To solve the problem step by step, we will follow the outlined process to find the vertex of the rhombus given the equations of two sides and the intersection point of the diagonals.
### Step 1: Identify the equations of the lines
The two sides of the rhombus are given by the equations:
1. \( x - y + 1 = 0 \) (let's call this Line 1)
2. \( 7x - y - 5 = 0 \) (let's call this Line 2)
### Step 2: Find the intersection point of the two lines
To find the intersection point of the two lines, we can solve them simultaneously.
From Line 1:
\[ y = x + 1 \]
Substituting \( y \) in Line 2:
\[ 7x - (x + 1) - 5 = 0 \]
\[ 7x - x - 1 - 5 = 0 \]
\[ 6x - 6 = 0 \]
\[ x = 1 \]
Now substituting \( x = 1 \) back into Line 1 to find \( y \):
\[ y = 1 + 1 = 2 \]
Thus, the intersection point of the two lines is \( (1, 2) \).
### Step 3: Find the midpoint of the diagonals
The diagonals of the rhombus intersect at the point \( (-1, -2) \).
### Step 4: Use the midpoint formula to find the vertices
Let’s denote the vertices of the rhombus as \( A, B, C, D \). We know that the midpoint \( M \) of the diagonal \( AC \) is given by:
\[ M = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right) = (-1, -2) \]
Let \( A = (1, 2) \) (found from the intersection of the lines) and \( C = (h, k) \). Therefore:
1. \( \frac{1 + h}{2} = -1 \)
2. \( \frac{2 + k}{2} = -2 \)
From the first equation:
\[ 1 + h = -2 \]
\[ h = -3 \]
From the second equation:
\[ 2 + k = -4 \]
\[ k = -6 \]
Thus, the coordinates of point \( C \) are \( (-3, -6) \).
### Step 5: Check the options for the vertex
Now we need to check which of the given options matches the coordinates of vertex \( C \):
1. \( (-3, -9) \)
2. \( (-3, -8) \)
3. \( \left(\frac{1}{3}, -\frac{8}{3}\right) \)
4. \( \left(-\frac{10}{3}, -\frac{7}{3}\right) \)
None of these options match \( (-3, -6) \). We need to find the other vertices \( B \) and \( D \).
### Step 6: Find the equation of the line parallel to \( AB \) for \( CD \)
The slope of line \( AB \) can be calculated from its equation \( x - y + 1 = 0 \):
- The slope of line \( AB \) is \( 1 \) (since it can be rearranged to \( y = x + 1 \)).
Since \( CD \) is parallel to \( AB \), it will also have a slope of \( 1 \). The equation of line \( CD \) passing through point \( C (-3, -6) \) is:
\[ y + 6 = 1(x + 3) \]
\[ y + 6 = x + 3 \]
\[ y = x - 3 \]
### Step 7: Find the intersection of line \( CD \) with line \( AD \)
The equation of line \( AD \) is:
\[ 7x - y - 5 = 0 \]
Rearranging gives:
\[ y = 7x - 5 \]
Now we can solve for the intersection of \( y = x - 3 \) and \( y = 7x - 5 \):
Set the equations equal to each other:
\[ x - 3 = 7x - 5 \]
\[ -3 + 5 = 7x - x \]
\[ 2 = 6x \]
\[ x = \frac{1}{3} \]
Substituting \( x = \frac{1}{3} \) back into \( y = x - 3 \):
\[ y = \frac{1}{3} - 3 = \frac{1}{3} - \frac{9}{3} = -\frac{8}{3} \]
Thus, the coordinates of point \( D \) are \( \left(\frac{1}{3}, -\frac{8}{3}\right) \).
### Step 8: Verify the options
Now we check the options again:
1. \( (-3, -9) \)
2. \( (-3, -8) \)
3. \( \left(\frac{1}{3}, -\frac{8}{3}\right) \) **(This matches)**
4. \( \left(-\frac{10}{3}, -\frac{7}{3}\right) \)
### Conclusion
The vertex of the rhombus is:
**Option (3) \( \left(\frac{1}{3}, -\frac{8}{3}\right) \)**.
