Question:
The locus of the incentre of the triangle formed by the lines
\[
xy-4x-4y+16=0
\]
and
\[
x+y=5
\]
is
The locus of the incentre of the triangle formed by the lines
\[
xy-4x-4y+16=0
\]
and
\[
x+y=5
\]
is
Show Hint
If a triangle is symmetric about a line, then important centres such as the incentre, circumcentre and centroid lie on that axis of symmetry.
Updated On: Jun 24, 2026
- \(x-y=0\)
- \(x+y=0\)
- \(x-2y=0\)
- \(2x-y=0\)
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The Correct Option is A
Solution and Explanation
Step 1: Factorize the given equation.
Given, \[ xy-4x-4y+16=0 \] Factorizing, \[ (x-4)(y-4)=0 \] Thus, the equation represents the pair of lines \[ x=4 \] and \[ y=4 \] The third line is \[ x+y=5 \] Hence, the triangle is formed by the lines \[ x=4,\quad y=4,\quad x+y=5 \]
Step 2: Observe the symmetry of the triangle.
The lines \[ x=4 \] and \[ y=4 \] are symmetric with respect to the line \[ x=y \] Also, the line \[ x+y=5 \] intersects them symmetrically.
Therefore, the triangle formed is symmetric about the line \[ x=y \]
Step 3: Use the property of incentre.
The incentre of a triangle always lies on the axis of symmetry of the triangle.
Hence, the locus of the incentre is \[ x=y \] That is, \[ x-y=0 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{x-y=0} \]
Given, \[ xy-4x-4y+16=0 \] Factorizing, \[ (x-4)(y-4)=0 \] Thus, the equation represents the pair of lines \[ x=4 \] and \[ y=4 \] The third line is \[ x+y=5 \] Hence, the triangle is formed by the lines \[ x=4,\quad y=4,\quad x+y=5 \]
Step 2: Observe the symmetry of the triangle.
The lines \[ x=4 \] and \[ y=4 \] are symmetric with respect to the line \[ x=y \] Also, the line \[ x+y=5 \] intersects them symmetrically.
Therefore, the triangle formed is symmetric about the line \[ x=y \]
Step 3: Use the property of incentre.
The incentre of a triangle always lies on the axis of symmetry of the triangle.
Hence, the locus of the incentre is \[ x=y \] That is, \[ x-y=0 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{x-y=0} \]
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