Question

show that (3,4) is equidistant from (1,7) and (5,1)

Answer 1

distance between 2 points (x1,y1) and (x2,y2)
is
$\sqrt{ {(x1 - x2)}^{2} + ( {y1 - y2)}^{2} }$
right,
distance between 3,4 and 5,1 is
$\sqrt{ {(3 - 5)}^{2} + ( {4 - 1)}^{2} }$
$\sqrt{ {-2}^{2} + {3}^{2} }}$
$\sqrt{ {4+9 }$
$\sqrt{ {13 }$
distance between 3,4 and 1,7 is
$\sqrt{ {(3 - 1)}^{2} + ( {4 - 7)}^{2} }$
$\sqrt{ {2}^{2} + {-3}^{2} }$
$\sqrt{ {4+9 }$
$\sqrt{ {13 }$
they both r equal