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

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Question

(b) Equation of line passing through point $(1, 1)$ is,

$y - 1 = m(x - 1)$......$(i)$

Line $(i)$ meets x-axis, so $y = 0$

$	herefore $ $\fra

Vedclass · Accepted Answer

$x + y - 2xy = 0$

Vedclass · Answer

(b) Equation of line passing through point $(1, 1)$ is,

$y - 1 = m(x - 1)$......$(i)$

Line $(i)$ meets x-axis, so $y = 0$

$	herefore $ $\frac{{ - 1}}{m} = x - 1 \Rightarrow x = 1 - \frac{1}{m}$

Line $(i)$ meets y-axis, so $x = 0$

$	herefore $ $y - 1 = - m \Rightarrow y = 1 - m$

Let mid point of $AB$ be $(h, k)$,

Then $h = \frac{{0 + (1 - (1/m))}}{2}$;$k = \frac{{0 + (1 - m)}}{2}$

$m = \frac{1}{{1 - 2h}}$ ; $m = 1 - 2k$

 $1 - 2k = \frac{1}{{1 - 2h}}$

==> $1 - 2k - 2h + 4hk = 1$

==> $ - 2h - 2k + 4hk = 0$

Hence the Locus of mid point 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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