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Chapter 3: Coordinate Geometry

    Introduction

    Additional Section: Distance Formula

    Cartesian System

    Exercise 3.1

    Exercise 3.2

    Summary

    Enhanced Curriculum Support

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Chapter 3: Coordinate Geometry > Additional Section: Distance Formula

Additional Section: Distance Formula

Consider the figure above. What is the distance from b to a?

Since both the points lie on the x-axis, we just need to walk on the x axis. So, the distance is nothing but ob - oa = units.

Now, what is the distance from d to c? These points are on the y-axis and their distance is units.

Can we find the distance between a and c? The distance between a and c is units. (hint: apply pythagoras theorem).

The easy parts are done :)

You have seen how we can find distances between points lying on the axis. What about points that are not on the axes like below?

This is also easy enough. We just need to visualize a scenario which we know. Draw a line from p to q.

The line is just hanging in the air. We still don't have enough information. Why dont you draw a perpendicular from p to x axis?

We are getting somewhere, but still we dont have any formation that we can recognize. Lets continue. Draw a perpendicular from q to x-axis.

Now we are getting somewhere. We just need one more final line segment. Can you guess and draw?

Yes! Now you have a right angle triangle and once we have the right angle triangle we can use our ever dependable Pythagoras theorem. Using the theorem we get the distance pq is .

Let us generalize this. If p and q are at (x1,y1) and (x2,y2) respectively then the distance between p and q can be calculated by

pq2=pt2+qt2=x2−x12+y2−y12

So pq = x2−x12+y2−y12

This is called as the distance formula.

Let's see with an example?

Let us apply this learning and see if we can solve some practical problem.

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

We are given two points and we have to find other points which are equidistant from these two points.

Distance of (x,y) from the first point (7,1) is

Distance of (x,y) from the second point (3,5) is

Since (x,y) is equidistant from both points we have: x−72+y−12=x−32+y−52

Solving for this we get the equation .

Plot the line on the graph and we observe that both lines intersect at (,).

We can conclude that point (5,3) is eqidistance from point p and point q.