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midpoint practice name: hart gladomski 1. find the midpoint of the line…

Question

midpoint practice
name: hart gladomski

  1. find the midpoint of the line - segment with the endpoints (7, 4) and (9, - 1).
  2. find the midpoint of the line - segment with the endpoints (8, - 9) and (0, 5).
  3. find the midpoint of the line - segment with the endpoints (1, - 7) and (1, - 12).
  4. find the midpoint of the line - segment with the endpoints (0, 4) and (-4, 12).
  5. find the other endpoint if one endpoint is (-1, 9) and the midpoint of the segment is (-9, -10).
  6. find the other endpoint if one endpoint is (2, 5) and the midpoint of the segment is (5, 1).
  7. find the other endpoint if one endpoint is (-6, 4) and the midpoint of the segment is (4, 8).
  8. find the other endpoint if one endpoint is (-9, 7) and the midpoint of the segment is (10, -3).
  9. find the other endpoint, c, if a is one endpoint and b is the midpoint of segment ac.

a(-1, 5)
b(-4, 4)

Explanation:

Step1: Recall mid - point formula

The mid - point formula for two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(M=(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\). If we know one endpoint \((x_1,y_1)\) and the mid - point \((M_x,M_y)\) and want to find the other endpoint \((x_2,y_2)\), we can use the equations \(M_x=\frac{x_1 + x_2}{2}\) and \(M_y=\frac{y_1 + y_2}{2}\), which can be rewritten as \(x_2 = 2M_x-x_1\) and \(y_2=2M_y - y_1\).

Step2: Solve problem 1

For endpoints \((7,4)\) and \((9, - 1)\), \(x_1 = 7,y_1 = 4,x_2 = 9,y_2=-1\). The mid - point is \((\frac{7 + 9}{2},\frac{4+( - 1)}{2})=(\frac{16}{2},\frac{3}{2})=(8,\frac{3}{2})\).

Step3: Solve problem 2

For endpoints \((8,-9)\) and \((0,5)\), the mid - point is \((\frac{8 + 0}{2},\frac{-9 + 5}{2})=(4,-2)\).

Step4: Solve problem 3

For endpoints \((1,-7)\) and \((1,-12)\), the mid - point is \((\frac{1+1}{2},\frac{-7+( - 12)}{2})=(1,\frac{-19}{2})\).

Step5: Solve problem 4

For endpoints \((0,4)\) and \((-4,12)\), the mid - point is \((\frac{0+( - 4)}{2},\frac{4 + 12}{2})=(-2,8)\).

Step6: Solve problem 5

Let one endpoint \((x_1,y_1)=(-1,9)\) and mid - point \((M_x,M_y)=(-9,-10)\). Then \(x_2=2\times(-9)-(-1)=-18 + 1=-17\) and \(y_2=2\times(-10)-9=-20 - 9=-29\). The other endpoint is \((-17,-29)\).

Step7: Solve problem 6

Let one endpoint \((x_1,y_1)=(2,5)\) and mid - point \((M_x,M_y)=(5,1)\). Then \(x_2=2\times5 - 2=8\) and \(y_2=2\times1 - 5=-3\). The other endpoint is \((8,-3)\).

Step8: Solve problem 7

Let one endpoint \((x_1,y_1)=(-6,4)\) and mid - point \((M_x,M_y)=(4,8)\). Then \(x_2=2\times4-(-6)=8 + 6 = 14\) and \(y_2=2\times8 - 4=12\). The other endpoint is \((14,12)\).

Step9: Solve problem 8

Let one endpoint \((x_1,y_1)=(-9,7)\) and mid - point \((M_x,M_y)=(10,-3)\). Then \(x_2=2\times10-(-9)=20 + 9=29\) and \(y_2=2\times(-3)-7=-6 - 7=-13\). The other endpoint is \((29,-13)\).

Step10: Solve problem 9

Let \(A(x_1,y_1)=(-1,5)\) and \(B(M_x,M_y)=(-4,4)\). Then \(x_2=2\times(-4)-(-1)=-8 + 1=-7\) and \(y_2=2\times4 - 5=3\). The point \(C\) is \((-7,3)\).

Answer:

  1. \((8,\frac{3}{2})\)
  2. \((4,-2)\)
  3. \((1,-\frac{19}{2})\)
  4. \((-2,8)\)
  5. \((-17,-29)\)
  6. \((8,-3)\)
  7. \((14,12)\)
  8. \((29,-13)\)
  9. \((-7,3)\)