Długość odcinka o końcach w punktach [tex]A=(x_{a} ,y_{a} )~~,~~B=(x_{b} ,y_{b})[/tex]
jest wyrażona wzorem: [tex]\mid AB \mid=\sqrt{(x_{b} -x_{a}) ^{2} +(y_{b} -y_{a})^{2} }[/tex]
[tex]A=(32,32)~~,B=(80,96)\\\\\mid AB \mid =\sqrt{(80-32)^{2} +(96-32)^{2} } =\sqrt{48^{2}+64^{2} } =\sqrt{2304+4096} =\sqrt{6400} =80\\\\\\C=(-32,144)~~,~~D=(96,48)\\\\\mid CD \mid =\sqrt{(96-(-32))^{2}+(48-144)^{2} } =\sqrt{(96+32)^{2}+(48-144)^{2} } =\sqrt{128^{2}+(-96)^{2} } =\sqrt{16384+9216} =\sqrt{25600} =160\\\\\\\mid AB \mid = 80~~\land ~~\mid CD \mid=160=2\cdot 80 ~~\Rightarrow ~~\mid CD \mid=2 \cdot \mid AB \mid~~\Rightarrow ~~\mid AB \mid=\dfrac{\mid CD \mid}{2} \\\\[/tex][tex]A.~~~~Tak~~~~poniewaz~~~~3.~~dlugosc~~odcinka~~\mid AB \mid = 80,~~a~~\mid CD \mid=160\\\\\\Odp:~~A.~3[/tex]
