Find the distance between two points whose coordinates are given:

Let P_{1} and P_{2} be the two given points, and let their coordinates be respectively (x_{1}, y_{1}) and (x_{2}, y_{2}).

Draw P_{1}M_{1} and P_{2}M_{2} parallel to OY, to meet OX in M_{1} and M_{2}. Draw P_{2}R parallel to OX to meet M_{1}P_{1} in R.

Then,

P_{2}R = M_{2}M_{1} = OM_{1}-OM_{2} = x_{1}-x_{2}

RP_{1} = M_{1}P_{1}– M_{2}P_{2} = y_{1}-y_{2}

And ∆P_{2}RP_{1} = ∆OM_{1}P_{1} = 180⁰ – P_{1}M_{1}X = 180⁰ – ω

We therefore, have

P_{1}P_{2}2 = P_{2}R2 + RP_{1}2 – 2P_{2}R . RP_{1} cos P_{2}RP_{1}

= (x_{1} – x_{2})^{2} + (y_{1} –y_{2})^{2} – 2(x_{1}-x_{2})(y_{1}-y_{2})cos(180⁰ – ω)

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

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

So that in rectangular coordinates the distance between the two points (x_{1}, y_{1}) and (x_{2} –y_{2}) is,

√((x_{1}-x_{2})^{2} + (y_{1} –y_{2})^{2} …..(2)

*Note: The distance of the point (x _{1}, y_{1}) from the origin is √(x_{1}-y_{1})^{2}, the axis being rectangular. This follows from (2) by making both x_{2} and y_{2} equal to zero.*

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

As a numerical example , Let P_{1} be the point (5, 6) and P_{2} be the point ( -7, -4 ), so that we have

x_{1} = 5, y_{1}= 6, x_{2}= -7, and y_{2} =-4

Then P_{2}R = M_{2}O+OM_{1} = 7+5

=-x_{2}+x_{1}

And RP_{1} = RM_{1} + M_{1}P_{1} = 4+6

= -y_{2} +y_{1}

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.