QJEEM15Mabc7e: JEE Main 2015 Mathematics - Class 11 JEEM15Mabc7e Solution: Locus of the image of the point in the line is a :

Question

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Locus of the image of the point

(2, 3)

in the line

\left( {2x - 3y + 4} \right) + k\left( {x - 2y + 3} \right) = 0,\,k \in R,

is a :

Accepted Answer

Intersection point of

2x - 3y + 4 = 0

and

x-2y+3=0

is

(1, 2)

Since,

P

is the fixed point for given family of lines

So,

PB=PA

{\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2}

{\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = 1 + 1 = 2

{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\sqrt 2 } \right)^2}

{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}

Therefore, given locus is a circle with center

(1, 2)

and radius

\sqrt 2 .

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