Question:
The equation of the straight line perpendicular to the straight line $3x+2y=0$ and passing through the point of intersection of the lines $x+3y-1=0$ and $x-2y+4=0$ is
The equation of the straight line perpendicular to the straight line $3x+2y=0$ and passing through the point of intersection of the lines $x+3y-1=0$ and $x-2y+4=0$ is
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A line perpendicular to $ax + by + c = 0$ is of the form $bx - ay + k = 0$.
Updated On: Apr 10, 2026
- $2x-3y+1=0$
- $2x-3y+3=0$
- $2x-3y+5=0$
- $2x-3y+7=0$
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The Correct Option is D
Solution and Explanation
Step 1: Point of Intersection
Solve $x + 3y = 1$ and $x - 2y = -4$.
Subtracting: $5y = 5 \Rightarrow y = 1$.
$x + 3(1) = 1 \Rightarrow x = -2$. The point is $(-2, 1)$.
Step 2: Line Slope
The given line is $3x + 2y = 0$ with slope $m_1 = -3/2$.
The perpendicular line has slope $m_2 = -1/m_1 = 2/3$.
Step 3: Line Equation
Equation of line: $y - 1 = \frac{2}{3}(x + 2)$.
$3y - 3 = 2x + 4 \Rightarrow 2x - 3y + 7 = 0$.
Final Answer: (d)
Solve $x + 3y = 1$ and $x - 2y = -4$.
Subtracting: $5y = 5 \Rightarrow y = 1$.
$x + 3(1) = 1 \Rightarrow x = -2$. The point is $(-2, 1)$.
Step 2: Line Slope
The given line is $3x + 2y = 0$ with slope $m_1 = -3/2$.
The perpendicular line has slope $m_2 = -1/m_1 = 2/3$.
Step 3: Line Equation
Equation of line: $y - 1 = \frac{2}{3}(x + 2)$.
$3y - 3 = 2x + 4 \Rightarrow 2x - 3y + 7 = 0$.
Final Answer: (d)
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