I sposób
[tex]C(x,y)[/tex]
[tex](x>0,y>0)[/tex]
Ponieważ A = (-2, -4) i B = (5, -4), czyli
[tex]y_A=y_B=-4[/tex]
więc odcinek AB będzie równoległy do osi OX, stąd
[tex]x_C=x_B=5[/tex]
Obliczam y
[tex]|BC|=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}[/tex]
[tex]|BC|=\sqrt{(x-5)^2+(y+4)^2}[/tex]
[tex]|BC|=\sqrt{(5-5)^2+(y+4)^2}[/tex]
[tex]|BC|=\sqrt{0^2+(y+4)^2}[/tex]
[tex]|BC|=\sqrt{0+(y+4)^2}[/tex]
[tex]|BC|=\sqrt{(y+4)^2}[/tex]
[tex]\sqrt{(y+4)^2}=6[/tex]
[tex]y+4=6[/tex]
[tex]y=6-4[/tex]
[tex]y=2[/tex]
[tex]C(5,2)[/tex]
==========================
II sposób
[tex]C(x,y)[/tex]
[tex](x>0,y>0)[/tex]
Obliczam [tex]|AB|[/tex]
[tex]|AB|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}[/tex]
[tex]|AB|=\sqrt{(5+2)^2+(-4+4)^2}[/tex]
[tex]|AB|=\sqrt{7^2+0^2}[/tex]
[tex]|AB|=\sqrt{7^2}[/tex]
[tex]|AB|=7[/tex]
Wyznaczam [tex]|AC|[/tex]
[tex]|AC|=\sqrt{(x_C-x_A)^2+(y_C-y_A)^2}[/tex]
[tex]|AC|=\sqrt{(x+2)^2+(y+4)^2}[/tex]
Wyznaczam [tex]|BC|[/tex]
[tex]|BC|=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}[/tex]
[tex]|BC|=\sqrt{(x-5)^2+(y+4)^2}[/tex]
Wyznaczam [tex](y+4)^2[/tex]
[tex]|BC|=6[/tex]
[tex]|BC|^2=6^2=36[/tex]
[tex]|BC|=\sqrt{(x-5)^2+(y+4)^2}[/tex]
[tex]|BC|^2=(x-5)^2+(y+4)^2[/tex]
[tex](x-5)^2+(y+4)^2=36[/tex]
[tex](y+4)^2=36-(x-5)^2[/tex]
Obliczam x
[tex]|AB|^2+|BC|^2=|AC|^2[/tex]
[tex]7^2+6^2=(x+2)^2+(y+4)^2[/tex]
[tex]49+36=(x+2)^2+36-(x-5)^2[/tex]
[tex](x+2)^2-(x-5)^2=49+36-36[/tex]
[tex]x^2+4x+4-x^2+10x-25=49[/tex]
[tex]4x+10x=49-4+25[/tex]
[tex]14x=70\ \ \ |:14[/tex]
[tex]x=5[/tex]
Obliczam y
[tex](y+4)^2=36-(x-5)^2[/tex]
[tex](y+4)^2=36-(5-5)^2[/tex]
[tex](y+4)^2=36-0^2[/tex]
[tex](y+4)^2=36-0[/tex]
[tex](y+4)^2=36[/tex]
[tex]y+4=\sqrt{36}[/tex]
[tex]y+4=6[/tex]
[tex]y=6-4[/tex]
[tex]y=2[/tex]
[tex]C(5,2)[/tex]
