Question

Find midsegment (EG) ̅ that is parallel to side (BC) ̅. Show all work to receive credit.

Answer 1

Answer:

Part 1) $E(-2,-1)$, $G(1,0)$  

Part 2) slope EG is equal to slope BC (EG is parallel to BC)

Part 3)  $EG=(1/2)BC$

Step-by-step explanation:

we have

$B(-3,1), C(3,3),D(-1,-3)$

Step 1

Find the midsegment (EG) ̅ that is parallel to side (BC)

Find the x-coordinate of point E

$Ex=\frac{Bx+Dx}{2}$

substitute

$Ex=\frac{-3-1}{2}=-2$

Find the y-coordinate of point E

$Ey=\frac{By+Dy}{2}$

substitute

$Ey=\frac{1-3}{2}=-1$

the coordinates of point E are $E(-2,-1)$

Find the x-coordinate of point G

$Gx=\frac{Cx+Dx}{2}$

substitute

$Gx=\frac{3-1}{2}=1$

Find the y-coordinate of point G

$Gy=\frac{Cy+Dy}{2}$

substitute

$Gy=\frac{3-3}{2}=0$

the coordinates of point G are $G(1,0)$

Step 2

Verifying EG is parallel to BC

we know that

If two lines are parallel, then their slopes are the same

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

Find the slope EG

we have

$E(-2,-1),G(1,0)$  

Substitute the values

$mEG=\frac{0+1}{1+2}$

$mEG=\frac{1}{3}$

Find the slope BC

we have

$B(-3,1), C(3,3)$

Substitute the values

$mBC=\frac{3-1}{3+3}$

$mBC=\frac{2}{6}=\frac{1}{3}$

therefore

$mEG=mBC$ -------> EG is parallel to BC

Step 3

Verifying $EG=(1/2)BC$

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance EG

$E(-2,-1),G(1,0)$  

Substitute the values

$dEG=\sqrt{(0+1)^{2}+(1+2)^{2}}$

$dEG=\sqrt{(1)^{2}+(3)^{2}}$

$dEG=\sqrt{10}\ units$

Find the distance BC

$B(-3,1), C(3,3)$

Substitute the values

$dBC=\sqrt{(3-1)^{2}+(3+3)^{2}}$

$dBC=\sqrt{(2)^{2}+(6)^{2}}$

$dBC=\sqrt{40}=2\sqrt{10}\ units$

Verifying

$EG=(1/2)BC$

substitute the values

$\sqrt{10}=(1/2)2\sqrt{10}$

$\sqrt{10}=\sqrt{10}$ -------> is true

therefore

$EG=(1/2)BC$