Discussion Forum

 Locus of the image of the point (2,3) in the line (2x-3y+4)+k(x-2y+3)=0 k ε  R  is a

Answer:

Explanation:

Central Idea      First of all find the point of intersection of the lines 2x-3y+4=0 and x-2y+3=0 (say   A) . Now  , the line

(2x-3y+4) +k(x-2y+3)=0 is the perpendicular bisector of the line joining points P(2,3) and image P ' (h,k). Now  AP= AP '  and simplify

  Given line is

   (2x-3y+4)+k(x-2y+3)=0 , k ε   R  ..... (i)

   This line will pass through the point of intersection of the lines

                  2x-3y+4= 0       .......(ii)

 and          x-2y+3= 0      ..........(iii)

 On solving Eqs (ii) and (iii), we get

  x=1, y=2

  $\therefore$ Point of intersection of lines (ii) and (iii)  is (1,2)

Let M be the mid-point  of PP' , then AM is perpendicular bisector of PP' (where A, is the point of intersection of given lines )

  $\therefore$       AP=AP'

   $\Rightarrow$     $\sqrt{(2-1)^{2}+(3-2)^{2}}=\sqrt{(h-1)^{2}+(k-2)^{2}}$

 $\Rightarrow$      $\sqrt{2}=\sqrt{h^{2}+k^{2}-2h-4k+1+4}$

 $\Rightarrow$      $\sqrt{2}=\sqrt{h^{2}+k^{2}-2h-4k+5}$

 $\Rightarrow$     $\sqrt{2}=\sqrt{h^{2}+k^{2}-2h-4k+5}$

 $\Rightarrow$    $h^{2}+k^{2}-2h-4k+3=0$

Thus  , the required locus is

              $x^{2}+y^{2}-2x-4y+3=0$

  which is a equation  of circle with

    radius  =   $\sqrt{1+4-3}=\sqrt{2}$

  Aliter

  (2x-3y+4)+k(x-2y+3)=0     is family of line passing through (1,2)  By congruency of triangles . We can prove that mirror image (h,k) and the point (2,3) will be equidistant  from (1,2)

                         $\therefore$  Locus of (h,k)  PR=PQ

             $\Rightarrow  (h-1)^{2}+(k-2)^{2}= (2-1)^{2}+(3-2)^{2}$

or       $(x-1)^{2}+(y-2)^{2}= 2$

  $\therefore$     Locus is a circle of radius =   $\sqrt{2}$