Question #249689

M​ is​ the​ mid​point​ of​ AC, find​ the​ coordinates​ of​ the​ missing​ end​point


1​ A​(-9,6) and​ M(5, 1)

2​ C​(1,-8) and​ M​ (-8,-4)


1
Expert's answer
2021-10-12T13:59:00-0400

Midpoint formula:

M(x,y)=x1+x22,y1+y22M(x,y) = \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}

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)

2.

(8,4)=x1+12,y1828=x1+1216=x1+1x1=174=y1828=y18y1=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)


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