Réitigh sqrt{(7-x)^2+(1-y)^2}=sqrt{(3-x)^2+(5-y)^2}

$$\sqrt { ( 7 - x ) ^ { 2 } + ( 1 - y ) ^ { 2 } } = \sqrt { ( 3 - x ) ^ { 2 } + ( 5 - y ) ^ { 2 } }$$

Réitigh do x.

$$x=y+2$$

Céimeanna Réitigh

Cearnaigh an dá thaobh den chothromóid.

$$\left(\sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(7-x\right)^{2} a leathnú.

$$\left(\sqrt{49-14x+x^{2}+\left(1-y\right)^{2}}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(1-y\right)^{2} a leathnú.

$$\left(\sqrt{49-14x+x^{2}+1-2y+y^{2}}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Suimigh 49 agus 1 chun 50 a fháil.

$$\left(\sqrt{50-14x+x^{2}-2y+y^{2}}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Ríomh cumhacht \sqrt{50-14x+x^{2}-2y+y^{2}} de 2 agus faigh 50-14x+x^{2}-2y+y^{2}.

$$50-14x+x^{2}-2y+y^{2}=\left(\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(3-x\right)^{2} a leathnú.

$$50-14x+x^{2}-2y+y^{2}=\left(\sqrt{9-6x+x^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(5-y\right)^{2} a leathnú.

$$50-14x+x^{2}-2y+y^{2}=\left(\sqrt{9-6x+x^{2}+25-10y+y^{2}}\right)^{2}$$

Suimigh 9 agus 25 chun 34 a fháil.

$$50-14x+x^{2}-2y+y^{2}=\left(\sqrt{34-6x+x^{2}-10y+y^{2}}\right)^{2}$$

Ríomh cumhacht \sqrt{34-6x+x^{2}-10y+y^{2}} de 2 agus faigh 34-6x+x^{2}-10y+y^{2}.

$$50-14x+x^{2}-2y+y^{2}=34-6x+x^{2}-10y+y^{2}$$

Cuir 6x leis an dá thaobh.

$$50-14x+x^{2}-2y+y^{2}+6x=34+x^{2}-10y+y^{2}$$

Comhcheangail -14x agus 6x chun -8x a fháil.

$$50-8x+x^{2}-2y+y^{2}=34+x^{2}-10y+y^{2}$$

Bain x^{2} ón dá thaobh.

$$50-8x+x^{2}-2y+y^{2}-x^{2}=34-10y+y^{2}$$

Comhcheangail x^{2} agus -x^{2} chun 0 a fháil.

$$50-8x-2y+y^{2}=34-10y+y^{2}$$

Bain 50 ón dá thaobh.

$$-8x-2y+y^{2}=34-10y+y^{2}-50$$

Dealaigh 50 ó 34 chun -16 a fháil.

$$-8x-2y+y^{2}=-16-10y+y^{2}$$

Cuir 2y leis an dá thaobh.

$$-8x+y^{2}=-16-10y+y^{2}+2y$$

Comhcheangail -10y agus 2y chun -8y a fháil.

$$-8x+y^{2}=-16-8y+y^{2}$$

Bain y^{2} ón dá thaobh.

$$-8x=-16-8y+y^{2}-y^{2}$$

Comhcheangail y^{2} agus -y^{2} chun 0 a fháil.

$$-8x=-16-8y$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$-8x=-8y-16$$

Roinn an dá thaobh faoi -8.

$$\frac{-8x}{-8}=\frac{-8y-16}{-8}$$

Má roinntear é faoi -8 cuirtear an iolrúchán faoi -8 ar ceal.

$$x=\frac{-8y-16}{-8}$$

Roinn -16-8y faoi -8.

$$x=y+2$$

Cuir y+2 in ionad x sa chothromóid \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}.

$$\sqrt{\left(7-\left(y+2\right)\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-\left(y+2\right)\right)^{2}+\left(5-y\right)^{2}}$$

Simpligh. An luach x=y+2 shásaíonn an gcothromóid.

$$\left(2y^{2}-12y+26\right)^{\frac{1}{2}}=\left(2y^{2}-12y+26\right)^{\frac{1}{2}}$$

Ag an chothromóid \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}} réiteach uathúil.

$$x=y+2$$

Réitigh do y.

$$y=x-2$$

Céimeanna Réitigh

Cearnaigh an dá thaobh den chothromóid.

$$\left(\sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(7-x\right)^{2} a leathnú.

$$\left(\sqrt{49-14x+x^{2}+\left(1-y\right)^{2}}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(1-y\right)^{2} a leathnú.

$$\left(\sqrt{49-14x+x^{2}+1-2y+y^{2}}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Suimigh 49 agus 1 chun 50 a fháil.

$$\left(\sqrt{50-14x+x^{2}-2y+y^{2}}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Ríomh cumhacht \sqrt{50-14x+x^{2}-2y+y^{2}} de 2 agus faigh 50-14x+x^{2}-2y+y^{2}.

$$50-14x+x^{2}-2y+y^{2}=\left(\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(3-x\right)^{2} a leathnú.

$$50-14x+x^{2}-2y+y^{2}=\left(\sqrt{9-6x+x^{2}+\left(5-y\right)^{2}}\right)^{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(5-y\right)^{2} a leathnú.

$$50-14x+x^{2}-2y+y^{2}=\left(\sqrt{9-6x+x^{2}+25-10y+y^{2}}\right)^{2}$$

Suimigh 9 agus 25 chun 34 a fháil.

$$50-14x+x^{2}-2y+y^{2}=\left(\sqrt{34-6x+x^{2}-10y+y^{2}}\right)^{2}$$

Ríomh cumhacht \sqrt{34-6x+x^{2}-10y+y^{2}} de 2 agus faigh 34-6x+x^{2}-10y+y^{2}.

$$50-14x+x^{2}-2y+y^{2}=34-6x+x^{2}-10y+y^{2}$$

Cuir 10y leis an dá thaobh.

$$50-14x+x^{2}-2y+y^{2}+10y=34-6x+x^{2}+y^{2}$$

Comhcheangail -2y agus 10y chun 8y a fháil.

$$50-14x+x^{2}+8y+y^{2}=34-6x+x^{2}+y^{2}$$

Bain y^{2} ón dá thaobh.

$$50-14x+x^{2}+8y+y^{2}-y^{2}=34-6x+x^{2}$$

Comhcheangail y^{2} agus -y^{2} chun 0 a fháil.

$$50-14x+x^{2}+8y=34-6x+x^{2}$$

Bain 50 ón dá thaobh.

$$-14x+x^{2}+8y=34-6x+x^{2}-50$$

Dealaigh 50 ó 34 chun -16 a fháil.

$$-14x+x^{2}+8y=-16-6x+x^{2}$$

Cuir 14x leis an dá thaobh.

$$x^{2}+8y=-16-6x+x^{2}+14x$$

Comhcheangail -6x agus 14x chun 8x a fháil.

$$x^{2}+8y=-16+8x+x^{2}$$

Bain x^{2} ón dá thaobh.

$$8y=-16+8x+x^{2}-x^{2}$$

Comhcheangail x^{2} agus -x^{2} chun 0 a fháil.

$$8y=-16+8x$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$8y=8x-16$$

Roinn an dá thaobh faoi 8.

$$\frac{8y}{8}=\frac{8x-16}{8}$$

Má roinntear é faoi 8 cuirtear an iolrúchán faoi 8 ar ceal.

$$y=\frac{8x-16}{8}$$

Roinn -16+8x faoi 8.

$$y=x-2$$

Cuir x-2 in ionad y sa chothromóid \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}.

$$\sqrt{\left(7-x\right)^{2}+\left(1-\left(x-2\right)\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-\left(x-2\right)\right)^{2}}$$

Simpligh. An luach y=x-2 shásaíonn an gcothromóid.

$$\left(2x^{2}-20x+58\right)^{\frac{1}{2}}=\left(2x^{2}-20x+58\right)^{\frac{1}{2}}$$

Ag an chothromóid \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}} réiteach uathúil.

$$y=x-2$$