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(D) Let the line $AB$ have slope $m$. Since $AB$ passes through $A(1, 1)$,its equation is $y - 1 = m(x - 1)$.
It intersects the $x$-axis at $B(1 - \f

Vedclass · Accepted Answer

$9xy - 6x - 6y + 5 = 0$

Vedclass · Answer

(D) Let the line $AB$ have slope $m$. Since $AB$ passes through $A(1, 1)$,its equation is $y - 1 = m(x - 1)$.
It intersects the $x$-axis at $B(1 - \frac{1}{m}, 0)$.
The line $AC$ is perpendicular to $AB$,so its slope is $-\frac{1}{m}$. Its equation is $y - 1 = -\frac{1}{m}(x - 1)$.
It intersects the $y$-axis at $C(0, 1 + m)$.
Let the centroid of $\Delta ABC$ be $G(h, k)$.
$h = \frac{1 + (1 - \frac{1}{m}) + 0}{3} = \frac{2 - \frac{1}{m}}{3} \implies 3h - 2 = -\frac{1}{m} \implies m = \frac{1}{2 - 3h}$.
$k = \frac{1 + 0 + (1 + m)}{3} = \frac{2 + m}{3} \implies 3k - 2 = m$.
Substituting $m$ in the second equation: $3k - 2 = \frac{1}{2 - 3h}$.
$(3k - 2)(2 - 3h) = 1 \implies 6k - 9hk - 4 + 6h = 1$.
$9hk - 6h - 6k + 5 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus is $9xy - 6x - 6y + 5 = 0$.

If two mutually perpendicular lines through the point $A(1, 1)$ intersect the $x$-axis and $y$-axis at points $B$ and $C$ respectively,then the locus of the centroid of $\Delta ABC$ is -

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