In the given graph, ABCD is a parallelogram.
(a) Write the coordinates of A, B, C and D.
(b) Calculate the coordinates of P, the intersection of diagonals AC and BD.
(c) Find the slopes of CB and DA and verify that they are parallel.
(d) Find the equation of diagonal AC.
(a) From the graph:
A(3, 3), B(0, −2), C(−4, −2) and D(−1, 3).
(b) Diagonals of a parallelogram bisect each other.
P is the midpoint of AC.
\mathrm{P}=\left(\frac{3+(-4)}{2},\frac{3+(-2)}{2}\right)⇒ P = (−1/2, 1/2)
⇒ P = (−0.5, 0.5)
(c) Slope of CB = \frac{-2-(-2)}{0-(-4)}=\frac{0}{4}=0
Slope of DA = \frac{3-3}{3-(-1)}=\frac{0}{4}=0
Since the slopes are equal, CB ∥ DA.
(d) Slope of AC = \frac{-2-3}{-4-3}=\frac{-5}{-7}=\frac{5}{7}
Using point A(3, 3):
\mathrm{y}-3=\frac{5}{7}(\mathrm{x}-3)⇒ 7y − 21 = 5x − 15
⇒ 5x − 7y + 6 = 0
Therefore, the equation of AC is 5x − 7y + 6 = 0.
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