Question

If the straight line \( 5x + y = k \) forms a triangle with the coordinate axes of area \(10\) sq. units, then the values of \(k\) are

• A. \( \pm 15 \)
• B. \( \pm 10 \)
• C. \( \pm 5 \)
• D. \( \pm 20 \)
• E. \( \pm 7 \)

Hint:

For intercept form problems, directly use \(\text{Area} = \frac{1}{2} \times \text{x-intercept} \times \text{y-intercept}\) — it saves time.

Answer 1

The Correct Option is B

Concept: A straight line forms a triangle with coordinate axes by cutting intercepts on \(x\)-axis and \(y\)-axis. For a line: \[ Ax + By = C \] Intercepts are: \[ x\text{-intercept} = \frac{C}{A}, \quad y\text{-intercept} = \frac{C}{B} \] Area of triangle formed with axes: \[ \text{Area} = \frac{1}{2} \times (\text{x-intercept}) \times (\text{y-intercept}) \]

Step 1: Write given line
\[ 5x + y = k \]

Step 2: Find intercepts
For \(x\)-intercept, put \(y=0\): \[ 5x = k \Rightarrow x = \frac{k}{5} \] For \(y\)-intercept, put \(x=0\): \[ y = k \] Thus: \[ x\text{-intercept} = \frac{k}{5}, \quad y\text{-intercept} = k \]

Step 3: Apply area formula
\[ \text{Area} = \frac{1}{2} \times \frac{k}{5} \times k \] \[ \text{Area} = \frac{k^2}{10} \]

Step 4: Use given condition
\[ \frac{k^2}{10} = 10 \]

Step 5: Solve equation
\[ k^2 = 100 \] \[ k = \pm 10 \]

Step 6: Interpretation
Both positive and negative values are valid because the line can cut intercepts in different quadrants, but area remains positive.

Step 7: Final Answer
\[ \boxed{\pm 10} \]