A straight line `x/a -y/b =1` passes through the point (8,6) and cuts off a triangle of area 12 units from the axes of coordinates . Find the equations of the straight line .

Options

A. 3x-2y =12
B. 4x-3y =12
C. 3x-8y+4=0
D. both (a) and (c)

Answer

To solve the problem step by step, we need to find the equations of the straight line given that it passes through the point (8, 6) and cuts off a triangle of area 12 units from the axes of coordinates. The equation of the line is given in the form \( \frac{x}{a} - \frac{y}{b} = 1 \).

### Step 1: Understand the area of the triangle
The area \( A \) of a triangle formed by the x-axis and y-axis intercepts can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is \( a \) (x-intercept) and the height is \( b \) (y-intercept). Given that the area is 12, we can write:
\[ \frac{1}{2} \times a \times b = 12 \]
This simplifies to:
\[ a \times b = 24 \quad \text{(Equation 1)} \]

### Step 2: Substitute the point (8, 6) into the line equation
The line passes through the point (8, 6). Substituting \( x = 8 \) and \( y = 6 \) into the line equation \( \frac{x}{a} - \frac{y}{b} = 1 \):
\[ \frac{8}{a} - \frac{6}{b} = 1 \quad \text{(Equation 2)} \]

### Step 3: Rearranging Equation 2
Rearranging Equation 2 gives:
\[ \frac{8}{a} - 1 = \frac{6}{b} \]
This can be rewritten as:
\[ \frac{8 - a}{a} = \frac{6}{b} \]
Cross-multiplying gives:
\[ b(8 - a) = 6a \]
This simplifies to:
\[ 8b - ab = 6a \quad \text{(Equation 3)} \]

### Step 4: Substitute \( b \) from Equation 1 into Equation 3
From Equation 1, we have \( b = \frac{24}{a} \). Substituting this into Equation 3:
\[ 8\left(\frac{24}{a}\right) - a\left(\frac{24}{a}\right) = 6a \]
This simplifies to:
\[ \frac{192}{a} - 24 = 6a \]
Multiplying through by \( a \) to eliminate the fraction:
\[ 192 - 24a = 6a^2 \]
Rearranging gives:
\[ 6a^2 + 24a - 192 = 0 \]
Dividing the entire equation by 6:
\[ a^2 + 4a - 32 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( A = 1, B = 4, C = -32 \):
\[ a = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1} \]
\[ = \frac{-4 \pm \sqrt{16 + 128}}{2} \]
\[ = \frac{-4 \pm \sqrt{144}}{2} \]
\[ = \frac{-4 \pm 12}{2} \]
This gives two possible values for \( a \):
1. \( a = \frac{8}{2} = 4 \)
2. \( a = \frac{-16}{2} = -8 \)

### Step 6: Find corresponding \( b \) values
Using \( a = 4 \):
\[ b = \frac{24}{4} = 6 \]
Using \( a = -8 \):
\[ b = \frac{24}{-8} = -3 \]

### Step 7: Write the equations of the lines
1. For \( a = 4 \) and \( b = 6 \):
\[ \frac{x}{4} - \frac{y}{6} = 1 \implies 6x - 4y = 24 \implies 3x - 2y = 12 \]

2. For \( a = -8 \) and \( b = -3 \):
\[ \frac{x}{-8} - \frac{y}{-3} = 1 \implies -3x + 8y = 24 \implies 3x - 8y = -24 \]

### Final Equations
The equations of the straight lines are:
1. \( 3x - 2y = 12 \)
2. \( 3x - 8y = -24 \)