Question

Find k so that the distance from (-3,-1) to (k, 5) is 10 units.

Answer 1

k = - 11, 5

Explanation:

By distance formula:

`\sqrt{ \{ {k - ( - 3) \}}^(2) + \{5 - ( - 1) \}^(2) } = 10 \\ \\ \sqrt{ {(k + 3)}^(2) + {(5 + 1)}^(2) } = 10 \\ \\ \sqrt{ {(k + 3)}^(2) + {(6)}^(2) } = 10 \\ \\ \sqrt{ {(k + 3)}^(2) + 36 } = 10 \\ \\ (k + 3)^(2) + 36 = {10}^(2) \\ ..(squaring \: both \: sides) \\ {k}^(2) + 6k + 9 + 36 = 100 \\ {k}^(2) + 6k + 45 - 100 = 0 \\ {k}^(2) + 6k - 55 = 0 \\ {k}^(2) + 11k - 5k - 55 = 0 \\ k(k + 11) - 5(k + 11) = 0 \\ (k + 11)(k - 5) = 0 \\ k + 11 = 0 \: or \: k - 5 = 0 \\ k = - 11 \: or \: k = 5 \\ \\ \huge \red { \boxed{k = \{ - 11, \: 5\}}}`