Use the graph of △ABC with midsegments DE, EF and DF. Show that EF is parallel to AC and that EF=1/2 AC

Mathematics

1 answer:

Vinvika [58]2 years ago

8 0

According to the midsegment theorem, the midsegments are parallel to

and half the length of the opposite side.

The completed statement are as follows;

Because the slopes of $\overline{EF}$ and $\overline{AC}$ are both -4, $\overline{EF}$ ║ $\overline{AC}$

EF = $\underline{\sqrt{17}}$ and AC = $\underline{2 \cdot \sqrt{17} }$

Because $\underline{\sqrt{17} }$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17} }$ , $EF = \frac{1}{2} \cdot AC$

Reasons:

From the given graph of ΔABC, we have;

Coordinates of the points A, B, and C are; A(-5, 2), B(1, -2), and C(-3, -6)

The coordinates of the point D and E on $\mathbf{\overline{DE}}$ are; D(-4, -2), and E(-2, 0)

The coordinates of the point F is; F(-1, -4)

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle Slope \ of \ line \ \overline{AC} = \mathbf{\frac{(-6) - 2}{-3 - (-5)} = \frac{-8}{2}} = -4$

$\displaystyle Slope \ of \ line \ \overline{EF} = \frac{0 - (-4)}{-2 - (-1)} = \frac{4}{-1} = -4$

$Length \ of \ segment,\ l = \sqrt{\left (x_{2}-x_{1} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}$

Length of EF = √((-1 - (-2))² + (-4 - 0)²) = √(17)

Length of AC = √((-3 - (-5))² + (-6 - 2)²) = √(4 × 17) = 2·√(17)

Therefore, we have;

Because the slope of $\mathbf{\overline {EF}}$ and $\mathbf{\overline {AC}}$ are both , -4, $\overline {EF}$ ║ $\overline {AC}$ . EF = $\underline{\sqrt{17}}$ , and AC

= $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ ,. Because $\underline{\sqrt{17}}$ = $\underline{\frac{1}{2} \cdot 2 \cdot \sqrt{17}}$ , $EF =\mathbf{ \frac{1}{2} AC}$