Answer:
(4,2) and (8,2)
Step-by-step explanation:
Using the image attached, you can observe that the midsegment parallel to BC has to be placed from the AB midpoint to the AC midpoint.
To find a midpoint we need to use the following formula:
[tex]m_{AB}=(\frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2} )[/tex]
Where [tex](x_{1},y_{1})[/tex] represents the coordinates of point A and [tex](x_{2},y_{2})[/tex] represents the coordinates of B.
From the image you can see that coordinates are [tex](7,6)[/tex] and [tex](1,-2)[/tex], using the formula, we have
[tex]m_{AB}=(\frac{7+1}{2} ,\frac{6+(-2)}{2} )\\m_{AB}=(\frac{8}{2} ,\frac{4}{2} )\\m_{AB}=(4 ,2 )[/tex]
This means that the beginning of the midsegment is at [tex]m_{AB}=(4 ,2 )[/tex]
Now, we do the same process to find the end point of the midsegment. The points that we are gonna use this time are [tex](7,6)[/tex] and [tex](9,-2)[/tex]. Using the formula we have
[tex]m_{AC}=(\frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2} )\\m_{AC}=(\frac{7+9}{2} ,\frac{6+(-2)}{2} )\\m_{AC}=(\frac{16}{2} ,\frac{4}{2} )\\m_{AC}=(8 ,2)[/tex]
Therefore, the endpoints of the midsegment for △ABC that is parallel to BC are at (4,2) and (8,2)
