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Distance between two points

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Find the distance between two points whose coordinates are given:

Let P1 and P2 be the two given points, and let their coordinates be respectively (x1, y1) and (x2, y2).



Draw P1M1 and P2M2 parallel to OY, to meet OX in M1 and M2. Draw P2R parallel to OX to meet M1P1 in R.

Then,

P2R = M2M1 = OM1-OM2 = x1-x2

RP1 = M1P1- M2P2 = y1-y2

And ∆P2RP1 = ∆OM1P1 = 180⁰ - P1M1X = 180⁰ - ω

We therefore, have

P1P22 = P2R2 + RP12 – 2P2R . RP1 cos P2RP1

= (x1 – x2)2 + (y1 –y2)2 – 2(x1-x2)(y1-y2)cos(180⁰ - ω)

= (x1 – x2)2 + (y1 –y2)2 – 2(x1-x2)(y1-y2)cos ω …(1)

If the axes be, as is generally the case, at right angles, we have ω = 90⁰ and hence, cos ω = 0.

The formula (1), then becomes

P1P22 = (x1-x2)2 + (y1 –y2)2

So that in rectangular coordinates the distance between the two points (x1, y1) and (x2 –y2) is,

√((x1-x2)2 + (y1 –y2)2  …..(2)

Note: The distance of the point  (x1, y1) from the origin is √(x1-y1)2, the axis being rectangular. This follows from (2) by making both x2 and y2 equal to zero.

To Find the distance between two points whose coordinates are given with example:




As a numerical example , Let P1 be the point (5, 6) and P2 be the point ( -7, -4 ), so that we have

x1 = 5, y1= 6, x2= -7, and y2 =-4

Then P2R = M2O+OM1 = 7+5

=-x2+x1

And RP1 = RM1 + M1P1 = 4+6

= -y2 +y1

The rest of the proof is as in the last article

Similarly, any other case could be considered to find the distance between two points whose coordinates are given.
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