A straight line makes on the co-ordinate axes positive intercepts whose sum is 5. If the line passes through the point P(-3, 4), find its equation.

Answer

To find the equation of the straight line that makes positive intercepts on the coordinate axes with a sum of 5 and passes through the point P(-3, 4), we can follow these steps:

### Step 1: Define the intercepts
Let the x-intercept be \( a \) and the y-intercept be \( b \). According to the problem, we know:
\[ a + b = 5 \]

### Step 2: Express one intercept in terms of the other
From the equation \( a + b = 5 \), we can express \( a \) in terms of \( b \):
\[ a = 5 - b \]

### Step 3: Write the equation of the line in intercept form
The equation of a line in intercept form is given by:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
Substituting \( a = 5 - b \) into this equation gives:
\[ \frac{x}{5 - b} + \frac{y}{b} = 1 \]

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

### Step 5: Clear the fractions
To eliminate the fractions, we can multiply through by \( b(5 - b) \):
\[ -3b + 4(5 - b) = b(5 - b) \]
Expanding both sides:
\[ -3b + 20 - 4b = 5b - b^2 \]
Combining like terms gives:
\[ 20 - 7b = 5b - b^2 \]

### Step 6: Rearranging the equation
Rearranging the equation to bring all terms to one side:
\[ b^2 - 12b + 20 = 0 \]

### Step 7: Factor the quadratic equation
We need to factor the quadratic:
\[ (b - 10)(b - 2) = 0 \]
This gives us two possible values for \( b \):
\[ b = 10 \quad \text{or} \quad b = 2 \]

### Step 8: Find corresponding values of \( a \)
Using \( a = 5 - b \):
1. If \( b = 10 \):
\[ a = 5 - 10 = -5 \]
2. If \( b = 2 \):
\[ a = 5 - 2 = 3 \]

### Step 9: Write the equations of the lines
Now we can write the equations of the lines using the intercepts:
1. For \( a = -5 \) and \( b = 10 \):
\[ \frac{x}{-5} + \frac{y}{10} = 1 \implies -2x + y = 10 \implies y = 2x + 10 \]
2. For \( a = 3 \) and \( b = 2 \):
\[ \frac{x}{3} + \frac{y}{2} = 1 \implies 2x + 3y = 6 \]

### Final Equations
The two equations of the lines are:
1. \( y = 2x + 10 \)
2. \( 2x + 3y = 6 \)