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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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mypstile | Student, College Freshman | eNotes Newbie

Posted July 24, 2011 at 11:10 PM via web

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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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samhouston | Middle School Teacher | (Level 1) Associate Educator

Posted July 24, 2011 at 11:19 PM (Answer #1)

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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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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted July 24, 2011 at 11:22 PM (Answer #2)

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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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giorgiana1976 | College Teacher | Valedictorian

Posted July 24, 2011 at 11:25 PM (Answer #3)

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To determine the vertices of the triangle whose sides are along the given lines, we'll have to determine the intercepting points of these lines.

We'll determine the intercepting point of the lines 2x+y-3=0 and x-y-6=0.

We'll solve the system of equations using elimination. We'll add the equations:

2x + y - 3 + x - y - 6 = 0

3x - 9 = 0

3x = 9 => x = 3

3 - y - 6 = 0 => -y = 6 - 3

y = -3

The first intercepting point and the 1st vertex  of the triangle is the pair (3 ; -3).

We'll determine the next intercepting point of the lines x-y-6=0; -2x+y-5=0.

We'll add the equations:

x - y - 6 - 2x + y - 5 = 0

-x - 11 = 0 => x = -11

-11 - y - 6 = 0

y = -17

The 2nd intercepting point and the 2nd vertex  of the triangle is the pair (-11 ; -17).

We'll determine the 3d intercepting point of the lines 2x+y-3=0; -2x+y-5=0.

We'll add the equations:

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

2y - 8 = 0

2y = 8 => y = 4

2x + 4 - 3 = 0

2x + 1 = 0

x = -1/2

The 3rd intercepting point and the 3rd vertex  of the triangle is the pair (-1/2 ; 4).

The vertices of triangle are represented by the following pairs: (3 ; -3) ; (-11 ; -17) ; (-1/2 ; 4).

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