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

Explanation:

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

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

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

Then:

$c=(\frac{3+5}{2},\frac{4+8}{2}) \\ \\ c=(\frac{8}{2},\frac{12}{2}) \\ \\ c=(4,6) \\ \\ \\ p=(\frac{3+7}{2},\frac{4+8}{2}) \\ \\ p=(\frac{10}{2},\frac{12}{2}) \\ \\ p=(5,6)$

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

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \\ m_{cp}=\frac{6-6}{5-4}=0$

And the slope for BC is:

$m_{BC}=\frac{8-8}{7-5}=0$

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