Question

If a straight line is perpendicular to \( 2x + 8y = 10 \) and meets the x-axis at \( (5,0) \), then it meets the y-axis at

• A. \( (0,-2) \)
• B. \( (0,-8) \)
• C. \( (0,-10) \)
• D. \( (0,-16) \)
• E. \( (0,-20) \)

Hint:

For perpendicular lines, just take negative reciprocal of slope — fastest method.

Answer 1

The Correct Option is C

Concept:
• Slope of line \( ax + by + c = 0 \) is \( -\frac{a}{b} \)
• Slopes of perpendicular lines satisfy: \[ m_1 \cdot m_2 = -1 \]
• Equation of line using point-slope form: \[ y - y_1 = m(x - x_1) \]

Step 1: Finding slope of given line.
Given: \[ 2x + 8y = 10 \] \[ 8y = -2x + 10 \Rightarrow y = -\frac{1}{4}x + \frac{5}{4} \] So slope: \[ m_1 = -\frac{1}{4} \]

Step 2: Finding slope of required line.
\[ m_2 = 4 \quad (\text{since } m_1 \cdot m_2 = -1) \]

Step 3: Equation of required line.
Passing through \( (5,0) \): \[ y - 0 = 4(x - 5) \] \[ y = 4x - 20 \]

Step 4: Finding y-intercept.
Put \( x = 0 \): \[ y = -20 \] Thus, point is \( (0,-20) \)

Step 5: Final Answer.
\[ \boxed{(0,-20)} \]