+884 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 \(M'\) ?
+9465 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' \(=\,\sqrt{(\text{difference in x values})^2+(\text{difference in y values})^2} \\~\\ =\,\sqrt{6^2+8^2} \\~\\ =\,\sqrt{100} \\~\\ =\,10\)
+9465 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' \(=\,\sqrt{(\text{difference in x values})^2+(\text{difference in y values})^2} \\~\\ =\,\sqrt{6^2+8^2} \\~\\ =\,\sqrt{100} \\~\\ =\,10\)
