How would you solve for the area of a parallelogram given the equation of its sides?

Question

Solve this;The sides of a parallelogram are on the line x - 3y + 20= 0, x + y + 6= 0, x - 3y - 10 = 0, and x + y - 2 = 0.Find its area..

Answer 1

No, the diagonales of a parallelogram do NOT intersect at 90 degrees...
1. Rewrite equations in the form of y = mx + b
a. y = 1/3x + 20/3
b. y = -x -6
c. y = 1/3x + 10/3
d. y = -x +2
2. eq a & c have the same m: 1/3. They are //
eq b & d have the same m: -1. They are //
3. Since m = dy / dx, eq b & d define lines at 45 deg slope downward. The distance between the two lines (or height of the parallelog) becomes:
h = (b[d] - b[b]) * cos 45 = (+2 - -6) cos(45) = 8 sqr(2).
4. Find intersection between [d] and [a] to get one corner.
x1 = (b[a] - b[d]) / (m[d] - m[a])
x1 = (20/3 - 2) / (-1 - 1/3)
x1 = -7/2 = -3.5
y1 = m[d]*x1 + b[d]
y1 = (-1 * -3.5) + 2 = +5.5
5. Find intersection between [d] and [c] to get the other.
x2 = (b[c] - b[d]) / (m[d] - m[c])
x2 = (10/3 - 2) / (-1 - 1/3)
x2 = -1
y2 = m[d]*x2 + b[d]
y2 = -1 * -1 + 2 = 3
6. Distance between these two corners = width of parallelog:
d² = dx² + dy² = (-3.5 - -1)² + (5.5 - 3)²
d = sqr (-2.5² + 2.5²) = 2.5 sqr(2).
7. Finally, area:
A = h x d = 8 sqr(2) x 2.5 sqr(2) = 8 * 2.5 * 2.
That, you can do!

Answer 2

If a superb triangle with hypotenuse a has aspects x and y+c, then x^2 + (y+c)^2 = a^2 => for the triangle with hypotenuse b, y^2 + (x+c)^2 = b^2 Subtracting the two eqns., 2c(y - x) = a^2 - b^2 next step is to seek 2 smallest and distinctive Pythagorian triples which would be suited to fulfill the above eqn. it is so some distance as i'd desire to realize this some distance. Will attempt the following day now. Edit: the form you have written the question is suited. I did have fake impact which you rightly found out, however the question is obvious. i develop into, in fact, scuffling with Pythagorian triples and gave up. After analyzing Quadrillatorer's answer and your added information, i found out my mistake. i does not have been waiting to think of ways Quadrillaterer does. i could be shocked if there's a sq. with area below he has got here across.

Answer 3

If you have no disriminant, no matrices, and no values for the variables, then it is a long problem...
the equation of the sides needs to be put into slope intercept form then graph each side. since the area of a parallelogram is a=bh, you have to make sure the lines are connect properly to form the correct angles. once you graph it you can get the angle using the law of sines. you have the angles and you can get you base and height, then the area.
this is the only way i can think to solve it. otherwise you end up with an algebraic expression as your final answer for the area. it is a combination of algebra, trig and calculus.
if you or anyone else finds and easier way to solve it, please let me know.

Answer 4

Basically, I'd also solve it by first solving for the corner points. But since eqn2 and eqn4 are relatively simple, u can see that they have intercepts -6 and 2 and make 45deg with the x axis. Thus the height h btw them is
|(-6-2)cos45| = 8/sqrt2. Now, solve eqn2 with first eqn3, then with eqn4 to get 2 corner points.
sqrt( (x1-x2)^2 + (y1-y2)^2) gives breadt b.
A=bh.

Answer 5

OK another equation to find the area of a parallelogram is

(length of diag. 1)(length of diag. 2)*.5

this is because each half is a triangle with the diagonals as their bases and heights, since the diags. meet at right angles.

1) find the four corners.
2) using the corners, find the lengths of the diagonals
3) plug them into the above formula and there you have it.