Question:medium

The equation of a straight line passes through the point (4,-5) and is perpendicular to the straight line 3x + 4y + 5 = 0.

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A quick shortcut: a line perpendicular to \(Ax + By + C = 0\) will have the form \(Bx - Ay + K = 0\). For the line \(3x + 4y + 5 = 0\), a perpendicular line is \(4x - 3y + K = 0\). Substitute the point (4, -5) to find K: \(4(4) - 3(-5) + K = 0 \Rightarrow 16 + 15 + K = 0 \Rightarrow 31 + K = 0 \Rightarrow K = -31\). So the equation is \(4x - 3y - 31 = 0\).
Updated On: Feb 18, 2026
  • 4x - 3y - 31 = 0
  • 4x - 3y - 1 = 0
  • 4x - 3y + 1 = 0
  • 4x - 3y + 31 = 0
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The Correct Option is A

Solution and Explanation

Step 1: Concept:
Determine the line equation given a point and a perpendicular line. The core concept involves the slope relationship between perpendicular lines.
Step 2: Method:
1. Find the slope (\(m_1\)) of the given line, using \(m = -A/B\) for \(Ax + By + C = 0\).
2. Calculate the perpendicular line's slope (\(m_2\)) using \(m_2 = -1/m_1\).
3. Apply the point-slope form: \(y - y_1 = m_2(x - x_1)\).
4. Convert to the general form: \(Ax + By + C = 0\).
Step 3: Detailed Solution:
1. Find the given line's slope.
Given line: \(3x + 4y + 5 = 0\).
Slope \(m_1 = -\frac{A}{B} = -\frac{3}{4}\).
2. Find the perpendicular line's slope.
Required line's slope, \(m_2\), is the negative reciprocal of \(m_1\). \[ m_2 = -\frac{1}{m_1} = -\frac{1}{(-3/4)} = \frac{4}{3} \] 3. Use the point-slope form.
The line passes through \((x_1, y_1) = (4, -5)\) with \(m_2 = 4/3\). \[ y - y_1 = m_2(x - x_1) \] \[ y - (-5) = \frac{4}{3}(x - 4) \] \[ y + 5 = \frac{4}{3}(x - 4) \] 4. Convert to general form.
Multiply by 3: \[ 3(y + 5) = 4(x - 4) \] \[ 3y + 15 = 4x - 16 \] Rearrange: \[ 0 = 4x - 3y - 16 - 15 \] \[ 4x - 3y - 31 = 0 \] Step 4: Answer:
The line equation is \(4x - 3y - 31 = 0\), which is option (1).
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