Answer:

The distance between two points
and
is given by the formula,

Proof: Let X'OX and YOY' be the x-axis and y-axis respectively. Then, O is the origin.
Let
and

be the given points.
Draw AL perpendicular to OX, BM perpendicular to OX and AN perpendicular to BM
Now,



In right angled triangle
,by Pythagorean theorem,
We have,



Thus ,the distance between the points A(x_1,y_1) and B(x_2,y_2) is given by,
