Two sides of a parallelogram are along the lines $$4x + 5y = 0$$ and $$7x + 2y = 0$$. If the equation of one of the diagonals of the parallelogram is $$11x + 7y = 9$$, then other diagonal passes through the point:

We are told that two adjacent sides of a parallelogram lie along the straight lines $$4x + 5y = 0$$ and $$7x + 2y = 0$$. Because the two lines intersect, their point of intersection is one vertex of the parallelogram.

Solving the simultaneous equations $$4x + 5y = 0 \quad\text{and}\quad 7x + 2y = 0$$ we first express one variable in terms of the other from the first equation:

$$4x + 5y = 0 \;\Longrightarrow\; 5y = -4x \;\Longrightarrow\; y = -\dfrac{4x}{5}.$$

Substituting this value of $$y$$ in the second equation gives

$$7x + 2\!\left(-\dfrac{4x}{5}\right)=0 \;\Longrightarrow\; 7x - \dfrac{8x}{5}=0 \;\Longrightarrow\; \dfrac{35x - 8x}{5}=0 \;\Longrightarrow\; \dfrac{27x}{5}=0 \;\Longrightarrow\; x = 0.$$

Putting $$x = 0$$ into $$y = -\dfrac{4x}{5}$$ yields $$y = 0$$. So the two lines meet at the origin $$O(0,0)$$, and we take this as the first vertex of the parallelogram.

Next, we need direction vectors for the two given sides. For $$4x + 5y = 0$$, set $$x = t$$; then $$y = -\dfrac{4}{5}t$$, giving the direction vector $$(5,-4).$$ For $$7x + 2y = 0$$, set $$x = s$$; then $$y = -\dfrac{7}{2}s$$, giving the direction vector $$(2,-7).$$

Let the adjacent sides from the origin be

$$\overrightarrow{OA} = a\,(5,-4)\quad\text{and}\quad \overrightarrow{OB} = b\,(2,-7),$$

where $$a\neq 0,\, b\neq 0.$$ Thus the coordinates of the next two vertices are

$$A(5a,\,-4a), \qquad B(2b,\,-7b).$$

The fourth vertex $$C$$ is obtained by the parallelogram law:

$$\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{OB},$$ so $$C\bigl(5a+2b,\,-4a-7b\bigr).$$

The two diagonals are $$\overline{OC}$$ and $$\overline{AB}$$. We are told that one diagonal satisfies $$11x + 7y = 9$$. Notice that $$O(0,0)$$ does not satisfy this equation because $$11\cdot0 + 7\cdot0 = 0 \neq 9$$. Therefore the diagonal that lies on $$11x + 7y = 9$$ cannot be $$\overline{OC}$$ (which passes through the origin); it must be the other diagonal $$\overline{AB}$$.

Because every point on $$\overline{AB}$$ satisfies $$11x + 7y = 9$$, both end-points $$A$$ and $$B$$ must satisfy it. So we substitute the coordinates of $$A$$ and $$B$$ into the equation.

For $$A(5a,\,-4a):$$ $$11(5a) + 7(-4a) = 55a - 28a = 27a.$$ Hence $$27a = 9 \;\Longrightarrow\; a = \dfrac{1}{3}.$$

For $$B(2b,\,-7b):$$ $$11(2b) + 7(-7b) = 22b - 49b = -27b.$$ Hence $$-27b = 9 \;\Longrightarrow\; b = -\dfrac{1}{3}.$$

With these values we now have explicit coordinates for all the vertices:

$$A\!\left(5\!\left(\dfrac13\right),\,-4\!\left(\dfrac13\right)\right)=\left(\dfrac{5}{3},\,-\dfrac{4}{3}\right),$$ $$B\!\left(2\!\left(-\dfrac13\right),\,-7\!\left(-\dfrac13\right)\right)=\left(-\dfrac{2}{3},\;\dfrac{7}{3}\right),$$ $$C = A + B = \left(\dfrac{5}{3}-\dfrac{2}{3},\,-\dfrac{4}{3}+\dfrac{7}{3}\right)=(1,1).$$

Now consider the other diagonal $$\overline{OC}$$ connecting $$O(0,0)$$ and $$C(1,1).$$ The two-point form of the equation of a straight line is $$(y - y_1) = \dfrac{y_2 - y_1}{x_2 - x_1}\,(x - x_1).$$ Using $$(x_1,y_1)=(0,0)$$ and $$(x_2,y_2)=(1,1)$$, we get

$$y - 0 = \dfrac{1 - 0}{1 - 0}\,(x - 0) \;\Longrightarrow\; y = x.$$

Therefore every point on the second diagonal satisfies $$y = x$$, or equivalently $$x - y = 0.$

We now check which of the four given points satisfies $$x = y$$:

• $$(1,2):\; 1 \neq 2$$ (reject) • $$(2,2):\; 2 = 2$$ (accept) • $$(2,1):\; 2 \neq 1$$ (reject) • $$(1,3):\; 1 \neq 3$$ (reject)

Only $$(2,2)$$ lies on the required diagonal.

Hence, the correct answer is Option 2.