I'll make up a problem like that.
Parallelogram ABCD has vertices A(-4,2), B(-5,-3), C(3,-1), D(4,4).
Find the slope and equation of of AB:
Slope formula:
m =
where (x1,y1) = A(-4,2)
and where (x2,y2) = B(-5,-3)
m =
m =
m =
m =
Point-slope formula:
y - y1 = m(x - x1)
where (x1,y1) = (-4,2)
y - (2) = (5)(x - (-4))
y - 2 = 5(x + 4)
y - 2 = 5x + 20
y = 5x + 22 <-- equation of AB, slope = 5
Using the same method find the equations of the other three sides:
<-- equation of BC, slope =
<-- equation of CD, slope = 5
<-- equation of AD, slope =
Notice that opposite sides AB and CD both have slope = 5
Notice that opposite sides BC and AD both have slope =
That demostrates that opposite sides of a parallelogram are parallel,
i.e., they have the same slope.
To find the area, find the area of the triangle ABC by the determinant
method and double it.
The determinant formula for the area of a triangle is
A =
Where the x's and y's are the three coordinates of the vertices
in order taken counterclockwise around the triangle.
A =
Area of the parallelogram = 2×19 = 38
Edwin