Two sides of a parallelogram are along the lines 4x + 5y = 0 and 7x + 2y = 0. If the equation of one

Q: Wait, why do we assume the intersection of the two sides is the origin? Is it always true?

Great question! Look at the general form of a line: $ax + by + c = 0$. If $c = 0$, the line passes through the origin. Since both your given side equations have $c = 0$, they are forced to meet at $(0, 0)$. It is not an assumption; it is a geometric necessity!

Q: Why can't we just use the slope of the sides to find the other diagonal?

You could, but that is a much longer path involving finding the equations of the other two sides first. The 'midpoint property' is the most elegant and direct route. In JEE, efficiency is just as important as accuracy.

Q: I got confused with the fractions when solving for the vertices. Is there a trick?

Fractions are just numbers in disguise! Don't let them intimidate you. When you have $11x + 7(-\frac{4x}{5}) = 9$, just multiply the entire equation by $5$ to clear the denominator immediately. It makes the algebra much cleaner and reduces the chance of a silly sign error.

Question

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:

Options:

(A) (1, 2)

(B) (2, 2)

(C) (2, 1)

(D) (1, 3)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to peel back the layers of a classic coordinate geometry problem. We are given two sides of a parallelogram: $4x + 5y = 0$ and $7x + 2y = 0$.

At first glance, these are just lines. However, notice that the constant term $c$ is missing in the general form $ax + by + c = 0$.

Because $c = 0$ for both equations, both lines are anchored to the origin. This is our first breakthrough: both lines pass through the point $A(0, 0)$, which serves as our anchor vertex.

### The Diagonal Mystery

Now, we are given a diagonal: $11x + 7y = 9$. The immediate question is: does this diagonal pass through our anchor point $A(0, 0)$?

If we substitute $x=0$ and $y=0$ into the diagonal equation, we get $0 = 9$, which is a mathematical impossibility. This is a moment of clarity.

Since the diagonal does not pass through $A$, it must connect the other two vertices, $B$ and $D$. We have successfully identified the role of this line as the bridge between the two vertices that are not the origin.

### The Intersection Dance

To find the coordinates of vertex $D$, we find where the side $4x + 5y = 0$ meets the diagonal $11x + 7y = 9$. We express $y$ as $y = -\frac{4x}{5}$ and substitute it into the diagonal equation:

$$11x + 7\left(-\frac{4x}{5}\right) = 9$$

Multiplying the entire equation by $5$ yields:

$$55x - 28x = 45$$

$$27x = 45 \Rightarrow x = \frac{5}{3}$$

Substituting $x = \frac{5}{3}$ back into the side equation, we find $y = -\frac{4}{3}$. Thus, vertex $D$ is located at $(\frac{5}{3}, -\frac{4}{3})$.

We repeat this exact logic for vertex $B$ using the side $7x + 2y = 0$. Substituting $y = -\frac{7x}{2}$ into the diagonal equation:

$$11x + 7\left(-\frac{7x}{2}\right) = 9$$

This simplifies to:

$$22x - 49x = 18 \Rightarrow -27x = 18 \Rightarrow x = -\frac{2}{3}$$

Substituting $x = -\frac{2}{3}$ back, we find $y = \frac{7}{3}$. Therefore, vertex $B$ is $(-\frac{2}{3}, \frac{7}{3})$.

### The Midpoint Magic

Here is where the beauty of geometry shines. A parallelogram is a perfectly balanced machine where the diagonals bisect each other. This means the midpoint $M$ of diagonal $BD$ is the same as the midpoint of the other diagonal $AC$.

We calculate the midpoint $M$ of $BD$ by averaging the coordinates:

$$M = \left( \frac{-\frac{2}{3} + \frac{5}{3}}{2}, \frac{\frac{7}{3} - \frac{4}{3}}{2} \right)$$

This simplifies beautifully to $M = (\frac{1}{2}, \frac{1}{2})$. The symmetry is satisfying, isn't it?

### The Final Line

Finally, we need the equation of the other diagonal. We know it passes through the origin $A(0, 0)$ and the midpoint $M(\frac{1}{2}, \frac{1}{2})$.

The slope $m$ is calculated as:

$$m = \frac{\frac{1}{2} - 0}{\frac{1}{2} - 0} = 1$$

The equation of the diagonal is simply **$y = x$**. Checking our options, any point where the coordinates are equal, such as $(2, 2)$, satisfies this perfectly. You have just navigated the logic of a parallelogram with precision.

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

Select Answer:

Visualized Solution

Visualizing the Given Sides

Analyzing the Given Diagonal

Setting up Intersection for Vertex $D$

Calculating Coordinates of $D$

Setting up Intersection for Vertex $B$

Calculating Coordinates of $B$

The Midpoint Property

Calculating Midpoint $M$

Equation of the Second Diagonal

Final Verification

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Diagonal Mystery

The Intersection Dance

The Midpoint Magic

The Final Line

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Select Answer:

Visualized Solution

Visualizing the Given Sides

Analyzing the Given Diagonal

Setting up Intersection for Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Calculating Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Setting up Intersection for Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Calculating Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B