The possible coordinates for the vertex D of the parallelogram could be (-4, 3)
It is given that A(1, –1), B (–1, 3), and C (4, –1) are the three vertices of the parallelogram.
Let us assume that the vertices are in the order A, C, B, and D where the coordinates of D are (a, b).
In this scenario, if we join AB and CD, they will become the diagonals of the parallelogram ABCD.
According to the properties of a parallelogram, diagonals bisect each other.
Hence, mid-point of AB = mid-point of CD
Now, according to the mid-point theorem, if mid-point of AB is given as (x,y), then,
x = (x₁ + x₂)/2 and y = (y₁ + y₂)/2
Here, for AC,
x₁ = 1, y₁ = -1
x₂ = -1, y₂ = 3
Then, x = ( 1 - 1)/2 and y = (-1 + 3)/2
(x, y) ≡ (0, 1) ............. (1)
Since (x, y) is also the mid-point of CD, we also have,
x₁ = 4, y₁ = -1
x₂ = a, y₂ = b
Then, x = (4 + a)/2 and y = (-1 + b)/2
(x, y) ≡ ((4 + a)/2, (-1 + b)/2) ................... (2)
From (1) and (2),
(4+a)/2 = 0 and (-1+b)/2 = 1
4+a = 0 and (-1+b) = 2
a = -4 and b = 3
Hence, the fourth vertex D of the parallelogram can be possibly located at (-4, 3)
Learn more about a parallelogram here:
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