Question

The locus of the incentre of the triangle formed by the lines \[ xy-4x-4y+16=0 \] and \[ x+y=5 \] is

• A. \(x-y=0\)
• B. \(x+y=0\)
• C. \(x-2y=0\)
• D. \(2x-y=0\)

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.

Answer 1

The Correct Option is A

Step 1: Find the three sides of the triangle.
Factor $xy - 4x - 4y + 16 = (x-4)(y-4) = 0$. So the sides are $x = 4$, $y = 4$, and $x + y = 5$.

Step 2: Find the vertices of the triangle.
Intersection of $x=4$ and $y=4$: vertex $A = (4, 4)$. Intersection of $x=4$ and $x+y=5$: $y = 1$, vertex $B = (4, 1)$. Intersection of $y=4$ and $x+y=5$: $x = 1$, vertex $C = (1, 4)$.

Step 3: Compute the side lengths.
$AB = |4 - 1| = 3$ (vertical side). $AC = |4 - 1| = 3$ (horizontal side). $BC = \sqrt{(4-1)^2 + (1-4)^2} = 3\sqrt{2}$ (hypotenuse).

Step 4: Note the symmetry.
The triangle is isosceles right-angled at $A = (4,4)$, with $AB = AC = 3$. The line $AB$ is vertical ($x = 4$) and the line $AC$ is horizontal ($y = 4$). By symmetry of an isosceles right triangle, the incentre lies on the angle bisector of the right angle at $A$, which has slope $-1$ through $(4,4)$: i.e., $y - 4 = -(x - 4) \Rightarrow y = -x + 8$.

Step 5: Find the incentre formula.
The incentre is $\left(\frac{a \cdot x_A + b \cdot x_B + c \cdot x_C}{a+b+c},\ \frac{a \cdot y_A + b \cdot y_B + c \cdot y_C}{a+b+c}\right)$ where $a, b, c$ are side lengths opposite vertices $A, B, C$. Here $a = BC = 3\sqrt{2}$, $b = AC = 3$, $c = AB = 3$.

Step 6: Compute and find the locus.
$I_x = \frac{3\sqrt{2}(4) + 3(4) + 3(1)}{3\sqrt{2}+6} = \frac{12\sqrt{2}+15}{3\sqrt{2}+6}$. $I_y = \frac{3\sqrt{2}(4) + 3(1) + 3(4)}{3\sqrt{2}+6} = \frac{12\sqrt{2}+15}{3\sqrt{2}+6}$. Since $I_x = I_y$ always for any position on these symmetric sides, the locus is $x = y$, i.e., $x - y = 0$.
\[ \boxed{x - y = 0} \]