Question:
The equation of a line is \(3x + 4y – 12 = 0\). What is the slope of a line perpendicular to it?
एक रेखा का समीकरण \(3x + 4y – 12 = 0\) है। इसके लंबवत (perpendicular) रेखा का ढाल (slope) क्या है?
Options:
- \(\dfrac{4}{3}\)
- \(-\dfrac{3}{4}\)
- \(\dfrac{3}{4}\)
- \(-\dfrac{4}{3}\)
- \(\dfrac{4}{3}\)
- \(-\dfrac{3}{4}\)
- \(\dfrac{3}{4}\)
- \(-\dfrac{4}{3}\)
Correct Answer: a)
Explanation:
The slope of the line \(Ax + By + C = 0\) is \(m_1 = -\dfrac{A}{B} = -\dfrac{3}{4}\).
The slope of a perpendicular line \(m_2\) satisfies \(m_1 \cdot m_2 = -1\).
\(m_2 = -\dfrac{1}{m_1} = -\dfrac{1}{-3/4} = \dfrac{4}{3}\).
रेखा \(3x + 4y – 12 = 0\) का ढाल \(m_1 = -\dfrac{3}{4}\) है।
लंबवत रेखा का ढाल \(m_2 = -\dfrac{1}{m_1} = \dfrac{4}{3}\) होता है।