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Given points P and Q, express the vector PQ as a column vector and find the coordinates of Q given PQ and P.

Mathematics

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2 parts
a

(−5−3)\begin{pmatrix} -5 \\ -3 \end{pmatrix}

b
(-2, 6)

Step-by-step solution

7 steps

Here are the solutions to the problems.

Part a· 3 steps

To express the vector PQ⃗\vec{PQ} as a column vector, subtract the coordinates of point P from the coordinates of point Q.

Given P=(2,4)P = (2, 4) and Q=(−3,1)Q = (-3, 1).

  1. Calculate the change in the x-coordinate.

    xQ−xP=−3−2=−5x_Q - x_P = -3 - 2 = -5

  2. Calculate the change in the y-coordinate.

    yQ−yP=1−4=−3y_Q - y_P = 1 - 4 = -3

  3. Write the vector PQ⃗\vec{PQ} as a column vector.

    PQ⃗=(−5−3)\vec{PQ} = \begin{pmatrix} -5 \\ -3 \end{pmatrix}

    The vector PQ⃗\vec{PQ} is (−5−3)\boxed{\begin{pmatrix} -5 \\ -3 \end{pmatrix}}.

Part b(-2, 6)

Given the vector PQ⃗=(−32)\vec{PQ} = \begin{pmatrix} -3 \\ 2 \end{pmatrix} and point P=(1,4)P = (1, 4). Let the coordinates of point Q be (x,y)(x, y).

  1. Set up the vector equation.

    The vector PQ⃗\vec{PQ} is found by subtracting the coordinates of P from Q:

    PQ⃗=(x−xPy−yP)=(x−1y−4)\vec{PQ} = \begin{pmatrix} x - x_P \\ y - y_P \end{pmatrix} = \begin{pmatrix} x - 1 \\ y - 4 \end{pmatrix}

    We are given PQ⃗=(−32)\vec{PQ} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}.
    So, we have the equation:

    (x−1y−4)=(−32)\begin{pmatrix} x - 1 \\ y - 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}

  2. Solve for the x-coordinate of Q.

    x−1=−3x - 1 = -3 x=−3+1x = -3 + 1 x=−2x = -2

  3. Solve for the y-coordinate of Q.

    y−4=2y - 4 = 2 y=2+4y = 2 + 4 y=6y = 6

  4. State the coordinates of point Q.

    The coordinates of point Q are (−2,6)\boxed{(-2, 6)}.

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