Question 163044This question is from textbook saxon algebra 2
: Graph and describe the shape that the lines form.
3x-2y=2
-3x-2y=-34
x-2y=6
find the vertices of the shape.
This question is from textbook saxon algebra 2
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Graph and describe the shape that the lines form.
3x-2y=2
-3x-2y=-34
x-2y=6
find the vertices of the shape.
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Since they're 3 straight lines, the "shape they form", or any space they enclose can only be a triangle. If 2 of them are parallel, they won't enclose any space.
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To find the 3 vertices, solve the eqns in pairs, ie, eqn 1&2, eqn 1&3, eqn 2&3.
1) 3x-2y=2
2) -3x-2y=-34
3) x-2y=6
Check the slopes of each to see if any are parallel:
1) m = 3/2
2) m = -3/2
3) m = 1/2
The slopes are different, so no 2 lines are parallel. Also, no 2 lines are perpendicular, so it's not a right triangle.
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1 & 2:
1) 3x-2y=2
2) -3x-2y=-34
Add them
0x -4y = -32
y = 8
Sub into 1:
3x - 2*8 = 2
3x = 18
x = 6
So the vertex is (6,8)
1 & 3:
1) 3x-2y=2
3) x-2y=6
Subtract 3 from 1
2x = -4
x = -2
3) -2 - 2y = 6
-2y = 8
y = -4
The 2nd vertex is (-2,-4)
2 & 3:
2) -3x-2y=-34
3) x-2y=6
Subtract 3 from 2
-4x = -40
x = 10
Sub into 3
10 - 2y = 6
-2y = -4
y = 2
Vertex #3 is (10,2)
So it is a triangle with vertices at (10,2), (-2,-4) and (6,8)
The lengths of the sides and the angles can be found, but it wasn't asked for.
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