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9.Straight Line
hard
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
A
$2xy + x + y = 0$
B
$x + y - 2xy = 0$
C
$x + y + 2 = 0$
D
$x + y - 2 = 0$
Solution
(b) Equation of line passing through point $(1, 1)$ is,
$y – 1 = m(x – 1)$……$(i)$
Line $(i)$ meets x-axis, so $y = 0$
$\therefore $ $\frac{{ – 1}}{m} = x – 1 \Rightarrow x = 1 – \frac{1}{m}$
Line $(i)$ meets y-axis, so $x = 0$
$\therefore $ $y – 1 = – m \Rightarrow y = 1 – m$
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, $x + y – 2xy = 0$.
Standard 11
Mathematics
