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Step-by-Step Solution
Step 1: Identify the given lines and the intersection of the diagonals
We have two lines given as sides of the rhombus:
Line 1: $x - y + 1 = 0$
Line 2: $7x - y - 5 = 0$
The diagonals of the rhombus intersect at the point $O(-1, -2)$ (the midpoint of both diagonals).
Step 2: Express points on each line in parametric form
Let the vertex where these two lines meet be $A$. Solve for the intersection of Line 1 and Line 2:
\[
\begin{aligned}
\text{Line 1: } & y = x + 1, \\
\text{Line 2: } & y = 7x - 5.
\end{aligned}
\]
Equate them:
\[
x + 1 = 7x - 5 \quad \Longrightarrow \quad 6x = 6 \quad \Longrightarrow \quad x = 1, \quad y = 2.
\]
Hence, $A$ is at $(1,\,2)$.
Now, any other point on Line 1 can be written as $B = (t,\,t+1)$, and any other point on Line 2 can be written as $D = (u,\,7u - 5)$.
Step 3: Use the midpoint condition of the diagonals
A rhombus is a special parallelogram, so the diagonals bisect each other. That means the midpoint of $BD$ should be the intersection point $O(-1, -2)$:
\[
O = \left(\frac{B + D}{2}\right) = (-1, -2).
\]
Hence,
\[
B + D = 2O = (-2, -4).
\]
Substitute $B=(t,\,t+1)$ and $D=(u,\,7u-5)$:
\[
(t+u,\; (t+1)+(7u-5)) = (-2,\; -4).
\]
This gives us two equations:
\[
\begin{aligned}
t + u &= -2, \\
(t + 1) + (7u - 5) &= -4 \quad \Longrightarrow \quad t + 7u - 4 = -4 \quad \Longrightarrow \quad t + 7u = 0.
\end{aligned}
\]
Step 4: Solve for the parameters
From the system:
\[
\begin{aligned}
t + u &= -2, \\
t + 7u &= 0,
\end{aligned}
\]
subtract the first equation from the second:
\[
(t + 7u) - (t + u) = 0 - (-2) \quad \Longrightarrow \quad 6u = 2 \quad \Longrightarrow \quad u = \frac{1}{3}.
\]
Then substitute $u = \tfrac{1}{3}$ into $t + u = -2$:
\[
t + \tfrac{1}{3} = -2 \quad \Longrightarrow \quad t = -2 - \tfrac{1}{3} = -\tfrac{7}{3}.
\]
Step 5: Find the specific vertices
Corresponding to $t = -\tfrac{7}{3}$ and $u = \tfrac{1}{3}$, the points on each line are:
$B = \bigl(t,\;t+1\bigr)
= \Bigl(-\tfrac{7}{3},\;-\tfrac{7}{3} + 1\Bigr)
= \Bigl(-\tfrac{7}{3},\;-\tfrac{4}{3}\Bigr).$
$D = \bigl(u,\;7u - 5\bigr)
= \Bigl(\tfrac{1}{3},\;7\cdot\tfrac{1}{3} - 5\Bigr)
= \Bigl(\tfrac{1}{3},\;-\tfrac{8}{3}\Bigr).$
Thus, one of the vertices of the rhombus is $D\Bigl(\tfrac{1}{3}, -\tfrac{8}{3}\Bigr)$.
Step 6: Verify with the given options
Among the given choices, the point $\Bigl(\tfrac{1}{3}, -\tfrac{8}{3}\Bigr)$ matches Option 1 and is indeed a vertex of the rhombus.
