The sides of a triangle are given by the equations 2x+y=5, x+2y=7 and x-y=1?

To find the vertices of the triangle, we need to solve the system of equations:

2x + y = 5
x + 2y = 7
x - y = 1

First, let's solve the system of equations by substitution or elimination method:

From the third equation, we can express x in terms of y:
x = y + 1

Now substitute x in the first equation:
2(y + 1) + y = 5
2y + 2 + y = 5
3y + 2 = 5
3y = 3
y = 1

Now substitute y back into x = y + 1:
x = 1 + 1
x = 2

Therefore, the first vertex of the triangle is (2, 1).

Now, substitute x and y into the second equation:
2 + 2y = 7
2y = 5
y = 5/2
y = 2.5

Now substitute y back into x = y + 1:
x = 2.5 + 1
x = 3.5

Therefore, the second vertex of the triangle is (3.5, 2.5).

Now, substitute x and y into the third equation:
3.5 - 2.5 = 1

Therefore, the third vertex of the triangle is (3.5, 2.5).

Therefore, the vertices of the triangle are (2, 1), (3.5, 2.5), and (3.5, 2.5).

Given the equations of the sides of a triangle are
2x - y = 5 ...(i)
x + 2y = 7 ...(ii)
x- y = 1 ...(iii)

Solving (ii) and (iii), we get
x = 4 and y = 3
Thus, the vertices of the triangle are A(4, 3), B(6, -1) and C(5, 4)
Area of triangle
Area = 1/2 - [x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
= 1/2 - [4(3+1) + 6(-1-3)+5(3+1)]
= 1/2 - [16 -24 + 20]
= 1/2 - 12
= 6 sq. units

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