M is the midpoint of AC, find the coordinates of the missing endpoint
1 A(-9,6) and M(5, 1)
2 C(1,-8) and M (-8,-4)
Midpoint formula:
M(x,y)=x1+x22,y1+y22M(x,y) = \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}M(x,y)=2x1+x2,2y1+y2
1.
(5,1)=−9+x22,6+y225=−9+x2210=−9+x2x2=191=6+y222=6+y2y2=−4C=(19,−4)(5,1) = \frac{-9+x_2}{2}, \frac{6+y_2}{2} \\ 5 = \frac{-9+x_2}{2} \\ 10 = -9 + x_2 \\ x_2 = 19 \\ 1 = \frac{6+y_2}{2} \\ 2 = 6+ y_2 \\ y_2 = -4 C =(19,-4)(5,1)=2−9+x2,26+y25=2−9+x210=−9+x2x2=191=26+y22=6+y2y2=−4C=(19,−4)
2.
(−8,−4)=x1+12,y1−82−8=x1+12−16=x1+1x1=−17−4=y1−82−8=y1−8y1=0A=(−17,0)(-8,-4) = \frac{x_1+1}{2}, \frac{y_1-8}{2} \\ -8 = \frac{x_1+1}{2} \\ -16 = x_1 + 1 \\ x_1 = -17 \\ -4 = \frac{y_1-8}{2} \\ -8 = y_1 -8 \\ y_1=0 \\ A = (-17,0)(−8,−4)=2x1+1,2y1−8−8=2x1+1−16=x1+1x1=−17−4=2y1−8−8=y1−8y1=0A=(−17,0)
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