Question

find the value of y for which the distance btw the points of A ( 3,-1 ) and B ( 11 , y ) is 10 units​

Answer 1

$10 = \sqrt{ {(11 - 3)}^{2} + {(y - ( - 1))}^{2} } \\ 10 = \sqrt{ {8}^{2} + {(y + 1)}^{2} } \\ 10 = \sqrt{64 + {y}^{2} + 2y + 1 } \\ 10 = \sqrt{ {y}^{2} + 2y + 65} \\ {10}^{2} = {y}^{2} + 2y + 65 \\ {y}^{2} + 2y + 65 = 100$

${y}^{2} + 2y + 65 - 100 = 0 \\ {y}^{2} + 2y - 35 = 0 \\ now \: solve \: by \: splitting \: the \: middle \\ term \\ {y}^{2} + 2y - 35 = 0 \\ {y}^{2} + (7 - 5)y - 35 = 0 \\ {y}^{2} + 7y - 5y - 35 = 0 \\ y(y + 7) - 5(y + 7) = 0 \\ (y + 7)(y - 5) = 0 \\ y + 7 = 0 \\ y = - 7 \\ \\ y - 5 = 0 \\ y = 5$

Hence, there can be 2 values of y: either -7 or 5.

Hope it will help you.