The straight lines represented by `(y-mx)^(2)=a^(2)(1+m^(2))and(y-nx)^(2)=a^(2)(1+n^(2))`form a

Options

A. rectangle
B. rhombus
C. trepezium
D. None of these

Explanation

To solve the problem, we need to analyze the equations of the straight lines given by:

1. \((y - mx)^2 = a^2(1 + m^2)\)
2. \((y - nx)^2 = a^2(1 + n^2)\)

### Step 1: Rewrite the equations in slope-intercept form

Starting with the first equation:

\[ (y - mx)^2 = a^2(1 + m^2) \]

Taking the square root of both sides, we get:

\[ y - mx = \pm a\sqrt{1 + m^2} \]

This gives us two equations:

\[ y = mx + a\sqrt{1 + m^2} \quad \text{(1)} \]
\[ y = mx - a\sqrt{1 + m^2} \quad \text{(2)} \]

### Step 2: Rewrite the second equation similarly

Now, for the second equation:

\[ (y - nx)^2 = a^2(1 + n^2) \]

Taking the square root of both sides, we have:

\[ y - nx = \pm a\sqrt{1 + n^2} \]

This gives us two more equations:

\[ y = nx + a\sqrt{1 + n^2} \quad \text{(3)} \]
\[ y = nx - a\sqrt{1 + n^2} \quad \text{(4)} \]

### Step 3: Identify the lines formed

From equations (1), (2), (3), and (4), we see that we have four lines:

1. \(y = mx + a\sqrt{1 + m^2}\)
2. \(y = mx - a\sqrt{1 + m^2}\)
3. \(y = nx + a\sqrt{1 + n^2}\)
4. \(y = nx - a\sqrt{1 + n^2}\)

### Step 4: Determine the nature of the lines

The lines represented by these equations can be analyzed based on their slopes and intercepts. The slopes are \(m\) and \(n\) for the first and second pairs of lines respectively. The vertical shifts are determined by \(a\sqrt{1 + m^2}\) and \(a\sqrt{1 + n^2}\).

### Step 5: Conclusion about the shape formed

Since we have two pairs of lines with different slopes and equal vertical distances between the pairs, these lines will intersect and form a rhombus.

### Final Answer

The straight lines represented by \((y - mx)^2 = a^2(1 + m^2)\) and \((y - nx)^2 = a^2(1 + n^2)\) form a rhombus.

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