Question

in exercises 27 - 30, the mid - point m and one endpoint of $overline{gh}$ are given. find the coordinates of the other endpoint. (see example 5.)

1. $g(5, - 6)$ and $m(4,3)$
2. $h(-3,7)$ and $m(-2,5)$
3. $h(-2,9)$ and $m(3,0)$
4. $g(-4,1)$ and $m(-\frac{13}{2}, - 6)$

Explanation:

1. Recall the mid - point formula:

• The mid - point formula for two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(M(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\). Let one endpoint be \((x_1,y_1)\), the mid - point be \((x_m,y_m)\), and the other endpoint be \((x_2,y_2)\). Then \(x_m=\frac{x_1 + x_2}{2}\) and \(y_m=\frac{y_1 + y_2}{2}\). We can solve for \(x_2\) and \(y_2\): \(x_2 = 2x_m−x_1\) and \(y_2 = 2y_m−y_1\).

1. For problem 27:

• Given \(G(5, - 6)\) and \(M(4,3)\). Let \(G(x_1,y_1)=(5, - 6)\) and \(M(x_m,y_m)=(4,3)\).
• First, find the \(x\) - coordinate of \(H\):
• Using \(x_2 = 2x_m−x_1\), substitute \(x_1 = 5\) and \(x_m = 4\). Then \(x_2=2\times4 - 5=8 - 5 = 3\).
• Second, find the \(y\) - coordinate of \(H\):
• Using \(y_2 = 2y_m−y_1\), substitute \(y_1=-6\) and \(y_m = 3\). Then \(y_2=2\times3-(-6)=6 + 6=12\). So the coordinates of \(H\) are \((3,12)\).

1. For problem 28:

• Given \(H(x_1,y_1)=(-3,7)\) and \(M(x_m,y_m)=(-2,5)\).
• First, find the \(x\) - coordinate of \(G\):
• Using \(x_2 = 2x_m−x_1\), substitute \(x_1=-3\) and \(x_m=-2\). Then \(x_2=2\times(-2)-(-3)=-4 + 3=-1\).
• Second, find the \(y\) - coordinate of \(G\):
• Using \(y_2 = 2y_m−y_1\), substitute \(y_1 = 7\) and \(y_m = 5\). Then \(y_2=2\times5-7 = 10 - 7=3\). So the coordinates of \(G\) are \((-1,3)\).

1. For problem 29:

• Given \(H(x_1,y_1)=(-2,9)\) and \(M(x_m,y_m)=(5,0)\).
• First, find the \(x\) - coordinate of \(G\):
• Using \(x_2 = 2x_m−x_1\), substitute \(x_1=-2\) and \(x_m = 5\). Then \(x_2=2\times5-(-2)=10 + 2 = 12\).
• Second, find the \(y\) - coordinate of \(G\):
• Using \(y_2 = 2y_m−y_1\), substitute \(y_1 = 9\) and \(y_m = 0\). Then \(y_2=2\times0-9=-9\). So the coordinates of \(G\) are \((12,-9)\).

1. For problem 30:

• Given \(G(x_1,y_1)=(-4,1)\) and \(M(x_m,y_m)=(-\frac{13}{2},-6)\).
• First, find the \(x\) - coordinate of \(H\):
• Using \(x_2 = 2x_m−x_1\), substitute \(x_1=-4\) and \(x_m=-\frac{13}{2}\). Then \(x_2=2\times(-\frac{13}{2})-(-4)=-13 + 4=-9\).
• Second, find the \(y\) - coordinate of \(H\):
• Using \(y_2 = 2y_m−y_1\), substitute \(y_1 = 1\) and \(y_m=-6\). Then \(y_2=2\times(-6)-1=-12 - 1=-13\). So the coordinates of \(H\) are \((-9,-13)\).

Answer:

1. \((3,12)\)
2. \((-1,3)\)
3. \((12,-9)\)
4. \((-9,-13)\)