Question:
A variable line passing through \((l,m)\) intersects the coordinate axes at the points \(A\) and \(B\). If the lines drawn parallel to \(y\)-axis through \(A\) and parallel to \(x\)-axis through \(B\) meet at \(P\), then the locus of \(P\) is
A variable line passing through \((l,m)\) intersects the coordinate axes at the points \(A\) and \(B\). If the lines drawn parallel to \(y\)-axis through \(A\) and parallel to \(x\)-axis through \(B\) meet at \(P\), then the locus of \(P\) is
Show Hint
For a line cutting intercepts \(a\) and \(b\) on the coordinate axes, use the intercept form:
\[
\frac{x}{a}+\frac{y}{b}=1
\]
Then substitute the fixed point through which the line passes.
Updated On: Jun 26, 2026
- \(\dfrac{l}{x}+\dfrac{m}{y}=1\)
- \(\dfrac{x}{l}+\dfrac{y}{m}=1\)
- \(\dfrac{m}{x}+\dfrac{l}{y}=1\)
- \(\dfrac{x}{m}+\dfrac{y}{l}=1\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Assume the intercepts of the variable line.
Let the variable line meet the \(x\)-axis at \[ A=(a,0) \] and the \(y\)-axis at \[ B=(0,b) \]
Step 2: Write the intercept form of the line.
The equation of the line with intercepts \(a\) and \(b\) is \[ \frac{X}{a}+\frac{Y}{b}=1 \]
Step 3: Use the condition that the line passes through \((l,m)\).
Since the line passes through \[ (l,m), \] we substitute \[ X=l,\quad Y=m \] Thus, \[ \frac{l}{a}+\frac{m}{b}=1 \]
Step 4: Find the coordinates of \(P\).
Through \(A=(a,0)\), a line parallel to the \(y\)-axis is \[ X=a \] Through \(B=(0,b)\), a line parallel to the \(x\)-axis is \[ Y=b \] These two lines meet at \[ P=(a,b) \]
Step 5: Replace \(a,b\) by the coordinates of \(P\).
Let \[ P=(x,y) \] Then, \[ a=x,\quad b=y \]
Step 6: Substitute in the relation.
From \[ \frac{l}{a}+\frac{m}{b}=1 \] we get \[ \frac{l}{x}+\frac{m}{y}=1 \]
Step 7: Final conclusion.
Therefore, the locus of \(P\) is \[ \boxed{\frac{l}{x}+\frac{m}{y}=1} \]
Let the variable line meet the \(x\)-axis at \[ A=(a,0) \] and the \(y\)-axis at \[ B=(0,b) \]
Step 2: Write the intercept form of the line.
The equation of the line with intercepts \(a\) and \(b\) is \[ \frac{X}{a}+\frac{Y}{b}=1 \]
Step 3: Use the condition that the line passes through \((l,m)\).
Since the line passes through \[ (l,m), \] we substitute \[ X=l,\quad Y=m \] Thus, \[ \frac{l}{a}+\frac{m}{b}=1 \]
Step 4: Find the coordinates of \(P\).
Through \(A=(a,0)\), a line parallel to the \(y\)-axis is \[ X=a \] Through \(B=(0,b)\), a line parallel to the \(x\)-axis is \[ Y=b \] These two lines meet at \[ P=(a,b) \]
Step 5: Replace \(a,b\) by the coordinates of \(P\).
Let \[ P=(x,y) \] Then, \[ a=x,\quad b=y \]
Step 6: Substitute in the relation.
From \[ \frac{l}{a}+\frac{m}{b}=1 \] we get \[ \frac{l}{x}+\frac{m}{y}=1 \]
Step 7: Final conclusion.
Therefore, the locus of \(P\) is \[ \boxed{\frac{l}{x}+\frac{m}{y}=1} \]
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