1.

Find the values of y for which the distance between the points p(2,-3) and q(10,y) is 10 units.

Answer»
\sqrt{(10 - 2) {}^{2} + (y + {3}^{2}) \: = 10 } \\ 64 + 9 + 6y + {y}^{2} = 100 \\ {y}^{2} + 6y - 27 = 0 \\ (y + 9) \: (y - 3) = 0 \\ y = - 9 \: and3
HOPE this HELPS.


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