The equations of two equal sides AB and AC of an isosceles triangle ABC are x + y = 5 and 7x - y = 3 respectively. What will be the equation of the side BC if area of triangle ABC is 5 square units.

Options

A. x + 3y - 1 = 0
B. x - 3y + 1 = 0
C. 2x - y = 5
D. x + 2y = 5

Answer

To find the equation of side BC of triangle ABC, given the equations of the equal sides AB and AC, and the area of the triangle, we can follow these steps:

### Step 1: Identify the equations of the lines
The equations of the equal sides are given as:
1. \( AB: x + y = 5 \)
2. \( AC: 7x - y = 3 \)

### Step 2: Convert the equations into slope-intercept form
We can rewrite these equations in the form \( y = mx + b \) to find their slopes.

For \( AB: x + y = 5 \):
\[ y = -x + 5 \]
Here, the slope \( m_1 = -1 \).

For \( AC: 7x - y = 3 \):
\[ y = 7x - 3 \]
Here, the slope \( m_2 = 7 \).

### Step 3: Find the coordinates of points A, B, and C
To find the vertices of the triangle, we need to find the intersection of the two lines.

Setting \( x + y = 5 \) equal to \( 7x - y = 3 \):
1. From \( x + y = 5 \), we have \( y = 5 - x \).
2. Substitute \( y \) into the second equation:
\[ 7x - (5 - x) = 3 \\ 7x - 5 + x = 3 \\ 8x - 5 = 3 \\ 8x = 8 \\ x = 1 \]
3. Substitute \( x = 1 \) back into \( y = 5 - x \):
\[ y = 5 - 1 = 4 \]
Thus, point A is \( (1, 4) \).

### Step 4: Find points B and C
To find points B and C, we can find the intersection of the lines with the x-axis (where \( y = 0 \)).

For line \( AB: x + y = 5 \):
\[ x + 0 = 5 \implies x = 5 \\ \text{Point B: } (5, 0) \]

For line \( AC: 7x - y = 3 \):
\[ 7x - 0 = 3 \implies x = \frac{3}{7} \\ \text{Point C: } \left(\frac{3}{7}, 0\right) \]

### Step 5: Calculate the area of triangle ABC
The area \( A \) of triangle ABC can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base BC can be calculated as the distance between points B and C.

Distance \( BC \):
\[ BC = 5 - \frac{3}{7} = \frac{35}{7} - \frac{3}{7} = \frac{32}{7} \]

The height from point A to line BC can be calculated as the y-coordinate of point A, which is 4.

Thus, the area is:
\[ A = \frac{1}{2} \times \frac{32}{7} \times 4 = \frac{64}{7} \]

### Step 6: Set the area equal to 5 square units
We know the area is given as 5 square units:
\[ \frac{64}{7} = 5 \implies \text{This is not correct.} \]

### Step 7: Find the correct equation of line BC
Since we need the area to be 5 square units, we can set up the equation:
\[ \frac{1}{2} \times \text{base} \times \text{height} = 5 \]
Let the height be \( h \):
\[ \frac{1}{2} \times \frac{32}{7} \times h = 5 \]
\[ h = \frac{5 \times 2 \times 7}{32} = \frac{70}{32} = \frac{35}{16} \]

### Step 8: Find the equation of line BC
The slope of line BC can be calculated using points B and C:
\[ \text{slope} = \frac{0 - 4}{\frac{3}{7} - 5} = \frac{-4}{\frac{3 - 35}{7}} = \frac{-4 \times 7}{-32} = \frac{28}{32} = \frac{7}{8} \]

Using point-slope form to find the equation of line BC:
\[ y - 0 = \frac{7}{8}(x - 5) \]
This simplifies to:
\[ y = \frac{7}{8}x - \frac{35}{8} \]

### Final Answer
The equation of side BC is:
\[ y = \frac{7}{8}x - \frac{35}{8} \]