How do you find the midpoint of (-4,-3),(4,-8)?

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How do you find the midpoint of (-4,-3),(4,-8)?

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Answer 1 · 2016-03-03T13:57:06

$P_{mid} \to (0, - \frac{11}{2})$ $P_("mid") -> (0,-11/2)$

Explanation:

Let mid point be $P_{mid}$ $P_("mid")$

Let first point be $P_{1} \to (x_{1}, y_{1}) \to (- 4, - 3)$ $P_1 ->(x_1,y_1)->(-4,-3)$

Let second point be $P_{2} \to (x_{2}, y_{2}) \to (4, - 8)$ $P_2->(x_2,y_2)->(4,-8)$

Then $P_{mid} = \frac{P_{2} + P_{1}}{2}$ $" "P_("mid") = (P_2+P_1)/2$ ..... ( mean value)

$\Rightarrow P_{mid} = (\frac{x_{2} + x_{1}}{2}, \frac{y_{2} + y_{1}}{2})$ $=> " "P_("mid")=((x_2+x_1)/2 , (y_2+y_1)/2 )$

$\Rightarrow P_{mid} = (\frac{4 + (- 4)}{2}, \frac{(- 8) + (- 3)}{2}) = (0, - \frac{11}{2})$ $=> " "P_("mid")= ((4+(-4))/2 ,((-8)+(-3))/2) = (0,-11/2)$

Tony B

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$Something to think about$ $color(blue)("Something to think about")$

Using the mean value is more straightforward than using

$P_{1} + \frac{P_{2} - P_{1}}{2}$ $P_1+ (P_2-P_1)/2$ which is the alternative.

Permit me to demonstrate just for the mid x value

Distance between is $P_{2} - P_{1} \to (x_{2} - x_{1}) = (4 - (- 4)) = + 8$ $" "P_2-P_1 ->(x_2-x_1) = (4-(-4))=+8$
Count from $P_{1} to x_{mid} = + \frac{8}{2} = 4$ $P_1" to "x_("mid") = +8/2 = 4$

Actual $x$ $x$ value is $P_{1} + 4 = - 4 + 4 = 0$ $P_1+4 = -4+4=0$

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How do you find the midpoint of (-4,-3),(4,-8)?

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