Question

A straight line makes positive intercepts on the coordinate axes whose sum is 5. If the line passes through the point \( P(-3,4) \) then the equation of the line is

• A. \( 2x - y + 10 = 0 \)
• B. \( 2x + 3y = 6 \)
• C. \( 3x + 2y = 6 \)
• D. \( x + 4y = 13 \)

Hint:

For intercept problems, always use \( \frac{x}{a}+\frac{y}{b}=1 \) and substitute given pointThen use given relation between \( a \) and \( b \).

Answer 1

The Correct Option is B

Step 1: Use intercept form of line.
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
where \( a \) and \( b \) are positive intercepts.

Step 2: Use given condition.
\[ a + b = 5 \]

Step 3: Use point \( (-3,4) \).
\[ \frac{-3}{a} + \frac{4}{b} = 1 \]

Step 4: Express \( b \) in terms of \( a \).
\[ b = 5 - a \]
Substitute into equation:
\[ \frac{-3}{a} + \frac{4}{5-a} = 1 \]

Step 5: Solve equation.
Multiply by \( a(5-a) \):
\[ -3(5-a) + 4a = a(5-a) \]
\[ -15 + 3a + 4a = 5a - a^2 \]
\[ -15 + 7a = 5a - a^2 \]

Step 6: Rearrange.
\[ a^2 + 2a - 15 = 0 \]
\[ (a+5)(a-3)=0 \]
\[ a=3 \; (\text{positive}) \]
\[ b=5-a=2 \]

Step 7: Final equation.
\[ \frac{x}{3} + \frac{y}{2} = 1 \]
Multiply by 6:
\[ 2x + 3y = 6 \]
\[ \boxed{2x + 3y = 6} \]