Certainly! Let's solve for the vertices of the triangle formed by the given lines:
The equations of the lines are:
1. [tex]\( 3x + y = 7 \)[/tex]
2. [tex]\( 3y - x = 1 \)[/tex]
3. [tex]\( 7y + x = -11 \)[/tex]
To find the vertices of the triangle, we need to determine the points of intersection of these lines pairwise.
### Finding the Intersection of [tex]\(3x + y = 7\)[/tex] and [tex]\(3y - x = 1\)[/tex]
We solve these two equations simultaneously:
1. [tex]\( 3x + y = 7 \)[/tex]
2. [tex]\( -x + 3y = 1 \)[/tex]
We can use the method of substitution or elimination. Let's opt for substitution here.
From the first equation:
[tex]\[ y = 7 - 3x \][/tex]
Substitute this into the second equation:
[tex]\[ -x + 3(7 - 3x) = 1 \][/tex]
[tex]\[ -x + 21 - 9x = 1 \][/tex]
[tex]\[ -10x + 21 = 1 \][/tex]
[tex]\[ -10x = 1 - 21 \][/tex]
[tex]\[ -10x = -20 \][/tex]
[tex]\[ x = 2 \][/tex]
Now substitute [tex]\( x = 2 \)[/tex] back into [tex]\( y = 7 - 3x \)[/tex]:
[tex]\[ y = 7 - 3(2) = 7 - 6 = 1 \][/tex]
Thus, the first vertex is:
[tex]\[ (x, y) = (2, 1) \][/tex]
### Finding the Intersection of [tex]\(3x + y = 7\)[/tex] and [tex]\(7y + x = -11\)[/tex]
We solve these two equations simultaneously:
1. [tex]\( 3x + y = 7 \)[/tex]
2. [tex]\( x + 7y = -11 \)[/tex]
Let's opt for substitution again.
From the second equation:
[tex]\[ x = -11 - 7y \][/tex]
Substitute this into the first equation:
[tex]\[ 3(-11 - 7y) + y = 7 \][/tex]
[tex]\[ -33 - 21y + y = 7 \][/tex]
[tex]\[ -33 - 20y = 7 \][/tex]
[tex]\[ -20y = 7 + 33 \][/tex]
[tex]\[ -20y = 40 \][/tex]
[tex]\[ y = -2 \][/tex]
Now substitute [tex]\( y = -2 \)[/tex] back into [tex]\( x = -11 - 7y \)[/tex]:
[tex]\[ x = -11 - 7(-2) \][/tex]
[tex]\[ x = -11 + 14 \][/tex]
[tex]\[ x = 3 \][/tex]
Thus, the second vertex is:
[tex]\[ (x, y) = (3, -2) \][/tex]
### Finding the Intersection of [tex]\(3y - x = 1\)[/tex] and [tex]\(7y + x = -11\)[/tex]
We solve these two equations simultaneously:
1. [tex]\( -x + 3y = 1 \)[/tex]
2. [tex]\( x + 7y = -11 \)[/tex]
Let's use the elimination method.
Add the two equations:
[tex]\[ (-x + 3y) + (x + 7y) = 1 + (-11) \][/tex]
[tex]\[ 10y = -10 \][/tex]
[tex]\[ y = -1 \][/tex]
Now substitute [tex]\( y = -1 \)[/tex] into the first equation:
[tex]\[ -x + 3(-1) = 1 \][/tex]
[tex]\[ -x - 3 = 1 \][/tex]
[tex]\[ -x = 1 + 3 \][/tex]
[tex]\[ -x = 4 \][/tex]
[tex]\[ x = -4 \][/tex]
Thus, the third vertex is:
[tex]\[ (x, y) = (-4, -1) \][/tex]
### Summary of the Vertices
The vertices of the triangle formed by the lines are:
[tex]\[ (2, 1), (3, -2), \text{ and } (-4, -1) \][/tex]
