Question

2. Mirror the triangle shown below along a 45°- axis Vt (1,3)P3 (3, 3) 2 450

Answer 1

The distance between P_{​​​​​} = (1,1) and Q= (1,3) is $\sqrt{(1-1)^2+(1-3)^2}=\sqrt{2^2}=2$ and that of Q= (1,3) and R= (3,3) is $\sqrt{(1-3)^2+(3-3)^2}=\sqrt{2^2}=2$ . Thus the ∆PQR is an isosceles triangle with PQ = QR. By the property of isosceles triangle angle QPR and PRQ are equal.

Now consider the point Q", the mirror image of the vertex Q and join the points P and R with Q". The line PQ" is parallel to X-axis. Since the mirror is kept at 45° axis which passes through vertices P and R, the angle Q"PR will be 45° as they are the corresponding exterior angles between parallel lines. Now the lines PQ" and QR are parallel and angles QRP(=QPR) = RPQ" = 45° as they are the corresponding interior angles of parallel lines. Therefore the angles PQR = 90° ( angle sum property of triangle PQR) , QPQ" = 90°( 45°+45°), QRQ"= 90°(45°+45°) and RQ"P = 90° (angle sum property of the parallelogram PQRQ"). Therefore PQRQ" is a square. Therefore side length PQ" = 2, thus the point Q" =(3,1) as PQ" is the part of the line y = 1 parallel to X-axis. Therefore the mirror image of the ∆PQR is ∆PQ"R with vertices P= (1,1), Q" = ( 3,1) and R = (3,3).