Respuesta :

You'll need to use pythagoras for this one. we know the distance between the 2 points is 5 (hypotenuse) and that the distance between the x coordinates is 4. to find the distance between the vertical parts of the coordinates we need to do 5^2-4^2 which is 9 and the square root of this is 3. We now know that the vertical distance between the two points is 3, so our y coordinate could be 0 or 6.
Panoyin
[tex]d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}[/tex]
[tex]5 = \sqrt{(3 - 7)^{2} + (3 - y)^{2}}[/tex]
[tex]5 = \sqrt{(-4)^{2} + (-y + 3)^{2}}[/tex]
[tex]5 = \sqrt{16 + (-y + 3)(-y + 3)}[/tex]
[tex]5 = \sqrt{16 + (-y(-y + 3) + 3(-y + 3)}[/tex]
[tex]5 = \sqrt{16 + (-y(-y) - y(3) + 3(-y) + 3(3)}[/tex]
[tex]5 = \sqrt{16 + (y^{2} - 3y - 3y + 9)}[/tex]
[tex]5 = \sqrt{16 + (y^{2} - 6y + 9)}[/tex]
[tex]5 = \sqrt{16 + y^{2} - 6y + 9}[/tex]
[tex]5 = \sqrt{y^{2} - 6y + 9 + 16}[/tex]
[tex]5 = \sqrt{y^{2} - 6y + 25}[/tex]
[tex]25 = y^{2} - 6y + 25[/tex]
[tex]0 = y^{2} - 6y[/tex]
[tex]0 = y(y) - y(6)[/tex]
[tex]0 = y(y - 6)[/tex]
[tex]0 = y[/tex]    [tex]or[/tex]    [tex]0 = y - 6[/tex]
[tex]0 = y[/tex]    [tex]or[/tex]    [tex]6 = y[/tex]

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