Two sides of a rhombus are along the lines, $$x - y + 1 = 0$$ and $$7x - y - 5 = 0$$. If its diagonals intersect at $$(-1, -2)$$, then which one of the following is a vertex of this rhombus?

We have a rhombus whose two adjacent sides lie on the straight lines

$$L_1:\;x-y+1=0 \qquad\text{and}\qquad L_2:\;7x-y-5=0.$$

For any straight line written as $$ax+by+c=0,$$ a direction vector is $$(b,\,-a).$$ Hence

$$\text{Direction along }L_1:\;(1,1),\qquad \text{Direction along }L_2:\;(1,7).$$

Let the side of the rhombus that is parallel to $$L_1$$ be represented by the vector

$$\vec v = a(1,1)=(a,a),$$

and the adjacent side parallel to $$L_2$$ be represented by

$$\vec w = b(1,7)=(b,7b).$$

Because every side of a rhombus has the same length, we impose

$$|\vec v| = |\vec w|.$$

First we write the magnitudes explicitly:

$$|\vec v|=\sqrt{a^{2}+a^{2}}=\sqrt{2\,a^{2}}=\sqrt2\,|a|,$$

$$|\vec w|=\sqrt{b^{2}+49\,b^{2}}=\sqrt{50\,b^{2}}=\sqrt{50}\,|b|=5\sqrt2\,|b|.$$

Equating the two gives

$$\sqrt2\,|a| = 5\sqrt2\,|b| \;\Longrightarrow\; |a| = 5|b|.$$

This can be written more simply as

$$a = 5b \quad\text{or}\quad a = -5b.$$

That is, the two side-vectors can point in the same sense ($$a=5b$$) or in opposite senses ($$a=-5b$$).

The diagonals of any parallelogram, and therefore of a rhombus, bisect each other. Hence if the point of intersection of the diagonals is $$O(-1,-2),$$ and if we take a vertex $$A$$ of the rhombus, then

$$\overrightarrow{OA} = -\dfrac{\vec v + \vec w}{2}.$$

We examine the two sign possibilities separately.

Case 1: $$a=5b$$ (both side vectors in the same general direction).

Here

$$\vec v = (5b,5b),\qquad \vec w = (b,7b).$$

Adding them,

$$\vec v+\vec w=(5b+b,\;5b+7b)=(6b,\;12b).$$

Therefore

$$\overrightarrow{OA}= -\dfrac{(6b,\,12b)}{2}=(-3b,\,-6b).$$

This vector has components in the ratio $$1:2.$$ None of the option vectors from $$O(-1,-2)$$ to the listed points is in that ratio, so this case produces no admissible vertex.

Case 2: $$a=-5b$$ (the two side vectors oppose each other).

Now

$$\vec v = (-5b,-5b),\qquad \vec w = (b,7b).$$

Add again:

$$\vec v+\vec w=(-5b+b,\,-5b+7b)=(-4b,\;2b).$$

Hence

$$\overrightarrow{OA}= -\dfrac{(-4b,\,2b)}{2}=(2b,\,-b).$$

This vector is proportional to $$(2,-1).$$ We now look for a choice of $$b$$ that places $$A$$ at one of the given options.

From each option we compute the displacement from $$O(-1,-2)$$.

Option A gives $$\left(\dfrac13,-\dfrac83\right)-(-1,-2)=\left(\dfrac13+1,\;-\dfrac83+2\right)=\left(\dfrac43,\;-\dfrac23\right).$$

This is precisely $$\left( \dfrac43,\;-\dfrac23 \right)=\left(2\cdot\dfrac23,\;-\dfrac23\right)=\bigl(2b,\,-b\bigr)$$ if we take

$$b=\dfrac23.$$

Because a consistent positive value of $$b$$ has been found, Option A indeed represents a vertex. (With $$b=\dfrac23$$ we have $$a=-5b=-\dfrac{10}{3},$$ which still satisfies $$|\vec v|=|\vec w|.$$)

No other option produces a vector proportional to $$(2,-1),$$ so no other choice can be a vertex.

The coordinate we obtained is therefore exactly the point in Option A, namely $$\left(\dfrac13,\,-\dfrac83\right).$$

Hence, the correct answer is Option A.