Question

Find the value of k for which the distance between the points (2,-3) and (k,5) is 10​

Answer 1

Answer:

$k = 8$

Step-by-step explanation:

$Let \: the \: points \: be \: A \: and \: B \\ A=(2,-3), B=(k, 5) \\ x1 = 2, \: x2=k, \: y1=(-3), \: y2=5\\ Distance \: between \: AB=10 \: units \\ \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2}} = 10 \\ \sqrt{ {(k - 2)}^{2} + {(5 + 3)}^{2}} = 10 \\ Squaring \: on \: both \: the \: sides \\ {( \sqrt{ {(k - 2)}^{2} + {(8)}^{2} } )}^{2} = {(10)}^{2} \\ {(k - 2)}^{2} + {(8)}^{2} = {(10)}^{2} \\ {k}^{2} - 4k + 4 + 64 = 100 \\ {k}^{2} - 4k = 100 - 4 - 64 \\ {k}^{2} - 4k = 32 \\ {k}^{2} - 4k - 32 = 0 \\ {k}^{2} - 8k + 4k - 32 = 0\\ ({k}^{2} - 8k) + (4k - 32) = 0 \\ k(k - 8) + 4(k - 8) = 0 \\ (k - 8)(k + 4) = 0 \\ (k - 8 = 0)(k + 4 = 0) \\ k = 8 \: and \: k = ( - 4) \\ Therefore, \: the \: value \: of \: k \: is \: 8.$