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Plot the triangle ABC with vertices A(5,7), B(3,4), and C(7,3). Then, translate the triangle by the vector (4, -2) to find the image triangle A'B'C'.

Mathematics

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Answer

e

−117-\frac{11}{7}

Step-by-step solution

11 steps

This problem involves plotting points and performing geometric transformations. Since I cannot draw graphs, I will provide the coordinates for each step.

a) The coordinates of △ABC\triangle ABC are given as A(5,7)A(5,7), B(3,4)B(3,4), and C(7,3)C(7,3). You would plot these points on graph paper using the specified scale (2cm to 2 units on each axis) and connect them to form the triangle.

Part b· 3 steps

To find the image △A′B′C′\triangle A'B'C' under the translation by the vector (4−2)\begin{pmatrix} 4 \\ -2 \end{pmatrix}, we add the translation vector to each coordinate.

  1. Calculate the coordinates of A′A'.

    A′=A+(4−2)=(57)+(4−2)=(5+47−2)=(95)A' = A + \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 5 \\ 7 \end{pmatrix} + \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 5+4 \\ 7-2 \end{pmatrix} = \begin{pmatrix} 9 \\ 5 \end{pmatrix}

    So, A′(9,5)A'(9,5).

  2. Calculate the coordinates of B′B'.

    B′=B+(4−2)=(34)+(4−2)=(3+44−2)=(72)B' = B + \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} + \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 3+4 \\ 4-2 \end{pmatrix} = \begin{pmatrix} 7 \\ 2 \end{pmatrix}

    So, B′(7,2)B'(7,2).

  3. Calculate the coordinates of C′C'.

    C′=C+(4−2)=(73)+(4−2)=(7+43−2)=(111)C' = C + \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 7 \\ 3 \end{pmatrix} + \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 7+4 \\ 3-2 \end{pmatrix} = \begin{pmatrix} 11 \\ 1 \end{pmatrix}

    So, C′(11,1)C'(11,1).
    The image △A′B′C′\triangle A'B'C' has coordinates A′(9,5)A'(9,5), B′(7,2)B'(7,2), and C′(11,1)C'(11,1).

Part c· 3 steps

To find the image △A′′B′′C′′\triangle A''B''C'' of △ABC\triangle ABC under a reflection in the line y=−2y = -2.

For a reflection in the line y=ky=k, a point (x,y)(x,y) maps to (x,2k−y)(x, 2k-y). Here k=−2k=-2.

  1. Calculate the coordinates of A′′A''.

    A′′=(5,2(−2)−7)=(5,−4−7)=(5,−11)A'' = (5, 2(-2)-7) = (5, -4-7) = (5, -11)

    So, A′′(5,−11)A''(5,-11).

  2. Calculate the coordinates of B′′B''.

    B′′=(3,2(−2)−4)=(3,−4−4)=(3,−8)B'' = (3, 2(-2)-4) = (3, -4-4) = (3, -8)

    So, B′′(3,−8)B''(3,-8).

  3. Calculate the coordinates of C′′C''.

    C′′=(7,2(−2)−3)=(7,−4−3)=(7,−7)C'' = (7, 2(-2)-3) = (7, -4-3) = (7, -7)

    So, C′′(7,−7)C''(7,-7).
    The image △A′′B′′C′′\triangle A''B''C'' has coordinates A′′(5,−11)A''(5,-11), B′′(3,−8)B''(3,-8), and C′′(7,−7)C''(7,-7).

Part d· 3 steps

To find the image △A′′′B′′′C′′′\triangle A'''B'''C''' of △ABC\triangle ABC under a rotation through 90∘90^\circ anticlockwise about the origin.

For a 90∘90^\circ anticlockwise rotation about the origin, a point (x,y)(x,y) maps to (−y,x)(-y,x).

  1. Calculate the coordinates of A′′′A'''.

    A′′′=(−7,5)A''' = (-7, 5)

    So, A′′′(−7,5)A'''(-7,5).

  2. Calculate the coordinates of B′′′B'''.

    B′′′=(−4,3)B''' = (-4, 3)

    So, B′′′(−4,3)B'''(-4,3).

  3. Calculate the coordinates of C′′′C'''.

    C′′′=(−3,7)C''' = (-3, 7)

    So, C′′′(−3,7)C'''(-3,7).
    The image △A′′′B′′′C′′′\triangle A'''B'''C''' has coordinates A′′′(−7,5)A'''(-7,5), B′′′(−4,3)B'''(-4,3), and C′′′(−3,7)C'''(-3,7).

Part e

−117-\frac{11}{7}

Find the gradient of line B′′B′′′B''B'''.

From part (c), B′′(3,−8)B''(3,-8).
From part (d), B′′′(−4,3)B'''(-4,3).
The formula for the gradient mm of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.

  1. Substitute the coordinates of B′′B'' and B′′′B''' into the gradient formula.

    Let (x1,y1)=(3,−8)(x_1, y_1) = (3, -8) and (x2,y2)=(−4,3)(x_2, y_2) = (-4, 3).

    m=3−(−8)−4−3m = \frac{3 - (-8)}{-4 - 3}

  2. Simplify the expression.

    m=3+8−7=11−7=−117m = \frac{3 + 8}{-7} = \frac{11}{-7} = -\frac{11}{7}

    The gradient of line B′′B′′′B''B''' is −117\boxed{-\frac{11}{7}}.

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