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# Cartesian

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On the Cartesian plane, the midpoint between two points $$A(a,b)$$ and $$B(c,d)$$ is $$M(m,n)$$ . If A is moved vertically upwards 20 units and horizontally to the right 14 units, and B is moved vertically downwards 4 units and horizontally to the left 2 units, then the new midpoint between $$A$$ and $$B$$ is $$M'$$. What is the distance between $$M$$ and $$M'$$ ?

ant101  Dec 29, 2017

#1
+5931
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M  =  midpoint of  (a, b)  and  (c, d)   $$=\,(\,\frac{a+c}{2},\frac{b+d}{2}\,)$$

M'  =  midpoint of  (a+14, b+20)  and  (c-2, d-4)   $$=\,(\,\frac{(a+14)+(c-2)}{2},\frac{(b+20)+(d-4)}{2}\,) \\~\\ =\,(\,\frac{a+c+12}{2},\frac{b+d+16}{2}\,) \\~\\ =\,(\,\frac{a+c}{2}+\frac{12}{2},\frac{b+d}{2}+\frac{16}{2}\,) \\~\\ =\,(\,\frac{a+c}{2}+6,\frac{b+d}{2}+8\,)$$

distance between  M  and  M'   $$=\,\sqrt{(\text{difference in x values})^2+(\text{difference in y values})^2} \\~\\ =\,\sqrt{6^2+8^2} \\~\\ =\,\sqrt{100} \\~\\ =\,10$$

hectictar  Dec 29, 2017
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#1
+5931
+1

M  =  midpoint of  (a, b)  and  (c, d)   $$=\,(\,\frac{a+c}{2},\frac{b+d}{2}\,)$$

M'  =  midpoint of  (a+14, b+20)  and  (c-2, d-4)   $$=\,(\,\frac{(a+14)+(c-2)}{2},\frac{(b+20)+(d-4)}{2}\,) \\~\\ =\,(\,\frac{a+c+12}{2},\frac{b+d+16}{2}\,) \\~\\ =\,(\,\frac{a+c}{2}+\frac{12}{2},\frac{b+d}{2}+\frac{16}{2}\,) \\~\\ =\,(\,\frac{a+c}{2}+6,\frac{b+d}{2}+8\,)$$

distance between  M  and  M'   $$=\,\sqrt{(\text{difference in x values})^2+(\text{difference in y values})^2} \\~\\ =\,\sqrt{6^2+8^2} \\~\\ =\,\sqrt{100} \\~\\ =\,10$$

hectictar  Dec 29, 2017
#2
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Wow! Very nice, hectictar!

ant101  Dec 29, 2017

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