Question

Prove that the point (-2, -11 ) is equidistance from the two points (-3, 7) and (4,6).

Answer 1

Step-by-step explanation:

$\blue{\mathfrak{\underline{\large{given}}}:} \\ points \: (-3, 7), \: (4,6) \: and \: (-2, \: -11)\\ \blue{\mathfrak{\underline{\large{to \: find}}}:} \\ point(-2, \: -11) \: is \: equidistant \: from \: point \: (-3, 7) \: and \: (4,6) \\ \blue{\mathfrak{\underline{\large{finding}}}:}\\ let \: A=(-3, 7), \: B=(4,6), \: C=(-2, \: -11) \\ \blue{\mathfrak{\underline{\large{Using \: distance \: formula }}}:} \\ distance = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} } \\ AC= \sqrt{ {( - 2 - ( - 3))}^{2} + {( - 11 - 7)}^{2} } \\ = \sqrt{ {(1)}^{2} + {( - 18)}^{2} } \\ = \sqrt{1 + 324} \\ = \sqrt{325} \\ = \sqrt{5 \times 5 \times 13} \\ = 5 \sqrt{13} \\ CB= \sqrt{ {(4 - ( - 2))}^{2} + {(6 - ( - 11))}^{2} } \\ = \sqrt{ {(6)}^{2} + {(17)}^{2} } \\ = \sqrt{36 + 289 } \\ = \sqrt{ 325} \\ = 5 \sqrt{13} \\ \blue{\mathfrak{\large{Since,\:AC=CB}}} \\ \blue{\mathfrak{\large{Hence,\:point \:C=(-2, \: -11) \: is \: equidistant \: from }}} \\ \blue{\mathfrak{\large{ points \:A=(-3, 7) \: and \: B=(4,6), }}} \\ \\ \red{\mathfrak{\underline{\large{hope \: it \: helps \: you}}}} \\ \green{\mathfrak{\underline{\large{mark \: me \: brainliest}}}}$