Locus of the image of the point (2, 3) in the line

(2x-3y+4)+k(x-2y+3)=0,k\, \epsilon \, R,\, is\, a\, :

  • Option 1)

    straight line parallel to x-axis.

  • Option 2)

    straight line parallel to y-axis

  • Option 3)

    circle of radius \sqrt{2}

  • Option 4)

    circle of radius \sqrt{3}

 

Answers (2)

As we learnt in

Family of straight lines -

Equation of lines that belongs to L_{1}+L_{2}  and  L_{3}+L_{4} can be found by joining the points of intersection of the two families.

- wherein

 

  

Distance formula -

The distance between the point A\left ( x_{1},y_{1} \right )\: and \: B\left ( x_{2},y_{2} \right )

is \sqrt{\left ( x_{1} -x_{2}\right )^{2}+\left ( y_{1} -y_{2}\right )^{2}}

- wherein

 

image of p\left ( 2, 3 \right ) in the line

\left ( 2x-3y+4 \right )+ k\left ( x-2y+3 \right )=0, \:k\epsilon R

Point of intersection is A\left ( 1, 2 \right )

Now if PF=RF

where

R is image of P, then

AP Should be equal to AR, i.e. AP=AR

\left ( h-1 \right )^2+\left ( k-2 \right )^2=2

Circle of Redlines = \sqrt{2} 

 


Option 1)

straight line parallel to x-axis.

Incorrect

Option 2)

straight line parallel to y-axis

Incorrect

Option 3)

circle of radius \sqrt{2}

Correct

Option 4)

circle of radius \sqrt{3}

Incorrect

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