Question

The value of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units.

Answer 1

Answer:

$10 = \sqrt{ {(y - ( - 3))}^{2} + (10 - 2) ^{2} } \\ {10}^{2} =( y + 3) ^{2} + {8}^{2} \\ 100 = {y}^{2} + 6y + 9 + 6 4\\ {y}^{2} + 6y - 27= 0 \\ {y}^{2} + 9y - 3y - 27 = 0 \\ (y + 9) - 3(y + 9) = 0 \\ (y + 9)(y - 3) = 0 \\ y = - 9 \: and \: y = 3$