4 = 13

∴  MN  = 13  unit …………….. (2)

LN2 = (8 – 2)2 + (9 – 5)= (6)2 + (4)2 = 36+ 16 = 52

∴  LN  = 52  =  213   unit …………….. (3)

From equations (1), (2) and (3) we have

LN = LM + MN

∴ L – M – N

Hence points L, M and N are collinear

• D(5, 1), E(1, -1) and F(11, 4)

Given points are D(5, 1), E(1, -1) and F(11, 4)

Using distance formula

DE2 = (1 – 5)2 + (-1 – 1)= (-4)2 + (-2)2 = 16 + 4 = 20

∴  DE  = 20  =  25   unit   …………….. (1)

EF2 = (11 – 1)2 + (4 + 1)= (10)2 + (5)2 = 100 + 25 = 125

∴  EF  = 125 =  55  unit …………….. (2)

DF2 = (11 – 5)2 + (4 – 1)= (6)2 + (3)2 = 36+ 9 = 45

∴  DF  = 45  =  35  unit …………….. (3)

From equations (1), (2) and (3) we have

EF = DE + DF

∴ E – D – F

Hence points D, E and F are collinear

• A(1, 5), B(2, 3) and C(– 2, – 11)

Given points are A(1, 5), B(2, 3) and C(-2, -11)

Using distance formula

AB2 = (2 – 1)2 + (3 – 5)= (1)2 + (-2)2 = 1 + 4 = 5

∴  AB  = 5   unit …………….. (1)

BC2 = (-2 – 2)2 + (-11 – 3)= (-4)2 + (-14)2 = 16 + 196 = 212

∴  BC  = 212    unit  …………….. (2)

AC2 = (-2 – 1)2 + (-11 – 5)= (-3)2 + (-16)2 = 9 + 256 = 265

∴  AC  = 265    unit …………….. (3)

From equations (1), (2) and (3) we have

AC ≠ AB + BC

Hence points A, B and C are not collinear

Example – 03:

• 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?

Given points are A(3, 1), B(6, 4) and C(8, 6)

Using distance formula

AB2 = (6 – 3)2 + (4 – 1)= (3)2 + (3)2 = 9 + 9 = 18

∴  AB  = 18  = 32 unit …………….. (1)

BC2 = (8 – 6)2 + (6 – 4)= (2)2 + (2)2 = 4 + 4 = 8

∴  BC  = 8  = 22   unit  …………….. (2)

AC2 = (8 – 3)2 + (6 – 1)= (5)2 + (5)2 = 25 + 25 = 50

∴  AC  = 50   = 52  unit …………….. (3)

From equations (1), (2) and (3) we have

AC = AB + BC

∴ A – B – C

Hence points A, B and C are collinear.

Thus Ashima, Bharti and Camella are seated in a line

Example – 04:

• The distance between the points (0, 0) and (x, 3) is 5. Find x.

Let A(0, 0) ≡ (x1, y1) and B(x, 3) ≡ (x2, y2) be the points. Thus AB = 5

By distance formula

AB2 = (x2 – x1)2 + (y2 – y1)2

∴  52 = (x – 0)2 + (3 – 0) = x2 + 9

∴   25 = x2 + 9

∴  x2 = 16

∴  x = ± 4

Example – 05:

• The distance between the points (a, 5) and (0, -3) is 45  unit. Find a.

Let A(a, 5) ≡ (x1, y1) and B(0, -3) ≡ (x2, y2) be the points. Thus AB = 45

By distance formula

AB2 = (x2 – x1)2 + (y2 – y1)2

∴  (45)2 = (0 – a)2 + (-3 – 5) = (-a)2 + (-8)2 = a2 + 64

∴  80 =  a2 + 64

∴  a2 = 16

∴  a = ± 4

Example – 06:

• The distance between the points (8, -7) and (-4, a) is 13. Find a.

Let A(8, -7) ≡ (x1, y1) and B(-4, a) ≡ (x2, y2) be the points. Thus AB = 5

By distance formula

AB2 = (x2 – x1)2 + (y2 – y1)2

∴  132 = (- 4 – 8)2 + (a + 7) = (-12)2 + a2 + 14a + 49

∴  169 =  144 + a2 + 14a + 49

∴  a2 + 14a + 197 – 169 = 0

∴  a2 + 14a + 24 = 0

∴  (a + 12)(a + 2) = 0

∴  a + 12 = 0 or a + 2 = 0

∴  a = -12 and a = -2

Example – 07:

• Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Given P(2, -3) ≡ (x1, y1) and Q(10, y) ≡ (x2, y2) be the points. Thus PQ = 10 units

By distance formula

PQ2 = (x2 – x1)2 + (y2 – y1)2

∴  102 = (10 – 2)2 + (y + 3) = (8)2 + y2 + 6y + 9

∴  100 = 64 + y2 + 6y + 9

∴  36 =  y2 + 6y + 9

∴  y2 + 6y + 9 – 36 = 0

∴  y2 + 6y + – 27 = 0

∴  (y + 9)(y – 3) = 0

∴  y + 9 = 0 or y – 3 = 0

∴  y = -9 and y = 3

Example – 08:

• Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Let P(2, –5) and Q(–2, 9) be the given points.

Let the point on the x-axis be A(a, 0)

Given PA = QA

PA2 = QA2

Using distance formula

(a – 2)2 + (0 + 5)2 = (a + 2)2 + (0 – 9)2

∴  a2 – 4a + 4 + 25  = a2 + 4a + 4 + 81

∴   – 4a + 25  = + 4a + 81

∴   – 8a =  56

∴  a = – 7

Hence the required point is (-7, 0)

Example – 09:

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

A(6, 5) and B(– 4, 3) are given points.

Let the point on the y-axis be P(0, b)

Given PA = PB

PA2 = PB2

Using distance formula

(0 – 6)2 + (b – 5)2 = (0 + 4)2 + (b – 3)2

∴  36 + b2 – 10 b + 25  = 16 + b2 – 6b + 9

∴  36  – 10 b + 25  = 16  – 6b + 9

∴  – 4b =  25 – 61 = -36

∴  b = 9

Hence the required point is (0, 9)

Example – 10:

• Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4)

Let P(3, 6) and Q(–3, 4) be the given points.

Given point A(x, y)

Given PA = QA

PA2 = QA2

Using distance formula

(x – 3)2 + (y – 6)2 = (x + 3)2 + (y – 4)2

∴  x2 – 6x + 9 + y2 – 12y + 36  = x2 + 6x + 9 + y2 – 8y + 16

∴  x2 – 6x + 9 + y2 – 12y + 36  – x2 – 6x – 9 – y2 + 8y – 16 = 0

∴   – 12x – 4y + 20  = 0

∴  3x + y – 5 = 0

Hence the requiredrelation is 3x + y – 5 = 0

Example – 11:

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

Let P(7, 1) and Q(3, 5) be the given points.

Given point A(x, y)

Given PA = QA

PA2 = QA2

Using distance formula

(x – 7)2 + (y – 1)2 = (x – 3)2 + (y – 5)2

∴  x2 – 14x + 49 + y2 – 2y + 1  = x2 – 6x + 9 + y2 – 10y + 25

∴  x2 – 14x + 49 + y2 – 2y + 1  – x2 + 6x – 9 – y2 + 10y – 25 = 0

∴   – 8x + 8y + 16  = 0

∴  x – y + 2 = 0

Hence the requiredrelation is x – y + 2 = 0

Example – 12:

• If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also, find the distances QR and PR.

Given P(5, -3) and R(x, 6) be the given points.

Given point Q(0, 1)

Given PQ = RQ

PQ2 = RQ2

Using distance formula

(5 – 0)2 + (-3 – 1)2 = (x – 0)2 + (6 – 1)2

∴  (5)2 + (-4)2 = (x)2 + (5)2

∴  25 + 16 = x2 + 25

∴  x2  = 16

∴  x = ± 4

When R(4, -3)

QR2 = (4 – 0)2 + (-3 – 1)= (4)2 + (-4)2 = 16 + 16 = 32

∴  QR  = 32  =  42   unit   …………….. (1)

PR2 = (4 – 5)2 + (-3 + 3)= (-1)2 + (0)2 = 1 + 0 = 1

∴  PR  = 1  =  1   unit   …………….. (1)

When R(-4, -3)

QR2 = (-4 – 0)2 + (-3 – 1)= (-4)2 + (-4)2 = 16 + 16 = 32

∴  QR  = 32  =  42   unit   …………….. (1)

PR2 = (-4 – 5)2 + (-3 + 3)= (-9)2 + (0)2 = 81 + 0 = 81

∴  PR  = 81  =  9   unit   …………….. (1)

Example – 13:

• If P (6, -1), Q(1, 3), and R(x, 8) are the points and PQ = QR,  find the values of x.

Given P(6, -1) and R(x, 8) be the given points.

Given point Q(1, 3)

Given PQ = RQ

PQ2 = RQ2

Using distance formula

(1 – 6)2 + (3 + 1)2 = (1 – x)2 + (3 – 8)2

∴  (-5)2 + (-4)2 = x2 – 2x + 1 + (-5)2

∴  25 + 16 = x2 – 2x + 1 +25

∴  x2  – 2x + 1 – 16 = 0

∴  x2  – 2x + – 15 = 0

∴  (x + 3)(x – 5) = 0

∴ x + 3 = 0 and x – 5 = 0

∴ x = – 3 and x = 5

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