Answer:
See Below.
Step-by-step explanation:
We are given a point P(x, y). It is equidistant from A(3, 6) and B(-3, 4).
So, let's first determine the distance from P to each point. We can use the distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Segment PA:
We will let P(x, y) be (x₂, y₂) and A(3, 6) be (x₁, y₁). By substitution:
[tex]d=\sqrt{(x-3)^2+(y-6)^2[/tex]
This represents the distance from P to A.
Segment PB:
Again, we will let P(x, y) be (x₂, y₂) and B(-3, 4) be (x₁, y₁). By the distance formula:
[tex]d=\sqrt{(x-(-3))^2+(y-4)^2}[/tex]
We may simplify:
[tex]d=\sqrt{(x+3)^2+(y-4)^2}[/tex]
Now, we know that the two distances are equivalent. Hence:
[tex]\sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2[/tex]
Simplify. Square both sides:
[tex](x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2[/tex]
Square:
[tex](x^2-6x+9)+(y^2-12x+36)=(x^2+6x+9)+(y^2-8x+16)[/tex]
Subtract all terms from the right:
[tex](x^2-6x+9)-(x^2+6x+9)+(y^2-12x+36)-(y^2-8x+16)=0[/tex]
Distribute the negative:
[tex][(x^2-6x+9)+(-x^2-6x-9)]+[(y^2-12y+36)+(-y^2+8y-16)]=0[/tex]
Rearrange:
[tex][(x^2-x^2)+(-6x-6x)+(9-9)]+[(y^2-y^2)+(-12y+8y)+(36-16)]=0[/tex]
Combine like terms;
[tex](-12x)+(-4y)+(20)=0[/tex]
We can divide both sides by -4:
[tex]3x+y-5=0[/tex]
Finally, adding 5 to both sides produces:
[tex]3x+y=5[/tex]
