The equation of the diagonals of a rectangle are `y+8x-17=0` and `y-8x+7=0`. If the area of the rectangle is `8 sq units` then find the sides of the rectangle

Options

A. `x = 1`
B. `x +y = 1`
C. `y = 9`
D. `x - 2y = 3`

Explanation

To solve the problem, we need to find the sides of a rectangle given the equations of its diagonals and the area. Let's go through the solution step by step.

### Step 1: Find the point of intersection of the diagonals

The equations of the diagonals are:
1. \( y + 8x - 17 = 0 \)
2. \( y - 8x + 7 = 0 \)

To find the point of intersection, we can solve these equations simultaneously.

From the first equation:
\[ y = 17 - 8x \]

Substituting this into the second equation:
\[ 17 - 8x - 8x + 7 = 0 \]
\[ 24 - 16x = 0 \]
\[ 16x = 24 \implies x = \frac{24}{16} = \frac{3}{2} \]

Now substituting \( x = \frac{3}{2} \) back into the first equation to find \( y \):
\[ y = 17 - 8 \left(\frac{3}{2}\right) = 17 - 12 = 5 \]

Thus, the point of intersection \( P \) is \( \left( \frac{3}{2}, 5 \right) \).

### Step 2: Determine the slopes of the diagonals

The slopes of the lines can be found by rearranging the equations into slope-intercept form \( y = mx + c \).

1. For \( y + 8x - 17 = 0 \):
\[ y = -8x + 17 \quad \text{(slope } m_1 = -8\text{)} \]

2. For \( y - 8x + 7 = 0 \):
\[ y = 8x - 7 \quad \text{(slope } m_2 = 8\text{)} \]

### Step 3: Calculate the angle between the diagonals

The angle \( \theta \) between the two lines can be calculated using the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting the slopes:
\[ \tan \theta = \left| \frac{-8 - 8}{1 + (-8)(8)} \right| = \left| \frac{-16}{1 - 64} \right| = \left| \frac{-16}{-63} \right| = \frac{16}{63} \]

### Step 4: Relate the sides of the rectangle to the area

Let the lengths of the sides of the rectangle be \( a \) and \( b \). The area \( A \) of the rectangle is given by:
\[ A = a \cdot b = 8 \text{ square units} \]

### Step 5: Use the relationship with the angle

From the geometry of the rectangle, we know:
\[ \tan \theta = \frac{a/2}{b/2} = \frac{a}{b} \]
Thus,
\[ \frac{a}{b} = \frac{16}{63} \implies a = \frac{16}{63}b \]

### Step 6: Substitute into the area equation

Substituting \( a \) into the area equation:
\[ \left(\frac{16}{63}b\right) b = 8 \]
\[ \frac{16}{63}b^2 = 8 \]
\[ b^2 = 8 \cdot \frac{63}{16} = \frac{504}{16} = \frac{63}{2} \]
\[ b = \sqrt{\frac{63}{2}} = \frac{\sqrt{126}}{2} \]

Now substituting back to find \( a \):
\[ a = \frac{16}{63}b = \frac{16}{63} \cdot \frac{\sqrt{126}}{2} = \frac{8\sqrt{126}}{63} \]

### Final Result

Thus, the sides of the rectangle are:
\[ a = \frac{8\sqrt{126}}{63}, \quad b = \frac{\sqrt{126}}{2} \]