A straight line through a fixed point (2,3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is:
Solutions · 1
Official solution
Let coordinate of point R = (h, k).
Equation of line PQ,
(y - 3) = m (x - 2).
Put y = 0 to get coordinate of point p,
0 - 3 = (x - 2)
\[ \Rightarrow \] x = 2 - \[{3 \over m}\]
\[\therefore\,\,\,\] p = (2 - \[{3 \over m}\], 0)
As p = (h, 0) then
h = 2 - \[{3 \over m}\]
\[ \Rightarrow \] \[{3 \over m}\] = 2 - h
\[ \Rightarrow \] m = \[{3 \over {2 - h}}\] . . . . . . (1)
Put x = 0 to get coordinate of point Q,
y - 3 = m (0 - 2)
\[ \Rightarrow \] y = 3 - 2m
\[\therefore\,\,\,\] point Q = (0, 3 - 2m)
And From the graph you can see Q = (0, k).
\[\therefore\,\,\,\] k = 3 - 2m
\[ \Rightarrow \] m = \[{{3 - k} \over 2}\] . . . . (2)
By comparing (1) and (2) get
\[{3 \over {2 - h}} = {{3 - k} \over 2}\]
\[ \Rightarrow \] (2 - h)(3 - k) = 6
\[ \Rightarrow \] 6 - 3h - 2K + hk = 6
\[ \Rightarrow \] 3 h + 2K = hk
\[\therefore\,\,\,\] locus of point R is 3x + 2y= xy