The locus of the image of the point 2-3 with respect to the line x -2 y -3-2 x -3 y -4-0- R A- x2-y2-3 x-4 y-4-0B- x2-y2-2 x-4 y-4-0C- x2-y2-2 x-4 y-4-0D- x2-y2-2 x-4 y-3-0

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Distinguishing between Conics from General Equation and Eccentricity

The locus of ...

Question

The locus of the image of the point (2, 3) with respect to the line
$(x - 2 y + 3) + λ (2 x - 3 y + 4) = 0 (λ ϵ R)$

x^{2} + y^{2} - 2 x - 4 y + 4 = 0

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x^{2} + y^{2} + 2 x - 4 y + 4 = 0

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x^{2} + y^{2} - 3 x - 4 y - 4 = 0

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x^{2} + y^{2} - 2 x - 4 y + 3 = 0

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Solution

The correct option is D $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
(1,2) lie on both the lines and locus is $(h - 1)^{2} + (k - 2)^{2} = (2 - 1)^{2} + (3 - 2)^{2}$

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The locus of the image of the point (2, 3) with respect to the line (x−2y+3)+λ(2x−3y+4)=0(λϵR)

The correct option is D x2+y2−2x−4y+3=0(1,2) lie on both the lines and locus is (h−1)2+(k−2)2=(2−1)2+(3−2)2

The locus of the image of the point (2, 3) with respect to the line
$(x - 2 y + 3) + λ (2 x - 3 y + 4) = 0 (λ ϵ R)$

The correct option is D $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
(1,2) lie on both the lines and locus is $(h - 1)^{2} + (k - 2)^{2} = (2 - 1)^{2} + (3 - 2)^{2}$