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

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

Whakaoti mō x

$$x=y+2$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(7-x\right)^{2}.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(1-y\right)^{2}.

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

Tāpirihia te 49 ki te 1, ka 50.

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

Tātaihia te \sqrt{50-14x+x^{2}-2y+y^{2}} mā te pū o 2, kia riro ko 50-14x+x^{2}-2y+y^{2}.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(3-x\right)^{2}.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(5-y\right)^{2}.

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

Tāpirihia te 9 ki te 25, ka 34.

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

Tātaihia te \sqrt{34-6x+x^{2}-10y+y^{2}} mā te pū o 2, kia riro ko 34-6x+x^{2}-10y+y^{2}.

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

Me tāpiri te 6x ki ngā taha e rua.

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

Pahekotia te -14x me 6x, ka -8x.

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

Tangohia te x^{2} mai i ngā taha e rua.

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

Pahekotia te x^{2} me -x^{2}, ka 0.

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

Tangohia te 50 mai i ngā taha e rua.

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

Tangohia te 50 i te 34, ka -16.

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

Me tāpiri te 2y ki ngā taha e rua.

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

Pahekotia te -10y me 2y, ka -8y.

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

Tangohia te y^{2} mai i ngā taha e rua.

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

Pahekotia te y^{2} me -y^{2}, ka 0.

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

He hanga arowhānui tō te whārite.

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

Whakawehea ngā taha e rua ki te -8.

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

Mā te whakawehe ki te -8 ka wetekia te whakareanga ki te -8.

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

Whakawehe -16-8y ki te -8.

$$x=y+2$$

Whakakapia te y+2 mō te x i te whārite \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}.

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

Whakarūnātia. Ko te uara x=y+2 kua ngata te whārite.

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

Ko te whārite \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}} he rongoā ahurei.

$$x=y+2$$

Whakaoti mō y

$$y=x-2$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(7-x\right)^{2}.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(1-y\right)^{2}.

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

Tāpirihia te 49 ki te 1, ka 50.

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

Tātaihia te \sqrt{50-14x+x^{2}-2y+y^{2}} mā te pū o 2, kia riro ko 50-14x+x^{2}-2y+y^{2}.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(3-x\right)^{2}.

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(5-y\right)^{2}.

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

Tāpirihia te 9 ki te 25, ka 34.

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

Tātaihia te \sqrt{34-6x+x^{2}-10y+y^{2}} mā te pū o 2, kia riro ko 34-6x+x^{2}-10y+y^{2}.

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

Me tāpiri te 10y ki ngā taha e rua.

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

Pahekotia te -2y me 10y, ka 8y.

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

Tangohia te y^{2} mai i ngā taha e rua.

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

Pahekotia te y^{2} me -y^{2}, ka 0.

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

Tangohia te 50 mai i ngā taha e rua.

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

Tangohia te 50 i te 34, ka -16.

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

Me tāpiri te 14x ki ngā taha e rua.

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

Pahekotia te -6x me 14x, ka 8x.

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

Tangohia te x^{2} mai i ngā taha e rua.

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

Pahekotia te x^{2} me -x^{2}, ka 0.

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

He hanga arowhānui tō te whārite.

$$8y=8x-16$$

Whakawehea ngā taha e rua ki te 8.

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

Mā te whakawehe ki te 8 ka wetekia te whakareanga ki te 8.

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

Whakawehe -16+8x ki te 8.

$$y=x-2$$

Whakakapia te x-2 mō te y i te whārite \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}.

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

Whakarūnātia. Ko te uara y=x-2 kua ngata te whārite.

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

Ko te whārite \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}} he rongoā ahurei.

$$y=x-2$$