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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)

best_answer

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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