Question

Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4), and (−2, 8), respectively. Then, the coordinates of the vertex D are

• A. (−4, 5)
• B. (4, 5)
• C. (−3, 4)
• D. (0, 11)

Answer 1

The Correct Option is A

The correct answer is :A
Since ABCD is a parallelogram, opposite sides are parallel and have the same length. The midpoint formula can be used to find the midpoint of a line segment given its endpoints.
First, let's find the midpoint of AB:
Midpoint of AB=\((\frac{(x1 + x2)}{2},\frac{(y1 + y2)}{2})=(\frac{(1 + 3)}{2},\frac{(1 + 4)}{2})=(2, 2.5)\)
Next, let's find the midpoint of BC:
Midpoint of BC=\((\frac{(x1 + x2)}{2},\frac{(y1 + y2)}{2})\)=\((\frac{(3 + (-2))}{2},\frac{(4 + 8)}{2})\)=(0.5, 6)
Now, since opposite sides of a parallelogram are parallel, the vector from A to B is equal to the vector from D to C. We can use this information to find the coordinates of point D.
Vector from A to B: (3-1, 4-1)=(2,3)
Coordinates of C: (-2, 8)
Now, we can add the vector AB to the coordinates of C to find D:
D = C + AB = (-2 + 2, 8 + 3) = (0, 11)
So, the correct answer is:
The coordinates of the vertex D are (0, 11).

Answer 2

When two diagonals in a parallelogram bisect one another, it means that the midpoints of both diagonals are the same.
Midpoint of AC =\((\frac{1-2}{2},\frac{1+8}{2})\)
Let vertex D's coordinates be \((x,y). \)
\((\frac{x+3}{2},\frac{y+4}{2})\)\(=\)\((\frac{1-2}{2},\frac{1+8}{2})\)
\(⇒ x = -4 \) and \(y = 5\)