A straight line through the point (1, 1) meet | vedclass

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(B) Let the equation of the line passing through $(1, 1)$ be $y - 1 = m(x - 1)$,where $m$ is the slope.For the $x$-axis intersection $(y = 0)$: $-1 =

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$x + y - 2xy = 0$

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(B) Let the equation of the line passing through $(1, 1)$ be $y - 1 = m(x - 1)$,where $m$ is the slope.For the $x$-axis intersection $(y = 0)$: $-1 = m(x - 1)$ $\Rightarrow x - 1 = -\frac{1}{m}$ $\Rightarrow x = 1 - \frac{1}{m}$. So,$A = (1 - \frac{1}{m}, 0)$.For the $y$-axis intersection $(x = 0)$: $y - 1 = m(-1) \Rightarrow y = 1 - m$. So,$B = (0, 1 - m)$.Let the mid-point of $AB$ be $(h, k)$.Then $h = \frac{1 - \frac{1}{m} + 0}{2} = \frac{m - 1}{2m}$ and $k = \frac{0 + 1 - m}{2} = \frac{1 - m}{2}$.From $k = \frac{1 - m}{2}$,we get $m = 1 - 2k$.Substitute $m$ into the expression for $h$: $2h = \frac{(1 - 2k) - 1}{1 - 2k} = \frac{-2k}{1 - 2k}$.$2h(1 - 2k) = -2k$ $\Rightarrow 2h - 4hk = -2k$ $\Rightarrow 2h + 2k - 4hk = 0$.Dividing by $2$,we get $h + k - 2hk = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x + y - 2xy = 0$.

$A$ straight line through the point $(1, 1)$ meets the $x$-axis at $A$ and the $y$-axis at $B$. The locus of the mid-point of $AB$ is

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