Solve sqrt{(x-2)^2+(y-5)^2}=sqrt{(x+1)^2+(y-4)^2}

$$\sqrt { ( x - 2 ) ^ { 2 } + ( y - 5 ) ^ { 2 } } = \sqrt { ( x + 1 ) ^ { 2 } + ( y - 4 ) ^ { 2 } }$$

Solve for x

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

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{\left(x-2\right)^{2}+\left(y-5\right)^{2}}\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{x^{2}-4x+4+\left(y-5\right)^{2}}\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{x^{2}-4x+4+y^{2}-10y+25}\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

Add 4 and 25 to get 29.

$$\left(\sqrt{x^{2}-4x+29+y^{2}-10y}\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

Calculate \sqrt{x^{2}-4x+29+y^{2}-10y} to the power of 2 and get x^{2}-4x+29+y^{2}-10y.

$$x^{2}-4x+29+y^{2}-10y=\left(\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$x^{2}-4x+29+y^{2}-10y=\left(\sqrt{x^{2}+2x+1+\left(y-4\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-4\right)^{2}.

$$x^{2}-4x+29+y^{2}-10y=\left(\sqrt{x^{2}+2x+1+y^{2}-8y+16}\right)^{2}$$

Add 1 and 16 to get 17.

$$x^{2}-4x+29+y^{2}-10y=\left(\sqrt{x^{2}+2x+17+y^{2}-8y}\right)^{2}$$

Calculate \sqrt{x^{2}+2x+17+y^{2}-8y} to the power of 2 and get x^{2}+2x+17+y^{2}-8y.

$$x^{2}-4x+29+y^{2}-10y=x^{2}+2x+17+y^{2}-8y$$

Subtract x^{2} from both sides.

$$x^{2}-4x+29+y^{2}-10y-x^{2}=2x+17+y^{2}-8y$$

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

$$-4x+29+y^{2}-10y=2x+17+y^{2}-8y$$

Subtract 2x from both sides.

$$-4x+29+y^{2}-10y-2x=17+y^{2}-8y$$

Combine -4x and -2x to get -6x.

$$-6x+29+y^{2}-10y=17+y^{2}-8y$$

Subtract 29 from both sides.

$$-6x+y^{2}-10y=17+y^{2}-8y-29$$

Subtract 29 from 17 to get -12.

$$-6x+y^{2}-10y=-12+y^{2}-8y$$

Subtract y^{2} from both sides.

$$-6x-10y=-12+y^{2}-8y-y^{2}$$

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

$$-6x-10y=-12-8y$$

Add 10y to both sides.

$$-6x=-12-8y+10y$$

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

$$-6x=-12+2y$$

The equation is in standard form.

$$-6x=2y-12$$

Divide both sides by -6.

$$\frac{-6x}{-6}=\frac{2y-12}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x=\frac{2y-12}{-6}$$

Divide -12+2y by -6.

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

Substitute -\frac{y}{3}+2 for x in the equation \sqrt{\left(x-2\right)^{2}+\left(y-5\right)^{2}}=\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}.

$$\sqrt{\left(-\frac{y}{3}+2-2\right)^{2}+\left(y-5\right)^{2}}=\sqrt{\left(-\frac{y}{3}+2+1\right)^{2}+\left(y-4\right)^{2}}$$

Simplify. The value x=-\frac{y}{3}+2 satisfies the equation.

$$\frac{1}{3}\left(y^{2}+9\left(y-5\right)^{2}\right)^{\frac{1}{2}}=\frac{1}{3}\left(225-90y+10y^{2}\right)^{\frac{1}{2}}$$

Equation \sqrt{\left(x-2\right)^{2}+\left(y-5\right)^{2}}=\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}} has a unique solution.

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

Solve for y

$$y=6-3x$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{\left(x-2\right)^{2}+\left(y-5\right)^{2}}\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{x^{2}-4x+4+\left(y-5\right)^{2}}\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{x^{2}-4x+4+y^{2}-10y+25}\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

Add 4 and 25 to get 29.

$$\left(\sqrt{x^{2}-4x+29+y^{2}-10y}\right)^{2}=\left(\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

Calculate \sqrt{x^{2}-4x+29+y^{2}-10y} to the power of 2 and get x^{2}-4x+29+y^{2}-10y.

$$x^{2}-4x+29+y^{2}-10y=\left(\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}\right)^{2}$$

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

$$x^{2}-4x+29+y^{2}-10y=\left(\sqrt{x^{2}+2x+1+\left(y-4\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-4\right)^{2}.

$$x^{2}-4x+29+y^{2}-10y=\left(\sqrt{x^{2}+2x+1+y^{2}-8y+16}\right)^{2}$$

Add 1 and 16 to get 17.

$$x^{2}-4x+29+y^{2}-10y=\left(\sqrt{x^{2}+2x+17+y^{2}-8y}\right)^{2}$$

Calculate \sqrt{x^{2}+2x+17+y^{2}-8y} to the power of 2 and get x^{2}+2x+17+y^{2}-8y.

$$x^{2}-4x+29+y^{2}-10y=x^{2}+2x+17+y^{2}-8y$$

Subtract y^{2} from both sides.

$$x^{2}-4x+29+y^{2}-10y-y^{2}=x^{2}+2x+17-8y$$

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

$$x^{2}-4x+29-10y=x^{2}+2x+17-8y$$

Add 8y to both sides.

$$x^{2}-4x+29-10y+8y=x^{2}+2x+17$$

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

$$x^{2}-4x+29-2y=x^{2}+2x+17$$

Subtract x^{2} from both sides.

$$-4x+29-2y=x^{2}+2x+17-x^{2}$$

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

$$-4x+29-2y=2x+17$$

Add 4x to both sides.

$$29-2y=2x+17+4x$$

Combine 2x and 4x to get 6x.

$$29-2y=6x+17$$

Subtract 29 from both sides.

$$-2y=6x+17-29$$

Subtract 29 from 17 to get -12.

$$-2y=6x-12$$

Divide both sides by -2.

$$\frac{-2y}{-2}=\frac{6x-12}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$y=\frac{6x-12}{-2}$$

Divide -12+6x by -2.

$$y=6-3x$$

Substitute 6-3x for y in the equation \sqrt{\left(x-2\right)^{2}+\left(y-5\right)^{2}}=\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}.

$$\sqrt{\left(x-2\right)^{2}+\left(6-3x-5\right)^{2}}=\sqrt{\left(x+1\right)^{2}+\left(6-3x-4\right)^{2}}$$

Simplify. The value y=6-3x satisfies the equation.

$$\left(5-10x+10x^{2}\right)^{\frac{1}{2}}=\left(5-10x+10x^{2}\right)^{\frac{1}{2}}$$

Equation \sqrt{\left(x-2\right)^{2}+\left(y-5\right)^{2}}=\sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}} has a unique solution.

$$y=6-3x$$