Answer :

dalendrk
[tex]A(x_A;\ y_A);\ B(x_B;\ y_B)\\\\The\ midpoint:M_{AB}\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right)\\\\\\M_{CD}(-9;-10);\ C(-1;\ 9);\ D(x;\ y)\\\\\dfrac{-1+x}{2}=-9\ \ \ \ |multiply\ both\ sides\ by\ 2\\-1+x=-18\ \ \ \ |add\ 1\ to\ both\ sides\\x=-17\\\\\dfrac{9+y}{2}=-10\ \ \ \ \ |multiply\ both\ sides\ by\ 2\\9+y=-20\ \ \ \ |subtract\ 9\ from\ both\ sides\\y=-29\\\\Answer{\boxed{C(-17;-29)}[/tex]
lifeinferno
From c to the midpoint the coordinates have changed from (-1;9) to (-9;-10). So the difference is (-9-(-1); -10-(9)) that equals (-8;-19). And you should add (-8;-19) to the coordinates of the middle point (-9;-10). So we get (-9+(-8); -10+(-19)) that is (-17;-29). This is the endpoint of d. Answer: d (-17;-29).

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