QJEEM18Ma7378: JEE Main 2018 Mathematics - Class 11 JEEM18Ma7378 Solution: A straight line through a fixed point (2, 3) intersects the…

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