Question

find the coordinates of the missing endpoint if e is the midpoint of df. 17. f(5, 8), e(4, 3) 18. f(2, 9), e(-1, 6) 19. d(-3, -8), e(1, -2)

Explanation:

1. For problem 17:

• Let \(E(x_m,y_m)=(4,3)\) be the mid - point and \(F(x_1,y_1)=(5,8)\). The mid - point formula is \(x_m=\frac{x_1 + x_2}{2}\) and \(y_m=\frac{y_1 + y_2}{2}\).
• Step 1: Find the x - coordinate of the other endpoint
• We know \(x_m = 4\), \(x_1=5\). Using \(x_m=\frac{x_1 + x_2}{2}\), we can solve for \(x_2\).
• \(4=\frac{5 + x_2}{2}\), then \(8 = 5+x_2\), so \(x_2=8 - 5=3\).
• Step 2: Find the y - coordinate of the other endpoint
• We know \(y_m = 3\), \(y_1 = 8\). Using \(y_m=\frac{y_1 + y_2}{2}\), we have \(3=\frac{8 + y_2}{2}\), then \(6=8 + y_2\), so \(y_2=6 - 8=-2\).

1. For problem 18:

• Let \(E(x_m,y_m)=(-1,6)\) be the mid - point and \(F(x_1,y_1)=(2,9)\).
• Step 1: Find the x - coordinate of the other endpoint
• Using \(x_m=\frac{x_1 + x_2}{2}\), with \(x_m=-1\) and \(x_1 = 2\), we get \(-1=\frac{2 + x_2}{2}\), then \(-2=2 + x_2\), so \(x_2=-2 - 2=-4\).
• Step 2: Find the y - coordinate of the other endpoint
• Using \(y_m=\frac{y_1 + y_2}{2}\), with \(y_m = 6\) and \(y_1=9\), we have \(6=\frac{9 + y_2}{2}\), then \(12=9 + y_2\), so \(y_2=12 - 9 = 3\).

1. For problem 19:

• Let \(E(x_m,y_m)=(1,-2)\) be the mid - point and \(D(x_1,y_1)=(-3,-8)\).
• Step 1: Find the x - coordinate of the other endpoint
• Using \(x_m=\frac{x_1 + x_2}{2}\), with \(x_m = 1\) and \(x_1=-3\), we get \(1=\frac{-3 + x_2}{2}\), then \(2=-3 + x_2\), so \(x_2=2 + 3=5\).
• Step 2: Find the y - coordinate of the other endpoint
• Using \(y_m=\frac{y_1 + y_2}{2}\), with \(y_m=-2\) and \(y_1=-8\), we have \(-2=\frac{-8 + y_2}{2}\), then \(-4=-8 + y_2\), so \(y_2=-4 + 8 = 4\).

Answer:

1. \((3,-2)\)
2. \((-4,3)\)
3. \((5,4)\)