Find the equation of the straight line which passes through the point (3,4) and has intercepts on the axes such that their sum is 14 :

Options

A. a. 4x+3y=24
B. b. x+y=7
C. c. 3x+7y=43
D. d. both (a) and (b)

Answer

To find the equation of the straight line that passes through the point (3, 4) and has intercepts on the axes such that their sum is 14, we can follow these steps:

### Step 1: Understand the intercept form of the equation of a line
The equation of a line in terms of its x-intercept \( a \) and y-intercept \( b \) is given by:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]

### Step 2: Set up the condition for the intercepts
According to the problem, the sum of the intercepts is 14:
\[ a + b = 14 \]

### Step 3: Substitute \( b \) in terms of \( a \)
From the equation \( a + b = 14 \), we can express \( b \) as:
\[ b = 14 - a \]

### Step 4: Substitute \( b \) into the line equation
Now, substituting \( b \) into the line equation gives:
\[ \frac{x}{a} + \frac{y}{14 - a} = 1 \]

### Step 5: Substitute the point (3, 4) into the equation
Since the line passes through the point (3, 4), we can substitute \( x = 3 \) and \( y = 4 \) into the equation:
\[ \frac{3}{a} + \frac{4}{14 - a} = 1 \]

### Step 6: Clear the fractions by multiplying through by \( a(14 - a) \)
Multiplying through by \( a(14 - a) \) gives:
\[ 3(14 - a) + 4a = a(14 - a) \]

### Step 7: Expand and rearrange the equation
Expanding both sides:
\[ 42 - 3a + 4a = 14a - a^2 \]
This simplifies to:
\[ 42 + a = 14a - a^2 \]
Rearranging gives:
\[ a^2 - 13a + 42 = 0 \]

### Step 8: Solve the quadratic equation
We can solve the quadratic equation \( a^2 - 13a + 42 = 0 \) using the factorization method:
\[ (a - 6)(a - 7) = 0 \]
Thus, we have two possible values for \( a \):
\[ a = 6 \quad \text{or} \quad a = 7 \]

### Step 9: Find corresponding values of \( b \)
Using \( b = 14 - a \):
- If \( a = 6 \), then \( b = 14 - 6 = 8 \).
- If \( a = 7 \), then \( b = 14 - 7 = 7 \).

### Step 10: Write the equations of the lines
Now we can write the equations of the lines:
1. For \( a = 6 \) and \( b = 8 \):
\[ \frac{x}{6} + \frac{y}{8} = 1 \quad \Rightarrow \quad 4x + 3y = 24 \]

2. For \( a = 7 \) and \( b = 7 \):
\[ \frac{x}{7} + \frac{y}{7} = 1 \quad \Rightarrow \quad x + y = 7 \]

### Final Result
The equations of the lines are:
1. \( 4x + 3y = 24 \)
2. \( x + y = 7 \)