Question

Prove that the distance between A(1,2) and
B(3,8) is double of the distance of C(3,-1)
from the origin.
solve the question pls​

Answer 1

Given ,

The distance between A(1,2) and B(3,8) is double of the distance of C(3,-1) from the origin

We know that , the distance b/w two points is given by

$\boxed{ \sf{D = \sqrt{ {( x_{2} -x_{1} )}^{2} + {(y_{2} -y_{1})}^{2} } }}$

Thus , the distance between A(1,2) and

B(3,8) will be

AB = √{(3 - 1)² + (8 - 2)²}

AB = √{4 + 36}

AB = √40

AB = 2√10 units

Now , the distance between C(3,-1)

from the origin or D(0,0) will be

CD = √{(0 - 3)² + (0 + 1)²}

CD = √{9 + 1}

CD = √10 units

It is observed that ,

CD = 2AB

Hence proved