Given A(1, 1) and AB is any line through it c | vedclass

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(A) Let the slope of line $AB$ be $m$. The equation of line $AB$ passing through $A(1, 1)$ is $y - 1 = m(x - 1)$.Since $AB$ cuts the $x-$ axis at $B$,

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$x + y = 1$

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(A) Let the slope of line $AB$ be $m$. The equation of line $AB$ passing through $A(1, 1)$ is $y - 1 = m(x - 1)$.Since $AB$ cuts the $x-$ axis at $B$,set $y = 0$: $-1 = m(x - 1) \implies x - 1 = -\frac{1}{m} \implies x = 1 - \frac{1}{m}$. So,$B = (1 - \frac{1}{m}, 0)$.Since $AC \perp AB$,the slope of $AC$ is $-\frac{1}{m}$. The equation of line $AC$ is $y - 1 = -\frac{1}{m}(x - 1)$.Since $AC$ meets the $y-$ axis at $C$,set $x = 0$: $y - 1 = -\frac{1}{m}(-1) = \frac{1}{m} \implies y = 1 + \frac{1}{m}$. So,$C = (0, 1 + \frac{1}{m})$.Let $P(h, k)$ be the midpoint of $BC$. Then $h = \frac{(1 - \frac{1}{m}) + 0}{2} = \frac{1}{2} - \frac{1}{2m}$ and $k = \frac{0 + (1 + \frac{1}{m})}{2} = \frac{1}{2} + \frac{1}{2m}$.Adding the two equations: $h + k = (\frac{1}{2} - \frac{1}{2m}) + (\frac{1}{2} + \frac{1}{2m}) = 1$.Replacing $(h, k)$ with $(x, y)$,the locus is $x + y = 1$.

Given $A(1, 1)$ and $AB$ is any line through it cutting the $x-$ axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y-$ axis in $C$,then the equation of the locus of the midpoint $P$ of $BC$ is

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