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

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

  1. Circle of radius 3
  2. Straight line parallel to x-axis.
  3. Straight line parallel to y-axis.
  4. Circle of radius 2

Solution

Let image is ( α,β )

∴ point ( 2+α 2 , 3+β 2 ) lies on the given variable line.

∴  2( 2+α 2 )−3( 3+β 2 )+4+k( 2+α 2 −2× 3+β 2 +3 )=0

⇒2α−3β+3+k(α−2β+2)=0    .......1 

again, product of two slopes =−1

⇒ 3−β 2−α × 2+k 3+2k =−1

⇒6−2β+k( 3−β )=−( 6−3α )−k( 4−2α )

⇒k(7−2α−β)=3α+2β−12

⇒k= 3α+2β−12 7−2α−β

Substituting value of k in ,1 we get

2α−3β+3+ 3α+2β+−12 7−2α−β ( α−2β+2 )=0

⇒14α−4 α 2 −2αβ−21β+6αβ+3 β 2 +21−6α−3β+3 α 2 −6αβ+6α+2αβ−4 β 2 +4β−12α+24β−24=0

⇒ α 2 + β 2 −2a−4β+3=0

∴ Locus is x 2 + y 2 −2x−4y+3=0

Which is a circle with radius = 1+4−3 = 2
 

Asked in: JEE Main 2015 (04 Apr)

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