Do the following vertices form a parallelogram? Prove using the theorem that allows you to know that IF opposite sides of a quadrilateral are congruent then the quadrilateral is a parallelogram.

M(−4,3), E(2,5), L(6,2), T(0,0)

(2 points)
Round to the tenths place as needed (2 decimal places).

ME¯¯¯¯¯¯¯¯¯=

EL¯¯¯¯¯¯¯=

LT¯¯¯¯¯¯¯=

MT¯¯¯¯¯¯¯¯¯=

Is this a parallelogram (Type 1 for Yes and Type 2 for No)?

1 answer

To check if the vertices M(−4,3), E(2,5), L(6,2), T(0,0) form a parallelogram, we will calculate the lengths of the opposite sides: ME, EL, LT, and MT.

We will use the distance formula for points \((x_1, y_1)\) and \((x_2, y_2)\): \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 1: Calculate lengths of the sides

Length of ME: \[ M(-4, 3) \quad E(2, 5) \] \[ ME = \sqrt{(2 - (-4))^2 + (5 - 3)^2} = \sqrt{(2 + 4)^2 + (5 - 3)^2} = \sqrt{(6)^2 + (2)^2} = \sqrt{36 + 4} = \sqrt{40} \approx 6.32 \]

Length of EL: \[ E(2, 5) \quad L(6, 2) \] \[ EL = \sqrt{(6 - 2)^2 + (2 - 5)^2} = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5.00 \]

Length of LT: \[ L(6, 2) \quad T(0, 0) \] \[ LT = \sqrt{(0 - 6)^2 + (0 - 2)^2} = \sqrt{(-6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} \approx 6.32 \]

Length of MT: \[ M(-4, 3) \quad T(0, 0) \] \[ MT = \sqrt{(0 - (-4))^2 + (0 - 3)^2} = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5.00 \]

Step 2: Record the lengths

\[ ME \approx 6.32 \] \[ EL = 5.00 \] \[ LT \approx 6.32 \] \[ MT = 5.00 \]

Step 3: Check if opposite sides are congruent

  • ME ≈ LT (6.32 ≈ 6.32)
  • EL = MT (5.00 = 5.00)

Since both pairs of opposite sides are congruent (ME = LT and EL = MT), we can conclude:

Is this a parallelogram? Type (1 for Yes or 2 for No): 1 (Yes)