Question

locus of the image of The. 2, 3 in the lines (2x - 3y + 4) + k(x - 2y + 3) = 0, k € R is a ?

Answer 1

P = (2,3)
Line L:   (2x - 3y + 4) + k (x - 2y +3) = 0    , k ∈ R

L passes through intersection point of  2x-3y +4 =0 and x-2y+3 = 0
Calculating that point:  C(1,2).

Now we are looking at image points Q (α, β) of P(2,3) wrt line L. 
Shift Origin from O(0,0) to C(1,2) with new coordinate system of X = x-1 and Y= y-2.

We need to find the images of P(X=2-1=1, Y=3-2=1) on lines passing thru C(X=0, Y=0).

Clearly we can see that points (X,Y)=(-1,-1), (-1, 1), (1,1), (1,-1) are image points wrt X=-Y, X=0, X= Y, Y= 0   respectively.

Clearly the locus is a circle of radius = CP = √(1²+1²) = √2
     Locus:   X² + Y² = 2     or   (x-1)² + (y-2)² = 2

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Another method for finding the image using formula was too long...

If (α, β) is the image of (2,3) wrt L:  (2+k)x - y(3+2k)+ (4+3k) = 0
Then we get:

(α-2) / (k+2) = (3-β) / (3+2k)
                  = -2 [2(2+k) - 3(3+2k) +(4+3k)] / [(2+k)²+(3+2k)²]

We need express k in terms of α & β.  Then eliminate it to get an equation in α, β  only.  Then replace them with (x,y) to get the locus.

    It simplifies to  (x - 1)² + (y-2)² = 2.