Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}) be two points in the Cartesian plane (or xy –plane), at a distance ‘d’ apart.

That is

d = PQ

By the definition of coordinates,

OM = x ON = x |
MP = y NQ = y |

**Step 1 : **

Now, (PR ⊥ NQ)

PR = MN

(Opposite sides of the rectangle MNRP)

**Step 2 : **

Find the length of MN.

MN = ON - OM

MN = x_{2 }- x_{1}

**Step 3 : **

Find the length of RQ.

RQ = NQ - NR

RQ = y_{2 }- y_{1}

**Step 4 : **

Triangle PQR is right angled at R. (PR ⊥ NQ)

By Pythagorean Theorem,

PQ^{2} = PR^{2} + RQ^{2 }

d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

By taking positive square root,

d = √[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]

Given two points (x_{1}, y_{1}) and (x_{2}, y_{2}), the distance between these points is given by the formula

**√[(x _{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]**

**Example 1 :**

Find the distance between the two points given below.

(-12, 3) and (2, 5)

**Solution:**

Formula for the distance between the two points is

√[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]

Substitute (x_{1}, y_{1}) = (-12, 3) and (x_{2}, y_{2}) = (2, 5).

**= **
√[(2 + 12)^{2} + (5 - 3)^{2}]

**= ** √[14^{2} + 2^{2}]

= ** **
√[196 + 4]

=** **
√200

= ** **
√(2 ⋅ 10 ⋅ 10)

= ** **10√2

So, the distance between the given points is 10√2 units.

**Example 2 :**

Find the distance between the two points given below.

(-2, -3) and (6, -5)

**Solution:**

Formula for the distance between the two points is

√[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]

Substitute (x_{1}, y_{1}) = (-2, -3) and (x_{2}, y_{2}) = (6, -5).

**= ** √[(6 + 2)^{2} + (-5 + 3)^{2}]

**= ** √[8^{2} + (-2)^{2}]

= ** **√[64 + 4]

=** ** √68

= ** **√(2 ⋅ 2 ⋅ 17)

= 2√17

So, the distance between the given points is 2√17 units.

**Example 3 :**

If the distance between the two points given below is 2√29, then find the value of k, given that k > 0.

(-7, 2) and (0, k)

**Solution:**

Distance between the above two points = 2√29

√[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}] = 2√29

Substitute (x_{1}, y_{1}) = (-7, 2) and (x_{2}, y_{2}) = (3, k).

√[(3 + 7)^{2} + (k - 2)^{2}] = 2√29

√[10^{2} + (k - 2)^{2}] = 2√29

√[100 + (k - 2)^{2}] = 2√29

Square both sides.

100 + (k - 2)^{2} = (2√29)^{2}

100 + k^{2} - 2(k)(2) + 2^{2} = 2^{2}(√29)^{2}

100 + k^{2} - 4k + 4 = 4(29)

k^{2} - 4k + 104 = 116

Subtract 116 from each side.

k^{2} - 4k - 12 = 0

(k - 6)(k + 2) = 0

k - 6 = 0 or k + 2 = 0

k = 6 or k = -2

Because k > 0, we have

k = 6

**Example 4 :**

Find the distance between the points A and B in the xy-pane shown below.

**Solution : **

Identify the points A and B in the xy-plane above.

Formula for the distance between the two points is

√[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]

To find the distance between the points A and B, substitute (x_{1}, y_{1}) = (2, -3) and (x_{2}, y_{2}) = (5, 5).

AB = √[(5 - 2)^{2} + (5 + 3)^{2}]

AB = √[3^{2} + 8^{2}]

AB = √(9 + 64)

AB = √73 units

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