A(2, 3), B(10, 4), C(12, 9), and D(4, 8) are four points - Junior Cycle Mathematics - Question 13 - 2014

To verify, we will calculate the lengths of sides AD and BC:

Using the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

For $|AD|$:
$|AD| = \sqrt{(4 - 2)^2 + (8 - 3)^2} = \sqrt{(2)^2 + (5)^2} = \sqrt{4 + 25} = \sqrt{29}$

For $|BC|$:
$|BC| = \sqrt{(12 - 10)^2 + (9 - 4)^2} = \sqrt{(2)^2 + (5)^2} = \sqrt{4 + 25} = \sqrt{29}$

Thus, we find $|AD| = |BC| = \sqrt{29}$, verifying that one pair of opposite sides are equal.

To find the midpoints:

• For midpoint E of [AC]: $E = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{2 + 12}{2}, \frac{3 + 9}{2}\right) = (7, 6)$

• For midpoint F of [BD]: $F = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{10 + 4}{2}, \frac{4 + 8}{2}\right) = (7, 6)$

Since points E and F coincide, we can conclude that the diagonals AC and BD bisect each other.

Yes, we can conclude that ABCD is a parallelogram.

Reason: In a quadrilateral, if the diagonals bisect each other, then the quadrilateral must be a parallelogram.