Question

Find the z such that the distance between the points (4, 5, z) and (7, 1, -6) is √146 units.

Answer 1

Answer:

$\sqrt{ {(7 - 4)}^{2} + {(1 - 5)}^{2} + {( - 6 - z)}^{2} } = \sqrt{146} \\ 9 + 16 + 36 + {z}^{2} + 12z = 146 \\ {z}^{2} + 12z + 61 - 146 = 0 \\ {z}^{2} + 12z - 85 = 0 \\ {z}^{2} + 17z - 5z - 85 = 0 \\ z(z + 17) - 5(z + 17) = 0 \\ (z + 17)(z - 5) = 0 \\ z = - 17 \: \: or \: \: 5 \\ ans.$