sqrt{{left({x}_{2}-{x}_{1}right)}^2+{left({y}_{2}-{y}_{1}right)}^2}=5sqrt{2} を解きます| Microsoft 数学ソルバー

$$\sqrt{ { \left( { x }_{ 2 } - { x }_{ 1 } \right) }^{ 2 } + { \left( { y }_{ 2 } - { y }_{ 1 } \right) }^{ 2 } } = 5 \sqrt{ 2 }$$

x_1 を解く (複素数の解)

$$x_{1}=\sqrt{50-y_{2}^{2}+2y_{1}y_{2}-y_{1}^{2}}+x_{2}$$

$$x_{1}=-\sqrt{50-y_{2}^{2}+2y_{1}y_{2}-y_{1}^{2}}+x_{2}$$

x_2 を解く (複素数の解)

$$x_{2}=-\sqrt{50-y_{2}^{2}+2y_{1}y_{2}-y_{1}^{2}}+x_{1}$$

$$x_{2}=\sqrt{50-y_{2}^{2}+2y_{1}y_{2}-y_{1}^{2}}+x_{1}$$

x_1 を解く

$$x_{1}=\sqrt{50-y_{2}^{2}+2y_{1}y_{2}-y_{1}^{2}}+x_{2}$$

$$x_{1}=-\sqrt{50-y_{2}^{2}+2y_{1}y_{2}-y_{1}^{2}}+x_{2}\text{, }y_{2}\geq y_{1}-5\sqrt{2}\text{ and }y_{2}\leq y_{1}+5\sqrt{2}$$

x_2 を解く

$$x_{2}=-\sqrt{50-y_{2}^{2}+2y_{1}y_{2}-y_{1}^{2}}+x_{1}$$

$$x_{2}=\sqrt{50-y_{2}^{2}+2y_{1}y_{2}-y_{1}^{2}}+x_{1}\text{, }y_{2}\geq y_{1}-5\sqrt{2}\text{ and }y_{2}\leq y_{1}+5\sqrt{2}$$