distance between the point of (-4,5) and (12,p) is 10unit find the value of p​

1 thought on “distance between the point of (-4,5) and (12,p) is 10unit find the value of p​”

1. SOLUTION

   GIVEN

   The distance between the point of ( 4, – 5) and (12,p) is 10 unit

   TO DETERMINE

   The value of p

   CONCEPT TO BE IMPLEMENTED

   For the given two points [tex] \sf{A( x_1 , y_1) \: \: and \: \: B( x_2 , y_2)}[/tex] the distance between the points

   [tex] = \sf{ \sqrt{ {(x_2 -x_1 )}^{2} + {(y_2 -y_1 )}^{2} } }[/tex]

   EVALUATION

   Here it is given that the distance between the point of (4,-5) and (12,p) is 10 unit

   So by the given condition

   [tex] \sf{ \sqrt{ {(12 – 4)}^{2} + {(p + 5)}^{2} } = 10}[/tex]

   [tex] \sf{ \implies \sqrt{ {(8)}^{2} + {(p + 5)}^{2} } = 10}[/tex]

   [tex] \sf{ \implies {(8)}^{2} + {(p + 5)}^{2} = {10}^{2} }[/tex]

   [tex] \sf{ \implies 64 + {(p + 5)}^{2} = 100 }[/tex]

   [tex] \sf{ \implies {(p + 5)}^{2} = 36 }[/tex]

   [tex] \sf{ \implies (p + 5) = \pm \: 6 }[/tex]

   Now p + 5 = 6 gives p = 1

   p + 5 = – 6 gives p = – 11

   FINAL ANSWER

   Hence the required value of p = 1 or – 11

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