Two opposite vertices of a rectangle are (1, 3) and (5, 1). If the equation of a diagonal of this rectangle is y = 2x + c. Then, the value of c is

Options

A. -1
B. -3
C. -4
D. -9

Answer

To find the value of \( c \) in the equation of the diagonal \( y = 2x + c \) of the rectangle with opposite vertices at \( (1, 3) \) and \( (5, 1) \), we can follow these steps:

### Step 1: Find the slope of the diagonal
The slope \( m \) of the line passing through the points \( (1, 3) \) and \( (5, 1) \) can be calculated using the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates:
\[ m = \frac{1 - 3}{5 - 1} = \frac{-2}{4} = -\frac{1}{2} \]

### Step 2: Compare the slopes
The equation of the diagonal is given as \( y = 2x + c \). The slope of this line is \( 2 \). Since the slope of the diagonal calculated from the vertices is \( -\frac{1}{2} \), we need to find the correct diagonal equation.

### Step 3: Identify the correct diagonal equation
Since the rectangle has two diagonals, we can find the equation of the diagonal that connects the points \( (1, 3) \) and \( (5, 1) \) using the point-slope form:
Using point \( (1, 3) \):
\[ y - 3 = -\frac{1}{2}(x - 1) \]
Expanding this:
\[ y - 3 = -\frac{1}{2}x + \frac{1}{2} \]
\[ y = -\frac{1}{2}x + \frac{1}{2} + 3 \]
\[ y = -\frac{1}{2}x + \frac{7}{2} \]

### Step 4: Set the two equations equal
Now we have two equations for the diagonals:
1. \( y = 2x + c \)
2. \( y = -\frac{1}{2}x + \frac{7}{2} \)

To find \( c \), we need to find the intersection point of these two lines. We can set them equal to each other:
\[ 2x + c = -\frac{1}{2}x + \frac{7}{2} \]

### Step 5: Solve for \( c \)
Rearranging gives:
\[ 2x + \frac{1}{2}x = \frac{7}{2} - c \]
Combining like terms:
\[ \frac{5}{2}x = \frac{7}{2} - c \]
To find \( c \), we can substitute \( x = 1 \) (the x-coordinate of one of the vertices):
\[ \frac{5}{2}(1) = \frac{7}{2} - c \]
\[ \frac{5}{2} = \frac{7}{2} - c \]
\[ c = \frac{7}{2} - \frac{5}{2} \]
\[ c = 1 \]

### Final Answer
Thus, the value of \( c \) is \( 1 \).

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