Whakaoti i te sqrt{(x-2)^2+(y-2)^2}=sqrt{(x-(-2))^2+(y-4)^2}

$$\sqrt { ( x - 2 ) ^ { 2 } + ( y - 2 ) ^ { 2 } } = \sqrt { ( x - ( - 2 ) ) ^ { 2 } + ( y - 4 ) ^ { 2 } }$$

Whakaoti mō x

$$x=\frac{y-3}{2}$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{\left(x-2\right)^{2}+\left(y-2\right)^{2}}\right)^{2}=\left(\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{x^{2}-4x+4+\left(y-2\right)^{2}}\right)^{2}=\left(\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{x^{2}-4x+4+y^{2}-4y+4}\right)^{2}=\left(\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

Tāpirihia te 4 ki te 4, ka 8.

$$\left(\sqrt{x^{2}-4x+8+y^{2}-4y}\right)^{2}=\left(\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

Tātaihia te \sqrt{x^{2}-4x+8+y^{2}-4y} mā te pū o 2, kia riro ko x^{2}-4x+8+y^{2}-4y.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

Ko te tauaro o -2 ko 2.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{\left(x+2\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{x^{2}+4x+4+\left(y-4\right)^{2}}\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(y-4\right)^{2}.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{x^{2}+4x+4+y^{2}-8y+16}\right)^{2}$$

Tāpirihia te 4 ki te 16, ka 20.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{x^{2}+4x+20+y^{2}-8y}\right)^{2}$$

Tātaihia te \sqrt{x^{2}+4x+20+y^{2}-8y} mā te pū o 2, kia riro ko x^{2}+4x+20+y^{2}-8y.

$$x^{2}-4x+8+y^{2}-4y=x^{2}+4x+20+y^{2}-8y$$

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

$$x^{2}-4x+8+y^{2}-4y-x^{2}=4x+20+y^{2}-8y$$

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

$$-4x+8+y^{2}-4y=4x+20+y^{2}-8y$$

Tangohia te 4x mai i ngā taha e rua.

$$-4x+8+y^{2}-4y-4x=20+y^{2}-8y$$

Pahekotia te -4x me -4x, ka -8x.

$$-8x+8+y^{2}-4y=20+y^{2}-8y$$

Tangohia te 8 mai i ngā taha e rua.

$$-8x+y^{2}-4y=20+y^{2}-8y-8$$

Tangohia te 8 i te 20, ka 12.

$$-8x+y^{2}-4y=12+y^{2}-8y$$

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

$$-8x-4y=12+y^{2}-8y-y^{2}$$

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

$$-8x-4y=12-8y$$

Me tāpiri te 4y ki ngā taha e rua.

$$-8x=12-8y+4y$$

Pahekotia te -8y me 4y, ka -4y.

$$-8x=12-4y$$

Whakawehea ngā taha e rua ki te -8.

$$\frac{-8x}{-8}=\frac{12-4y}{-8}$$

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

$$x=\frac{12-4y}{-8}$$

Whakawehe 12-4y ki te -8.

$$x=\frac{y-3}{2}$$

Whakakapia te \frac{y-3}{2} mō te x i te whārite \sqrt{\left(x-2\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}.

$$\sqrt{\left(\frac{y-3}{2}-2\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(\frac{y-3}{2}-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}$$

Whakarūnātia. Ko te uara x=\frac{y-3}{2} kua ngata te whārite.

$$\frac{1}{2}\left(65-30y+5y^{2}\right)^{\frac{1}{2}}=\frac{1}{2}\left(65-30y+5y^{2}\right)^{\frac{1}{2}}$$

Ko te whārite \sqrt{\left(x-2\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(y-4\right)^{2}+\left(x-\left(-2\right)\right)^{2}} he rongoā ahurei.

$$x=\frac{y-3}{2}$$

Whakaoti mō y

$$y=2x+3$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{\left(x-2\right)^{2}+\left(y-2\right)^{2}}\right)^{2}=\left(\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{x^{2}-4x+4+\left(y-2\right)^{2}}\right)^{2}=\left(\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{x^{2}-4x+4+y^{2}-4y+4}\right)^{2}=\left(\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

Tāpirihia te 4 ki te 4, ka 8.

$$\left(\sqrt{x^{2}-4x+8+y^{2}-4y}\right)^{2}=\left(\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

Tātaihia te \sqrt{x^{2}-4x+8+y^{2}-4y} mā te pū o 2, kia riro ko x^{2}-4x+8+y^{2}-4y.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

Ko te tauaro o -2 ko 2.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{\left(x+2\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{x^{2}+4x+4+\left(y-4\right)^{2}}\right)^{2}$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(y-4\right)^{2}.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{x^{2}+4x+4+y^{2}-8y+16}\right)^{2}$$

Tāpirihia te 4 ki te 16, ka 20.

$$x^{2}-4x+8+y^{2}-4y=\left(\sqrt{x^{2}+4x+20+y^{2}-8y}\right)^{2}$$

Tātaihia te \sqrt{x^{2}+4x+20+y^{2}-8y} mā te pū o 2, kia riro ko x^{2}+4x+20+y^{2}-8y.

$$x^{2}-4x+8+y^{2}-4y=x^{2}+4x+20+y^{2}-8y$$

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

$$x^{2}-4x+8+y^{2}-4y-y^{2}=x^{2}+4x+20-8y$$

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

$$x^{2}-4x+8-4y=x^{2}+4x+20-8y$$

Me tāpiri te 8y ki ngā taha e rua.

$$x^{2}-4x+8-4y+8y=x^{2}+4x+20$$

Pahekotia te -4y me 8y, ka 4y.

$$x^{2}-4x+8+4y=x^{2}+4x+20$$

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

$$-4x+8+4y=x^{2}+4x+20-x^{2}$$

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

$$-4x+8+4y=4x+20$$

Me tāpiri te 4x ki ngā taha e rua.

$$8+4y=4x+20+4x$$

Pahekotia te 4x me 4x, ka 8x.

$$8+4y=8x+20$$

Tangohia te 8 mai i ngā taha e rua.

$$4y=8x+20-8$$

Tangohia te 8 i te 20, ka 12.

$$4y=8x+12$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4y}{4}=\frac{8x+12}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

$$y=\frac{8x+12}{4}$$

Whakawehe 8x+12 ki te 4.

$$y=2x+3$$

Whakakapia te 2x+3 mō te y i te whārite \sqrt{\left(x-2\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(y-4\right)^{2}}.

$$\sqrt{\left(x-2\right)^{2}+\left(2x+3-2\right)^{2}}=\sqrt{\left(x-\left(-2\right)\right)^{2}+\left(2x+3-4\right)^{2}}$$

Whakarūnātia. Ko te uara y=2x+3 kua ngata te whārite.

$$\left(5+5x^{2}\right)^{\frac{1}{2}}=\left(5+5x^{2}\right)^{\frac{1}{2}}$$

Ko te whārite \sqrt{\left(x-2\right)^{2}+\left(y-2\right)^{2}}=\sqrt{\left(y-4\right)^{2}+\left(x-\left(-2\right)\right)^{2}} he rongoā ahurei.

$$y=2x+3$$