To prove that parallelogram MATH is a rhombus, we need to show that all four sides of the parallelogram are equal in length.
Given the vertices of parallelogram MATH:
- M (-7, -2)
- A (0, 4)
- T (9, 2)
- H (2, -4)
We will calculate the lengths of all four sides: [tex]\( \overline{MA} \)[/tex], [tex]\( \overline{AT} \)[/tex], [tex]\( \overline{TH} \)[/tex], and [tex]\( \overline{HM} \)[/tex].
### Step 1: Calculate the distance between each pair of vertices
1. Distance [tex]\( \overline{MA} \)[/tex]:
The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]
For points M (-7, -2) and A (0, 4):
[tex]\[
MA = \sqrt{(0 - (-7))^2 + (4 - (-2))^2}
\][/tex]
Calculate:
[tex]\[
MA = \sqrt{(7)^2 + (6)^2} = \sqrt{49 + 36} = \sqrt{85} \approx 9.22
\][/tex]
2. Distance [tex]\( \overline{AT} \)[/tex]:
For points A (0, 4) and T (9, 2):
[tex]\[
AT = \sqrt{(9 - 0)^2 + (2 - 4)^2}
\][/tex]
Calculate:
[tex]\[
AT = \sqrt{(9)^2 + (-2)^2} = \sqrt{81 + 4} = \sqrt{85} \approx 9.22
\][/tex]
3. Distance [tex]\( \overline{TH} \)[/tex]:
For points T (9, 2) and H (2, -4):
[tex]\[
TH = \sqrt{(2 - 9)^2 + (-4 - 2)^2}
\][/tex]
Calculate:
[tex]\[
TH = \sqrt{(-7)^2 + (-6)^2} = \sqrt{49 + 36} = \sqrt{85} \approx 9.22
\][/tex]
4. Distance [tex]\( \overline{HM} \)[/tex]:
For points H (2, -4) and M (-7, -2):
[tex]\[
HM = \sqrt{(-7 - 2)^2 + (-2 - (-4))^2}
\][/tex]
Calculate:
[tex]\[
HM = \sqrt{(-9)^2 + (2)^2} = \sqrt{81 + 4} = \sqrt{85} \approx 9.22
\][/tex]
### Step 2: Compare the lengths of all sides
- [tex]\( \overline{MA} \approx 9.22 \)[/tex]
- [tex]\( \overline{AT} \approx 9.22 \)[/tex]
- [tex]\( \overline{TH} \approx 9.22 \)[/tex]
- [tex]\( \overline{HM} \approx 9.22 \)[/tex]
Since all four sides of parallelogram MATH are equal in length, we can conclude that parallelogram MATH is indeed a rhombus.
Thus, the proof is complete, and parallelogram MATH is confirmed to be a rhombus.
