Locus of the image of the point ( 2, 3 ) in the line (2 x - 3 y +...

Let image is $( \alpha, \beta )$

$∴$ point $( \dfrac{2 + \alpha}{2} , \dfrac{3 + \beta}{2} )$ lies on the given variable line.

$∴ 2( \dfrac{2 + \alpha}{2} ) - 3( \dfrac{3 + \beta}{2} ) + 4 + k( \dfrac{2 + \alpha}{2} - 2 \times \dfrac{3 + \beta}{2} + 3 ) = 0$

$⇒ 2\alpha - 3\beta + 3 + k( \alpha - 2\beta + 2 ) = 0 . . . . . . . \left(1\right) <br /><br />again, product$ of two slopes= - 1$<br /><br />$⇒ $\dfrac{3 - \beta}{2 - \alpha}$ \times $\dfrac{2 + k}{3 + 2k} =$- 1$<br /><br />$⇒ 6 - 2\beta + k( 3 - \beta ) = - ( 6 - 3\alpha ) - k( 4 - 2\alpha )$<br /><br />$⇒ k ( 7 - 2\alpha - \beta ) = 3\alpha + 2\beta - 12$<br /><br />$⇒ k = $\dfrac{3\alpha + 2\beta - 12}{7 - 2\alpha - \beta}<br /><br />Substituting value$ of k$in ,\left(1\right) we get<br /><br />$2\alpha - 3\beta + 3 + $\dfrac{3\alpha + 2\beta + - 12}{7 - 2\alpha - \beta}$ ( \alpha - 2\beta + 2 ) = 0$<br /><br />$⇒ 14\alpha - 4{\alpha}^{2} - 2\alpha\beta - 21\beta + 6\alpha\beta + 3{\beta}^{2} + 21 - 6\alpha - 3\beta + 3{\alpha}^{2} - 6\alpha\beta + 6\alpha + 2\alpha\beta - 4{\beta}^{2} + 4\beta - 12\alpha + 24\beta - 24 = 0$<br /><br />$⇒ {\alpha}^{2} + {\beta}^{2} - 2a - 4\beta + 3 = 0$<br /><br />$∴$Locus is${x}^{2} + {y}^{2} - 2x - 4y + 3 = 0$<br /><br />Which is a circle with radius$= \sqrt{1 + 4 - 3} = \sqrt{2}$<br />$