Find the vertices of triangle which sides are on the lines 2x+y-3=0; x-y-6=0; -2x+y-5=0?

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The vertexes of the triangle are at the points of intersection of each pair of the three lines:

2x+y-3=0 ...(1)

x-y-6=0 ...(2)

-2x+y-5=0 ...(3)

The point of intersection of (1) and (2) can be determined by substituting x = y + 6 from (2) in (1)

=> 2(y + 6) + y - 3 = 0

=> 2y + 12 + y - 3 = 0

=> 3y = -9

=> y = -3

x = 3

One of the vertexes is (3, -3)

The point of intersection of (2) and (3) can be determined by substituting x = y + 6 from (2) in (3)

=> -2(y + 6) + y - 5 = 0

=> -2y - 12 + y - 5 = 0

=> -y = 17

=> y = -17

x = -11

The second vertex is at (-11, -17)

Add (1) and (3)

=> 2x + y - 3 - 2x + y - 5 = 0

=> 2y - 8 = 0

=> y = 4

2x = -4 + 3

=> x = -1/2

The vertexes of the triangle are at (-1/2, 4), (-11, -17) and (3, -3).

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First, rewrite each equation in slope-intercept form.

2x + y - 3 = 0          y = -2x + 3

x - y - 6 = 0             y = x - 6

-2x + y - 5 = 0         y = 2x + 5

Now graph the three equations and find the 3 points of intersection.  These will be the vertices of the triangle.

The vertices are (3, -3), (-0.5, 4), and (-11, -17).

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