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A triangle XYZ with vertices X(1, 2), Y(-1, -2) and Z(5, 0) is translated by a vector 1 2 . The image so formed is reflected on the x-axis. Write the co-ordinate

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4. A triangle XYZ with vertices X(1, 2), Y(-1, -2) and Z(5, 0) is translated by a vector (12)\begin{pmatrix} 1 \\ 2 \end{pmatrix}. The image so formed is reflected on the x-axis. Write the co-ordinates of the vertices of the images thus obtained and represent the △XYZ\triangle XYZ and its images on the same graph paper.

  • Step 1: Translation The translation vector is T=(12)T = \begin{pmatrix} 1 \\ 2 \end{pmatrix}. The formula for translation is P′(x′,y′)=P(x,y)+TP'(x', y') = P(x, y) + T. For X(1, 2):

    X′=(1+1,2+2)=(2,4)X' = (1+1, 2+2) = (2, 4)

    For Y(-1, -2):

    Y′=(−1+1,−2+2)=(0,0)Y' = (-1+1, -2+2) = (0, 0)

    For Z(5, 0):

    Z′=(5+1,0+2)=(6,2)Z' = (5+1, 0+2) = (6, 2)

    The coordinates after translation are X′(2,4),Y′(0,0),Z′(6,2)\boxed{X'(2, 4), Y'(0, 0), Z'(6, 2)}.

  • Step 2: Reflection on the x-axis The formula for reflection on the x-axis is (x,y)→(x,−y)(x, y) \to (x, -y). For X'(2, 4):

    X′′=(2,−4)X'' = (2, -4)

    For Y'(0, 0):

    Y′′=(0,0)Y'' = (0, 0)

    For Z'(6, 2):

    Z′′=(6,−2)Z'' = (6, -2)

    The coordinates after reflection are X′′(2,−4),Y′′(0,0),Z′′(6,−2)\boxed{X''(2, -4), Y''(0, 0), Z''(6, -2)}.

  • Step 3: Graph Representation Plot △XYZ\triangle XYZ with vertices X(1, 2), Y(-1, -2), Z(5, 0). Plot △X′Y′Z′\triangle X'Y'Z' with vertices X'(2, 4), Y'(0, 0), Z'(6, 2). Plot △X′′Y′′Z′′\triangle X''Y''Z'' with vertices X''(2, -4), Y''(0, 0), Z''(6, -2). This step should be performed on graph paper.

5. △PQR\triangle PQR with vertices P(2, 1), Q(2, 4) and R(5, 2) is translated by the vector (1−2)\begin{pmatrix} 1 \\ -2 \end{pmatrix} and then the image so formed is rotated about the origin through 90∘90^\circ in positive direction. Write down the co-ordinates of the vertices of the images thus obtained and present △PQR\triangle PQR and its images in the same graph paper.

  • Step 1: Translation The translation vector is T=(1−2)T = \begin{pmatrix} 1 \\ -2 \end{pmatrix}. For P(2, 1):

    P′=(2+1,1−2)=(3,−1)P' = (2+1, 1-2) = (3, -1)

    For Q(2, 4):

    Q′=(2+1,4−2)=(3,2)Q' = (2+1, 4-2) = (3, 2)

    For R(5, 2):

    R′=(5+1,2−2)=(6,0)R' = (5+1, 2-2) = (6, 0)

    The coordinates after translation are P′(3,−1),Q′(3,2),R′(6,0)\boxed{P'(3, -1), Q'(3, 2), R'(6, 0)}.

  • Step 2: Rotation about the origin through 90∘90^\circ in positive direction (counter-clockwise) The formula for rotation about the origin 90∘90^\circ counter-clockwise is (x,y)→(−y,x)(x, y) \to (-y, x). For P'(3, -1):

    P′′=(−(−1),3)=(1,3)P'' = (-(-1), 3) = (1, 3)

    For Q'(3, 2):

    Q′′=(−2,3)Q'' = (-2, 3)

    For R'(6, 0):

    R′′=(−0,6)=(0,6)R'' = (-0, 6) = (0, 6)

    The coordinates after rotation are P′′(1,3),Q′′(−2,3),R′′(0,6)\boxed{P''(1, 3), Q''(-2, 3), R''(0, 6)}.

  • Step 3: Graph Representation Plot △PQR\triangle PQR with vertices P(2, 1), Q(2, 4), R(5, 2). Plot △P′Q′R′\triangle P'Q'R' with vertices P'(3, -1), Q'(3, 2), R'(6, 0). Plot △P′′Q′′R′′\triangle P''Q''R'' with vertices P''(1, 3), Q''(-2, 3), R''(0, 6). This step should be performed on graph paper.

**6. A triangle LMN with vertices L(-3, 3), M(-6, 3) and N(-3, 9) respectively is reflected in y-axis and then image formed is rotated about the origin through 180∘180^\circ. Write down the co-ordinates of the

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