in the parallelogram with vertices (2,3),(4,1),(8,2) AND (6,4), find the equation of its (a)sides (b) diagonal

Answer 1

It seems like the previous answers have given the complete detail of the solution, so I'll discuss how to get it. The equation of the line will be handy in this problem.

First, plot these points in the Cartesian Plane and form the parallelogram. (b) From the figure, the diagonals should have the vertices (6,4) - (4,1) and (2,3) - (8,2). Since you are given two points, then you can find the equations...(deal with the pair of vertices separately).

For example, (2,3) - (8,2) should give you a slope m = -1/6. Then, using the slope-intercept form, y = mx + b (where b is the y-intercept), use either (2,3) or (8,2) and find b... so here, if I choose (2,3), then the equation is 3 = (-1/6)(2) + b... hence b = 10/3. So, in general, y = (-1/6)x + 10/3 or in the general form,
x + 6y -20 = 0. (This is one of the equations required)

Do the same thing to the other pair.


For (a), do the same thing, but this time, the sides are now considered.

Answer 2

I'm going to start by labeling the vertexes of the parallelogram
A = (2, 3)
B = (4, 1)
C = (8, 2)
D = (6, 4)

I'm going to do all of these the same way.
Calculate the slope (m) by using (Y1 - Y2)/(X1 - X2). Write equation in the form y = mx + b. Calulate b, by substituting a value of x and y into the equation.

a)
Equation of line AB -> y = -x + 5
Equation of line BC -> y = (1/4)x
Equation of line CD -> y = -x + 10
Equation of line DA -> y = (1/4)x + 5/2

b)
There are two diagonals
Equation of line AC -> y = (-1/6)x + 10/3
Equation of line BD ->y = (3/2)x - 5

Answer 3

Name the paralleogram's vertices:
- A(2,3)
- B(4,1)
- C(8,2)
- D(6,4)
Write the form of the lines' equations: y=ax+b
To calculate a and b we'll have to solve each pair of equations:
For the line AB does go through vertex A and vertex B and so x(A), y(A) and x(B), y(B), hence we replace them (x, y) with their values,
We have this pair of equations:
3=2a+b
1=4a+b
<=> 6=4a+2b (we got 5 multiplied with each side of this equation)
1=4a+b
<=> 6=4a+2b (1)
1=4a+b (2)
(1) minus (2)
=> b=5
=> 6=4a+2*5
=> a=-1
=> the equation of line => AB: y= -x+5
For the rest, I similarly apply the method above...
a) The equation of line BC => y=0.25x
The equation of line CD => y= -x+10
The equation of line AD => y= 0.25x+2.5
b) The 2 diagonals:
The equation of line AC => y =(-1/6)x+10/3
The equation of line BD =>y =1.5x-5