# DISTANCE BETWEEN TWO POINTS

Let P(x1, y1) and Q(x2, y2) 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  =  x1ON  =  x2 MP  =  y1NQ  =  y2

Step 1 :

Now, (PR  NQ)

PR  =  MN

(Opposite sides of the rectangle MNRP)

Step 2 :

Find the length of MN.

MN  =  ON - OM

MN  =  x2 - x1

Step 3 :

Find the length of RQ.

RQ  =  NQ - NR

RQ  =  y- y1

Step 4 :

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

By Pythagorean Theorem,

PQ2  =  PR2 + RQ

d2  =  (x2 - x1)2 + (y2 - y1)2

By taking positive square root,

d  =  √[(x2 - x1)2 + (y2 - y1)2]

Given two points (x1, y1) and (x2, y2), the distance between these points is given by the formula

√[(x2 - x1)2 + (y2 - y1)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

√[(x2 - x1)2 + (y2 - y1)2]

Substitute (x1, y1)  =  (-12, 3) and (x2, y2)  =  (2, 5).

√[(2 + 12)2 + (5 - 3)2]

√[142 + 22]

=   √[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

√[(x2 - x1)2 + (y2 - y1)2]

Substitute (x1, y1)  =  (-2, -3) and (x2, y2)  =  (6, -5).

√[(6 + 2)2 + (-5 + 3)2]

√[82 + (-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 (3, k)

Solution :

Distance between the above two points  =  2√29

√[(x2 - x1)2 + (y2 - y1)2]  =  2√29

Substitute (x1, y1)  =  (-7, 2) and (x2, y2)  =  (3, k).

√[(3 + 7)2 + (k - 2)2]  =  2√29

√[102 + (k - 2)2]  =  2√29

√[100 + (k - 2)2]  =  2√29

Square both sides.

100 + (k - 2)2  =  (2√29)2

100 + k2 - 2(k)(2) +  22  =  22(√29)2

100 + k2 - 4k +  4  =  4(29)

k2 - 4k + 104  =  116

Subtract 116 from each side.

k2 - 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

√[(x2 - x1)2 + (y2 - y1)2]

To find the distance between the points A and B, substitute (x1, y1)  =  (2, -3) and (x2, y2)  =  (5, 5).

AB  =  √[(5 - 2)2 + (5 + 3)2]

AB  =  √[32 + 82]

AB  =  √(9 + 64)

AB  =  √73 units

Kindly mail your feedback to v4formath@gmail.com

## Recent Articles

1. ### SAT Math Resources (Videos, Concepts, Worksheets and More)

Jul 11, 24 05:20 PM

SAT Math Resources (Videos, Concepts, Worksheets and More)

2. ### SAT Math Videos (Part - 21)

Jul 11, 24 05:14 PM

SAT Math Videos (Part - 21)