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use distance formula to find the center of a circle whose endpoints of the diameter are (4,-3) and (-6,-5)

Sagot :

[tex] x_{1} =4[/tex]   
[tex] x_{2} [/tex] = -6
[tex] y_{1} [/tex] = -3
[tex] y_{2} [/tex] = -5

d = [tex] \sqrt{(x_{2}- x_{1})^{2} + (y_{2} - y_{1})^{2} } [/tex]
d = [tex] \sqrt{(4-(-6))^{2} + (-3- (-5)) ^{2} } [/tex]
d = [tex] \sqrt{(-10)^{2}+ (-2) ^{2} } [/tex]
d = [tex] \sqrt{100+4} [/tex]
d = [tex] \sqrt{104} [/tex]
d = [tex] \sqrt{4 · 26} [/tex]
d = 2[tex] \sqrt{26} [/tex]
I don't see the use of using the distance formula so I'm going to use the midpoint formula

The midpoint formula is:
[tex]( \frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )[/tex]

So we plug in the values with (4,-3) being x₁ and y
₁ and (-6,-5) being x₂ and y₂.
[tex]( \frac{4-6}{2},\frac{-3-5}{2} ) \\\\ ( \frac{-2}{2},\frac{-8}{2} )\\\\(-1,-4)[/tex]

The coordinates of the center of the diameter of the circle with the given points is (-1,-4)