Question

if A(1,7),B(4,2),C(-1,-1) are the vertices of a square ABCD,then show that AC=BD
(co-ordinate geometry)​

Answer 1

Answer:

We have to first find Coordinates of D

Let coordinates of D be (x, y).

using that duagonal bisect each other in square.

So midpoint of AC = midpoint of BD

$( \frac{1 - 1}{2}, \frac{7 - 1}{2} ) = ( \frac{4 + x}{2} , \frac{2 + y}{2} ) \\ (0 , 3) = ( \frac{4 + x}{2} , \frac{2 + y}{2} ) \\ so \: x + 4 = 0 \: and \: y + 2 = 6 \\ = > \: x = - 4 \: and \: y = 4$

$AC= \sqrt{ {( - 1 - 1)}^{2} + {( - 1 - 7)}^{2} } = \sqrt{4 + 64} = \sqrt{68} \\ BD = \sqrt{ {( - 4 - 4)}^{2} + {(4 - 2)}^{2} } = \sqrt{64 + 4} = \sqrt{68}$

Hence, AC = BD

Answer 2

Answer:

this is correct................