Question

A(3,4),B(5,8),C(7,8) are the vertices of ABC, prove that the line passing through the mid point of AB and AC is parallel to BC

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Answer 1

Step-by-step explanation:

`The \ midpoint \ formula \ of \ the \ line \ segment \\ that \ joins \ the \ two \ points \ (x_(1),y_(1)) \ and \ (x_(2),y_(2)) \ is \\ given \ by \ the \ following \ Midpoint \ Formula:\\ \\ Midpoint=((x_(1)+x_(2))/(2),(y_(1)+y_(2))/(2))`

• Let's name the mid point of
  `AB` as
  `c`

• Let's name the mid point of
  `AC` as
  `p`

Then:

`c=((3+5)/(2),(4+8)/(2)) \\ \\ c=((8)/(2),(12)/(2)) \\ \\ c=(4,6) \\ \\ \\ p=((3+7)/(2),(4+8)/(2)) \\ \\ p=((10)/(2),(12)/(2)) \\ \\ p=(5,6)`

So the slope of the line that passes through
`c \ and \ p` is:

`m=(y_(2)-y_(1))/(x_(2)-x_(1)) \\ \\ m_(cp)=(6-6)/(5-4)=0`

And the slope for BC is:

`m_(BC)=(8-8)/(7-5)=0`

As you can see, both slopes are zero, so these are horizontal lines.