I have attempted to solve the following problems numerous times, but I have yet to find the solution. I strongly believe that they involve families of lines.
1. Two vertices of a rectangle are (1,4) and (-6,-5). Two sides are parallel to 3x + 2y = 0. Find the equations of each side.
2. The equations of two sides and a diagonal of a parallelogram are 5x + 2y -1 = 0, x + 2y - 13 = 0, and x - 6y + 19 = 0 respectively. Find the coordinates of the vertices and the equations of the other two sides.
Thank-You for your time.
1. Two vertices of a rectangle are (1,4) and (-6,-5). Two sides are parallel to 3x + 2y = 0. Find the equations of each side.
2. The equations of two sides and a diagonal of a parallelogram are 5x + 2y -1 = 0, x + 2y - 13 = 0, and x - 6y + 19 = 0 respectively. Find the coordinates of the vertices and the equations of the other two sides.
Thank-You for your time.
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1. Vertices at (1, 4) and (-6, -5). For 3x + 2y = 0, the slope is m = -3/2, so the slope of the other lines must be 2/3 (negative reciprocals). We get the equations by using point-slope 4 times:
a. y - 4 = -3/2 (x - 1)
b. y + 5 = -3/2 (x + 6)
c. y - 4 = 2/3 (x - 1)
d. y + 5 = 2/3 (x + 6)
You can clean up those equations if you wish.
2. (a) 2y = -5x + 1 ==> y = -5/2 x + 1/2
(b) 2y = -x + 13 ==> y = -1/2 x + 13/2
(c) 6y = x + 19 ==> y = 1/6 x + 19/6
By lining up these three slopes, you can tell that (a) and (c) are sides, and (b) is the diagonal. The slope of the diagonal must be in the middle. (-5/2 < -1/2 < 1/6)
I have a hunch this can best be solved with vectors, but we'll just use algebra. First, let's see where (a) and (c) intersect:
-5/2 x + 1/2 = 1/6 x + 19/6 ==> -15x + 3 = x + 19 ==> 16x = -16 ==> x = -1 ==> y = 3
(-1, 3) is a vertex. (Answer)
Now we find the intersection of (a) and (b):
-5/2 x + 1/2 = -1/2 x + 13/2 ==> 2x = -6 ==> x = -3 ==> y = 8. (-3, 8) is a vertex. (Answer)
And the intersection of (b) and (c):
-1/2 x + 13/2 = 1/6 x + 19/6 ==> -3x + 39 = x + 19 ==> 4x = 20 ==> x = 5 ==> y = 4
(5, 4) is a vertex. (Answer)
It's a good idea to plot these three points to see where we are ...
Use the slope of (a) with the vertex (5, 4) to draw a line:
(d) y - 4 = -5/2 (x - 5) (Equation of a side) (Answer)
Use the slope of (c) with the vertex (-3, 8) to draw a line:
(e) y - 8 = 1/6 (x + 3) (Equation of a side) (Answer)
To get the last vertex, we solve (d) and (e) simultaneously:
-5/2 (x - 5) + 4 = 1/6 (x + 3) + 8 ==> -15 (x - 5) + 24 = x + 3 + 48 ==> 16x = 99 - 51 = 48
==> x = 3 ==> y = 9. (3, 9) is a vertex. (Answer)
You have all your answers. You may wish to clean up a couple of equations.
a. y - 4 = -3/2 (x - 1)
b. y + 5 = -3/2 (x + 6)
c. y - 4 = 2/3 (x - 1)
d. y + 5 = 2/3 (x + 6)
You can clean up those equations if you wish.
2. (a) 2y = -5x + 1 ==> y = -5/2 x + 1/2
(b) 2y = -x + 13 ==> y = -1/2 x + 13/2
(c) 6y = x + 19 ==> y = 1/6 x + 19/6
By lining up these three slopes, you can tell that (a) and (c) are sides, and (b) is the diagonal. The slope of the diagonal must be in the middle. (-5/2 < -1/2 < 1/6)
I have a hunch this can best be solved with vectors, but we'll just use algebra. First, let's see where (a) and (c) intersect:
-5/2 x + 1/2 = 1/6 x + 19/6 ==> -15x + 3 = x + 19 ==> 16x = -16 ==> x = -1 ==> y = 3
(-1, 3) is a vertex. (Answer)
Now we find the intersection of (a) and (b):
-5/2 x + 1/2 = -1/2 x + 13/2 ==> 2x = -6 ==> x = -3 ==> y = 8. (-3, 8) is a vertex. (Answer)
And the intersection of (b) and (c):
-1/2 x + 13/2 = 1/6 x + 19/6 ==> -3x + 39 = x + 19 ==> 4x = 20 ==> x = 5 ==> y = 4
(5, 4) is a vertex. (Answer)
It's a good idea to plot these three points to see where we are ...
Use the slope of (a) with the vertex (5, 4) to draw a line:
(d) y - 4 = -5/2 (x - 5) (Equation of a side) (Answer)
Use the slope of (c) with the vertex (-3, 8) to draw a line:
(e) y - 8 = 1/6 (x + 3) (Equation of a side) (Answer)
To get the last vertex, we solve (d) and (e) simultaneously:
-5/2 (x - 5) + 4 = 1/6 (x + 3) + 8 ==> -15 (x - 5) + 24 = x + 3 + 48 ==> 16x = 99 - 51 = 48
==> x = 3 ==> y = 9. (3, 9) is a vertex. (Answer)
You have all your answers. You may wish to clean up a couple of equations.
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The question is where did I learn math. I took separate classes in college algebra, analytic geometry, differential calculus, and integral calculus. Also a class in vector analysis, and a semester of solid geometry in high school. My H.S. plane geometry class was strictly Euclid, lots of proofs.
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