직각삼각형이 되는 좌표 구하기

$\displaystyle \overline{\mathrm{AB}} ^{2} = \left( 5 - 3 \right) ^{2} + \left( 5 - 1 \right) ^{2} = 20$ $\displaystyle \overline{\mathrm{AC}} ^{2} = \left( a - 3 \right) ^{2} + \left( - 1 - 1 \right) ^{2} = \left( a - 3 \right) ^{2} + 4$ $\displaystyle \overline{\mathrm{BC}} ^{2} = \left( a - 5 \right) ^{2} + \left( - 1 - 5 \right) ^{2} = \left( a - 5 \right) ^{2} + 36$ $\mathrm{A}$가 직각이므로 $\displaystyle \overline{\mathrm{AB}} ^{2} + \overline{\mathrm{AC}} ^{2} = \overline{\mathrm{BC}} ^{2}$에서 $20 + \left( a - 3 \right) ^{2} + 4 = \left( a - 5 \right) ^{2} + 36$ $a ^{2} - 6 a + 33 = a ^{2} - 10 a + 61$ $4 a = 28$ 따라서 $a = 7$