The line segment joining A (3, 0), B(5, 2) is rotated about a point A in anticlockwise sense through an angle $\frac{\pi }{4}andB$ move to C. If a point D be the reflection of C in y-axis then D=

 Given A (3, 0)  B(5, 2)

$\begin{array}{l}SlopeofAB=\frac{2-0}{5-3}=1\\ \therefore \overline{AB}makingangle\frac{\pi }{4}withx-axis\\ Let\theta =Anglemadeby\overline{AC}withx-axis\\ \theta ={45}^0+{45}^0={90}^0\\ r=\overline{AC}=\overline{AB}=\sqrt{{\left(5-3\right)}^2+{\left(2-0\right)}^2}=\sqrt{4+4}=2\sqrt{2}\\ \therefore c=\left(x_1+r\cos \theta ,y_1+r\sin \theta \right)=\left(3+2\sqrt{2}\cos {90}^0,0+2\sqrt{2}\sin {90}^0\right)\\ =\left(3+0,0+2\sqrt{2}\right)\\ =\left(3,2\sqrt{2}\right)\in Q_1\\ \mathrm{Re}flectionofCw.r.toy-axisD\in Q_2\\ \therefore D=\left(-3,2\sqrt{2}\right)\end{array}$