SOLUTION: I need help with a math project. Its about using Linear Equations to make a shape. The Guidelines are: 1. The Design/Picture has to be at least 10 line segments & can only use s

Question 1029129: I need help with a math project. Its about using Linear Equations to make a shape. The Guidelines are:
1. The Design/Picture has to be at least 10 line segments & can only use straight lines in all 4 quadrants. 2 must be vertical, 2 horizontal, 3 positive lines and 3 negative lines. There must be three systems of equations in the picture Infinate, No solution & one solution
2. Quad 1 has to have a negative slope. Quad 2 has to have a undefined slope. Quad 3 a zero slope and positive slope in quadrant 4.Label three systems of equations on the coordiane plane
3. In a paragraph explain how to find the slope of any given line segment and the equation of a line segment given two points
4. Slopes and equation for one segment in each quadrant. (Cant be zero or undefined) List the four segments that you have giving the coordinates of the endpoints. Find the slope & equation of these segments
I know this is alot but I tried this project twice and my teacher told me it was ungradeable twice. This is my thrid and final time and I really want to understand this material. Thank you

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
y=-x+4. That gives a negative slope in the first quadrant' make y=-x+6,. Those two lines have no solution. They are parallel. The slope is the value in front of the x, and the y-intercept is the number at the end, the b in y=mx +b.
Now say 2y=-2x+12. That is the same as y=-x+6. That line can have those two labels, and it has infinite solutions. Extend these lines into the second quadrant, so you have a line with non-infinite or non-zero slope in both quadrants.
x=-2. That gives an undefined slope in the second quadrant.
y=-3 That gives a slope of 0 in the third quadrant.
y=x-4. That gives a positive slope in the fourth quadrant. Extend that line into the third quadrant so that you can say you have a line with that slope in the third quadrant.
The -x+4 and x-4 will intersect.
y=-x+6
y=x-4
2y=2, y=1. When y=1, x=5. So the lines intersect at (5,1).
I can't draw in here x=-2, but it is vertical along the x=-2 line.
Those fulfill the basic requirements for the quadrants.
Those lines may be extended or not.
To make the shape with 6 more lines, you need one more vertical line, and you can put it anywhere. The equation is x= whatever it is on the x-axis.
For the horizontal line, you may put it anywhere, and its equation is y- whatever it is.

This graph needs more lines to make 10. You can draw any shape you want and erase the parts of the lines that are going to infinity. For each line, where it crosses the y-axis is the intercept or b. The slope of the line is positive if it goes up and to the right, and it is the change in y divided by the change in x. This is one of those instances where it would be easier to draw the line on graph paper and figure out the equation from the intercept and the slope.
For any two points, you can determine the slope: change in y/change in x.
(2,3) and (4,9), the change in y is 9-3 or 6 divided by the change in x, which is 4-2 or 2. The slope is 3.
To get the equation of the line, use the point slope formula with slope 3.
y-3=3(x-2) using the first point. y-3=3x-6; y=3x-3. This line crosses the y-axis at -3 and has a slope of 3, going up 3 for every 1 it goes to the right.
using the other point gives you the same answer.
y-9=3(x-4)
y-9=3x-12
y=3x-3.
See if that is enough to put in other lines and give their equation. I'm not sure I understand the whole question, but this to me seems to fulfill the basic parts of it.
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