Question

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

Answer 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 $\sf{A( x_1 , y_1) \: \: and \: \: B( x_2 , y_2)}$ the distance between the points

$= \sf{ \sqrt{ {(x_2 -x_1 )}^{2} + {(y_2 -y_1 )}^{2} } }$

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

$\sf{ \sqrt{ {(12 - 4)}^{2} + {(p + 5)}^{2} } = 10}$

$\sf{ \implies \sqrt{ {(8)}^{2} + {(p + 5)}^{2} } = 10}$

$\sf{ \implies {(8)}^{2} + {(p + 5)}^{2} = {10}^{2} }$

$\sf{ \implies 64 + {(p + 5)}^{2} = 100 }$

$\sf{ \implies {(p + 5)}^{2} = 36 }$

$\sf{ \implies (p + 5) = \pm \: 6 }$

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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