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

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

Solve for x

$$x=y+2$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-x\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-y\right)^{2}.

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

Add 49 and 1 to get 50.

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

Calculate \sqrt{50-14x+x^{2}-2y+y^{2}} to the power of 2 and get 50-14x+x^{2}-2y+y^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-y\right)^{2}.

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

Add 9 and 25 to get 34.

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

Calculate \sqrt{34-6x+x^{2}-10y+y^{2}} to the power of 2 and get 34-6x+x^{2}-10y+y^{2}.

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

Add 6x to both sides.

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

Combine -14x and 6x to get -8x.

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

Subtract x^{2} from both sides.

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

Combine x^{2} and -x^{2} to get 0.

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

Subtract 50 from both sides.

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

Subtract 50 from 34 to get -16.

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

Add 2y to both sides.

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

Combine -10y and 2y to get -8y.

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

Subtract y^{2} from both sides.

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

Combine y^{2} and -y^{2} to get 0.

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

The equation is in standard form.

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

Divide both sides by -8.

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

Dividing by -8 undoes the multiplication by -8.

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

Divide -16-8y by -8.

$$x=y+2$$

Substitute y+2 for x in the equation \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}.

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

Simplify. The value x=y+2 satisfies the equation.

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

Equation \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}} has a unique solution.

$$x=y+2$$

Solve for y

$$y=x-2$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-x\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-y\right)^{2}.

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

Add 49 and 1 to get 50.

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

Calculate \sqrt{50-14x+x^{2}-2y+y^{2}} to the power of 2 and get 50-14x+x^{2}-2y+y^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-y\right)^{2}.

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

Add 9 and 25 to get 34.

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

Calculate \sqrt{34-6x+x^{2}-10y+y^{2}} to the power of 2 and get 34-6x+x^{2}-10y+y^{2}.

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

Add 10y to both sides.

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

Combine -2y and 10y to get 8y.

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

Subtract y^{2} from both sides.

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

Combine y^{2} and -y^{2} to get 0.

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

Subtract 50 from both sides.

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

Subtract 50 from 34 to get -16.

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

Add 14x to both sides.

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

Combine -6x and 14x to get 8x.

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

Subtract x^{2} from both sides.

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

Combine x^{2} and -x^{2} to get 0.

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

The equation is in standard form.

$$8y=8x-16$$

Divide both sides by 8.

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

Dividing by 8 undoes the multiplication by 8.

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

Divide -16+8x by 8.

$$y=x-2$$

Substitute x-2 for y in the equation \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}}.

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

Simplify. The value y=x-2 satisfies the equation.

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

Equation \sqrt{\left(7-x\right)^{2}+\left(1-y\right)^{2}}=\sqrt{\left(3-x\right)^{2}+\left(5-y\right)^{2}} has a unique solution.

$$y=x-2$$