Find k such that P ( 0 , 2 ) is equidistant from ( 3 , k ) and ( k , 5 ) . Why can't we use the Midpoint Formula?

Question

Find k such that   P  (  0  ,  2  ) is equidistant from   (  3  ,  k  ) and   (  k  ,  5  ) . Why can&#039;t we use the Midpoint Formula?

Zayden Andrade · Accepted Answer

Step 1Guide:You can&#039;t do that because p need not be on the straight line connecting   (  3  ,  k  ) and   (  k  ,  5  ) .Instead solve for  (  3  −  0      )    2    +  (  k  −  2      )    2    =  (  k  −  0      )    2    +  (  5  −  2      )    2

Petrovcic2x · Accepted Answer

Step 1Let&#039;s calculate the distance from P to   (  3  ,  k  ) and   (  k  ,  5  ) . By definition, the distances are  d  (  (  0  ,  2  )  ,  (  3  ,  k  )  )  =      (    0    −    3          )      2        +    (    k    −    2          )      2        =      9    +    (    k    −    2          )      2      and  d  (  (  0  ,  2  )  ,  (  k  ,  5  )  )  =      (    0    −    k          )      2        +    (    2    −    5          )      2        =      (    −    k          )      2        +    9    =            k      2        +    9  These distances are equal, therefore      9    +    (    k    −    2          )      2        =            k      2        +    9  Squaring, we get that  9  +  (  k  −  2      )    2    =      k    2    +  9or  (  k  −  2      )    2    =      k    2  This is equivalent to       k    2    −  4  k  +  4  =      k    2   or   −  4  k  +  4  =  0 . Subtracting 4 from both sides we get   −  4  k  =  −  4 or                               k          =          1

Find k such that P ( 0 , 2 ) is equidistant from ( 3 , k ) an

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