Question:
A line passes through \(\mathrm{P}(-4,1)\) and meets the co-ordinate axes at points A and B . If P divides the segment AB internally in the ratio \(1: 2\), then the equation of the line is
A line passes through \(\mathrm{P}(-4,1)\) and meets the co-ordinate axes at points A and B . If P divides the segment AB internally in the ratio \(1: 2\), then the equation of the line is
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Let the intercept points be A(a, 0) and B(0, b) and apply the section formula for P.
Updated On: Nov 13, 2026
- \(x-2y+6=0\)
- \(x+10y-6=0\)
- \(2x+y+4=0\)
- \(x-y+5=0\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
Let the line meet the x-axis at \(A(a,0)\) and the y-axis at \(B(0,b)\).
Step 2: Section formula
P divides AB in the ratio \(1:2\), so \(P=\left(\frac{1\cdot0+2a}{3},\frac{1\cdot b+2\cdot0}{3}\right)=\left(\frac{2a}3,\frac b3\right)\).
Step 3: Find intercepts
\(\frac{2a}3=-4\Rightarrow a=-6\) and \(\frac b3=1\Rightarrow b=3\).
Step 4: Equation
\(\frac x{-6}+\frac y3=1\Rightarrow-x+2y=6\Rightarrow x-2y+6=0\), option (A).
Final Answer:
The line is \(x-2y+6=0\). \[ \boxed{\text{(A) }x-2y+6=0} \]
Let the line meet the x-axis at \(A(a,0)\) and the y-axis at \(B(0,b)\).
Step 2: Section formula
P divides AB in the ratio \(1:2\), so \(P=\left(\frac{1\cdot0+2a}{3},\frac{1\cdot b+2\cdot0}{3}\right)=\left(\frac{2a}3,\frac b3\right)\).
Step 3: Find intercepts
\(\frac{2a}3=-4\Rightarrow a=-6\) and \(\frac b3=1\Rightarrow b=3\).
Step 4: Equation
\(\frac x{-6}+\frac y3=1\Rightarrow-x+2y=6\Rightarrow x-2y+6=0\), option (A).
Final Answer:
The line is \(x-2y+6=0\). \[ \boxed{\text{(A) }x-2y+6=0} \]
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