Question

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

Answer 1

$\sqrt{(10 - 2) {}^{2} + (y + {3}^{2}) \: = 10 } \\ 64 + 9 + 6y + {y}^{2} = 100 \\ {y}^{2} + 6y - 27 = 0 \\ (y + 9) \: (y - 3) = 0 \\ y = - 9 \: and3$
Hope this helps.

Answer 2

Answer

$\begin{lgathered}\sqrt{(10 - 2) {}^{2} + (y + {3}^{2}) \: = 10 } \\</em></strong><strong><em>(</em></strong><strong><em> 64 + 9 + 6y + {y}^{2} = 100 </em></strong><strong><em>)</em></strong><strong><em>\\ {y}^{2}</em></strong><strong><em>(</em></strong><strong><em> + 6y - 27 = 0 \\ (y + 9) \:(y-3)=0\\y=-9\: and3\end {lgathered}$