# SOLUTION: 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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 Click here to see ALL problems on Inequalities 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(30993)   (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. ------------------- 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. -------------- 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. ---------- 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.