Solve sqrt{left(-3+yright)^2+left(6-yright)^2}=sqrt{left(2+yright)^2+left(1-yright)^2}

$$\sqrt{ { \left(-3+y \right) }^{ 2 } + { \left(6-y \right) }^{ 2 } } = \sqrt{ { \left(2+y \right) }^{ 2 } + { \left(1-y \right) }^{ 2 } }$$

Solve for y

$$y=2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{\left(-3+y\right)^{2}+\left(6-y\right)^{2}}\right)^{2}=\left(\sqrt{\left(2+y\right)^{2}+\left(1-y\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{9-6y+y^{2}+\left(6-y\right)^{2}}\right)^{2}=\left(\sqrt{\left(2+y\right)^{2}+\left(1-y\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6-y\right)^{2}.

$$\left(\sqrt{9-6y+y^{2}+36-12y+y^{2}}\right)^{2}=\left(\sqrt{\left(2+y\right)^{2}+\left(1-y\right)^{2}}\right)^{2}$$

Add 9 and 36 to get 45.

$$\left(\sqrt{45-6y+y^{2}-12y+y^{2}}\right)^{2}=\left(\sqrt{\left(2+y\right)^{2}+\left(1-y\right)^{2}}\right)^{2}$$

Combine -6y and -12y to get -18y.

$$\left(\sqrt{45-18y+y^{2}+y^{2}}\right)^{2}=\left(\sqrt{\left(2+y\right)^{2}+\left(1-y\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{45-18y+2y^{2}}\right)^{2}=\left(\sqrt{\left(2+y\right)^{2}+\left(1-y\right)^{2}}\right)^{2}$$

Calculate \sqrt{45-18y+2y^{2}} to the power of 2 and get 45-18y+2y^{2}.

$$45-18y+2y^{2}=\left(\sqrt{\left(2+y\right)^{2}+\left(1-y\right)^{2}}\right)^{2}$$

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

$$45-18y+2y^{2}=\left(\sqrt{4+4y+y^{2}+\left(1-y\right)^{2}}\right)^{2}$$

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

$$45-18y+2y^{2}=\left(\sqrt{4+4y+y^{2}+1-2y+y^{2}}\right)^{2}$$

Add 4 and 1 to get 5.

$$45-18y+2y^{2}=\left(\sqrt{5+4y+y^{2}-2y+y^{2}}\right)^{2}$$

Combine 4y and -2y to get 2y.

$$45-18y+2y^{2}=\left(\sqrt{5+2y+y^{2}+y^{2}}\right)^{2}$$

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

$$45-18y+2y^{2}=\left(\sqrt{5+2y+2y^{2}}\right)^{2}$$

Calculate \sqrt{5+2y+2y^{2}} to the power of 2 and get 5+2y+2y^{2}.

$$45-18y+2y^{2}=5+2y+2y^{2}$$

Subtract 2y from both sides.

$$45-18y+2y^{2}-2y=5+2y^{2}$$

Combine -18y and -2y to get -20y.

$$45-20y+2y^{2}=5+2y^{2}$$

Subtract 2y^{2} from both sides.

$$45-20y+2y^{2}-2y^{2}=5$$

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

$$45-20y=5$$

Subtract 45 from both sides.

$$-20y=5-45$$

Subtract 45 from 5 to get -40.

$$-20y=-40$$

Divide both sides by -20.

$$y=\frac{-40}{-20}$$

Divide -40 by -20 to get 2.

$$y=2$$

Substitute 2 for y in the equation \sqrt{\left(-3+y\right)^{2}+\left(6-y\right)^{2}}=\sqrt{\left(2+y\right)^{2}+\left(1-y\right)^{2}}.

$$\sqrt{\left(-3+2\right)^{2}+\left(6-2\right)^{2}}=\sqrt{\left(2+2\right)^{2}+\left(1-2\right)^{2}}$$

Simplify. The value y=2 satisfies the equation.

$$17^{\frac{1}{2}}=17^{\frac{1}{2}}$$

Equation \sqrt{\left(y-3\right)^{2}+\left(6-y\right)^{2}}=\sqrt{\left(y+2\right)^{2}+\left(1-y\right)^{2}} has a unique solution.

$$y=2$$