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To find the fourth line that intersects to form a square, we need to ensure that the lines are perpendicular to each other and form a square. The given equations are:

1. \( y = -\frac{4}{15}x \)
2. \( y = \frac{15}{4}x \)
3. \( y = \frac{15}{4}x - \frac{241}{2} \)

We know that for a square, the slopes of the lines must form right angles. Lines that are perpendicular have slopes that are negative reciprocals of each other. The first and second lines given are:

- Slope of \( y = -\frac{4}{15}x \) is \(-\frac{4}{15}\).
- Slope of \( y = \frac{15}{4}x \) is \(\frac{15}{4}\).

These slopes are negative reciprocals of each other, so they are perpendicular.

The third line \( y = \frac{15}{4}x - \frac{241}{2} \) has the same slope as the second line (\(\frac{15}{4}\)), indicating it is parallel to the second line. For a square, the fourth line must be perpendicular to the third line and parallel to the first line.

Since the fourth line must be parallel to the line \( y = -\frac{4}{15}x \), it must have the same slope. The equation of the fourth line will thus be of the form:

\[ y = -\frac{4}{15}x + c \]

To find the specific value of \( c \), we need to find a point of intersection between the third and fourth lines. Notice the third line intersects the first line:

1. Third line: \( y = \frac{15}{4}x - \frac{241}{2} \)
2. Fourth line: \( y = -\frac{4}{15}x + c \)

By equating the y-values of the third and fourth lines and solving for \( x \) and \( c \):

\[ \frac{15}{4}x - \frac{241}{2} = -\frac{4}{15}x + c \]

Let's solve for \( x \) and \( c \):

First, solve for \( x \):

\[ \frac{15}{4}x + \frac{4}{15}x = c + \frac{241}{2} \]

Multiply through by 60 (LCM of 4 and 15) to clear the fractions:

\[ 15 \cdot 15x + 4 \cdot 4x = 60c + 60 \cdot \frac{241}{2} \]
\[ 225x + 16x = 60c + 30 \cdot 241 \]
\[ 241x = 60c + 7230 \]

Now we need to find the value of \( x \) using the intersection point with one of the given lines. Since the exact intersection points are not given, we'll assume that the lines intersect at points forming a square.

Since the lines are already forming perpendiculars and we know the slopes, we can deduce that the fourth line must intersect at the same point but parallel to \( y = -\frac{4}{15}x \).

Hence, the fourth line is:

\[ y = -\frac{4}{15}x + k \]

By geometry and assuming the square is correctly aligned, we can leave \( k \) as a constant where lines intersect. Without loss of generality, we keep \( c = k \) from symmetrical properties.

So, the fourth equation is \( y = -\frac{4}{15}x + c \), where \( c \) is to be determined by the actual intersection point values.

Would you like further details or have any specific questions? Here are some questions you might find useful:

1. How do we find the exact value of \( c \) in the fourth line equation?
2. How can we verify that four lines form a perfect square?
3. What are the properties of lines that form squares?
4. Can we solve similar problems using geometric transformations?
5. How do negative reciprocal slopes help in determining perpendicular lines?

**Tip:** Remember that perpendicular lines have slopes that are negative reciprocals of each other. This can be particularly useful when determining angles in geometric shapes such as squares and rectangles.

Learn how to determine the fourth equation that forms a square with three given lines in a coordinate plane. This problem involves understanding perpendicular lines and equations of lines. Suitable for advanced high school students and above.

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