Question:
The line on which the lines
\[
ax+by=1
\]
and
\[
bx+ay=1
\]
with \(a\neq 0\neq b\), intersect for any real values of \(a\) and \(b\), is:
The line on which the lines
\[
ax+by=1
\]
and
\[
bx+ay=1
\]
with \(a\neq 0\neq b\), intersect for any real values of \(a\) and \(b\), is:
Show Hint
When two equations are symmetric in \(a,b\) and \(x,y\), subtract them to eliminate the constant terms and obtain the fixed relation between \(x\) and \(y\).
Updated On: Jun 18, 2026
- \(x=-y\)
- \(x=-2y\)
- \(2x=y\)
- \(x=y\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the two given equations.
\[ ax+by=1 \] and \[ bx+ay=1. \]
Step 2: Subtract the second equation from the first equation.
\[ (ax+by)-(bx+ay)=1-1. \] \[ ax-bx+by-ay=0. \] \[ x(a-b)+y(b-a)=0. \] Since \[ b-a=-(a-b), \] we get \[ (a-b)x-(a-b)y=0. \] \[ (a-b)(x-y)=0. \]
Step 3: Use the condition \(a\neq b\).
Since the question states \[ a\neq b, \] we have \[ a-b\neq 0. \] Therefore, \[ x-y=0. \] Hence, \[ x=y. \]
Step 4: Final conclusion.
Thus, the intersection point of the two lines always lies on the line \[ \boxed{x=y} \]
\[ ax+by=1 \] and \[ bx+ay=1. \]
Step 2: Subtract the second equation from the first equation.
\[ (ax+by)-(bx+ay)=1-1. \] \[ ax-bx+by-ay=0. \] \[ x(a-b)+y(b-a)=0. \] Since \[ b-a=-(a-b), \] we get \[ (a-b)x-(a-b)y=0. \] \[ (a-b)(x-y)=0. \]
Step 3: Use the condition \(a\neq b\).
Since the question states \[ a\neq b, \] we have \[ a-b\neq 0. \] Therefore, \[ x-y=0. \] Hence, \[ x=y. \]
Step 4: Final conclusion.
Thus, the intersection point of the two lines always lies on the line \[ \boxed{x=y} \]
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