Question

find the coordinates of the trisenction whose points are equidistant (1,2),(-3,-4)​

Answer 1

let ( x1,x1)=(1,2)

(×2,×2)=(-3,-4)

d=

$\sqrt{(x2 - x1)2 \: + \: (y2 - y1)2}$

=

$\sqrt{ (- 3 -1)2 \ \: + \: ( - 4 - 2)2 \ }$

=

$\sqrt{( - 4)2 \: + \: ( - 6)2}$

=

$\sqrt{16 \: + \: 36}$

=

$\sqrt{52}$

Answer 2

Answer:

let ( x1,x1)=(1,2)

(×2,×2)=(-3,-4)

d=

\sqrt{(x2 - x1)2 \: + \: (y2 - y1)2}

(x2−x1)2+(y2−y1)2

=

\sqrt{ (- 3 -1)2 \ \: + \: ( - 4 - 2)2 \ }

(−3−1)2 +(−4−2)2

=

\sqrt{( - 4)2 \: + \: ( - 6)2}

(−4)2+(−6)2

=

\sqrt{16 \: + \: 36}

16+36

=

\sqrt{52}

52