Answer:
[tex](4-\sqrt{14}, 2)\text{ or } (4+\sqrt{14}, 2)[/tex]
Step-by-step explanation:
We have the two points: (4, -3) and (x, 2).
We know that the distance between them is √39.
To find our x-coordinate of the second point, we can use the distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2[/tex]
Let (4, -3) be (x₁, y₁) and let (x, 2) be (x₂, y₂). We also know that the distance d is √39. Substitute:
[tex]\sqrt{39}=\sqrt{(x-4)^2+(2-(-3))^2[/tex]
Solve for x. First, let's square both sides. This cancels out the square roots:
[tex]39=(x-4)^2+(2+3)^2[/tex]
Evaluate the second term on the right:
[tex]39=(x-4)^2+(5)^2[/tex]
Square:
[tex]39=(x-4)^2+25[/tex]
Subtract 25 from both sides:
[tex]14=(x-4)^2[/tex]
Take the square root of both sides. Since we're taking an even root, we need plus/minus. So:
[tex]x-4=\pm\sqrt{14}[/tex]
Add 4 to both sides:
[tex]x=4\pm\sqrt{14}[/tex]
Therefore, there exists two possibilities for our second point. It can either be:
[tex](4-\sqrt{14}, 2)\text{ or } (4+\sqrt{14}, 2)[/tex]
And we're done!
