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

justaguide | Certified Educator

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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...

(The entire section contains 2 answers and 185 words.)

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samhouston | Certified Educator

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giorgiana1976 | Student

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.

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.

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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