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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 \(M'\) ?

 Dec 29, 2017

Best Answer 

 #1
avatar+9470 
+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'   \(=\,\sqrt{(\text{difference in x values})^2+(\text{difference in y values})^2} \\~\\ =\,\sqrt{6^2+8^2} \\~\\ =\,\sqrt{100} \\~\\ =\,10\)

 Dec 29, 2017
 #1
avatar+9470 
+1
Best Answer

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
avatar+884 
+3

Wow! Very nice, hectictar! 

 Dec 29, 2017

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