Question

Three vertices of parallelogram ABCD are A(−5, −1), B(3, −1) and C(1, −6). Use graphical method to find the co-ordinates of fourth vertex D.

Answer 1

Given:

Vertices of parallelogram ABCD are A(−5, −1), B(3, −1) and C(1, −6)

Formula:

Diagonals of a parallelogram bisect each other, so the midpoint of diagonal AC is equal to the midpoint of diagonal BD:

\[\left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right) = \left(\frac{x_2 + x_4}{2}, \frac{y_2 + y_4}{2}\right)\]

Solution:

\[\text{Midpoint of AC} = \left(\frac{-5 + 1}{2}, \frac{-1 + (-6)}{2}\right)\]  [Substituting the coordinates of vertices A and C]

Midpoint of AC = (−2, −3.5)  [Simplifying the midpoint coordinates]

\[\frac{3 + x_D}{2} = -2 \quad [\text{Equating the x-coordinate of the midpoint of BD to } -2]\]

\[3 + x_D = -4 \quad [\text{Multiplying both sides by 2}]\]

\[\therefore x_D = -7 \quad [\text{Solving for the x-coordinate of vertex D}]\]

\[\frac{-1 + y_D}{2} = -3.5 \quad [\text{Equating the y-coordinate of the midpoint of BD to } -3.5]\]

\[-1 + y_D = -7 \quad [\text{Multiplying both sides by 2}]\]

\[\therefore y_D = -6 \quad [\text{Solving for the y-coordinate of vertex D}]\]

\[\therefore {D(-7, -6)}\]