The locus of the incentre of the triangle for | vedclass

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(A) The given equation is $xy-4x-4y+16=0$. Factoring this,we get $x(y-4)-4(y-4)=0$,which implies $(x-4)(y-4)=0$. Thus,the two lines are $L_1: x=4$ and

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$x-y=0$

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(A) The given equation is $xy-4x-4y+16=0$. Factoring this,we get $x(y-4)-4(y-4)=0$,which implies $(x-4)(y-4)=0$. Thus,the two lines are $L_1: x=4$ and $L_2: y=4$. The third line is $L_3: x+y=5$. The vertices of the triangle are the intersection points of these lines: $A = L_1 \cap L_2 = (4, 4)$ $B = L_1 \cap L_3 = (4, 1)$ $C = L_2 \cap L_3 = (1, 4)$ The side lengths are: $c = AB = \sqrt{(4-4)^2 + (4-1)^2} = 3$ $a = BC = \sqrt{(4-1)^2 + (1-4)^2} = \sqrt{3^2 + (-3)^2} = 3\sqrt{2}$ $b = CA = \sqrt{(1-4)^2 + (4-4)^2} = 3$ The incentre $I(x, y)$ is given by $\left(\frac{ax_A+bx_B+cx_C}{a+b+c}, \frac{ay_A+by_B+cy_C}{a+b+c}\right)$. Substituting the values: $x = \frac{3\sqrt{2}(4) + 3(4) + 3(1)}{3\sqrt{2}+3+3} = \frac{12\sqrt{2}+15}{3\sqrt{2}+6} = \frac{3(4\sqrt{2}+5)}{3(\sqrt{2}+2)} = \frac{4\sqrt{2}+5}{\sqrt{2}+2}$ $y = \frac{3\sqrt{2}(4) + 3(1) + 3(4)}{3\sqrt{2}+3+3} = \frac{12\sqrt{2}+15}{3\sqrt{2}+6} = \frac{4\sqrt{2}+5}{\sqrt{2}+2}$ Since $x=y$,the locus of the incentre is $x-y=0$.

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

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