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WAEC 2012 · Paper 2 · Q16

  1. (a)

    Given that p=(35)\mathbf p = \begin{pmatrix} 3 \\ 5 \end{pmatrix}, q=(2−1)\mathbf q = \begin{pmatrix} 2 \\ -1 \end{pmatrix} and r=(517)\mathbf r = \begin{pmatrix} 5 \\ 17 \end{pmatrix}, express r\mathbf r in terms of p\mathbf p and q\mathbf q.

  2. (b)

    In the quadrilateral ABCDABCD, AB→=(−5−1)\overrightarrow{AB} = \begin{pmatrix} -5 \\ -1 \end{pmatrix}, AC→=(−6−9)\overrightarrow{AC} = \begin{pmatrix} -6 \\ -9 \end{pmatrix} and BD→=(4−7)\overrightarrow{BD} = \begin{pmatrix} 4 \\ -7 \end{pmatrix}. Show that ABCDABCD is a parallelogram.

    Model answer

    BC→=AC→−AB→=(−1−8)\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB} = \begin{pmatrix} -1 \\ -8 \end{pmatrix} and AD→=AB→+BD→=(−1−8)\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD} = \begin{pmatrix} -1 \\ -8 \end{pmatrix}, so AD∥BCAD \parallel BC and AD=BCAD = BC. Also DC→=AC→−AD→=(−5−1)=AB→\overrightarrow{DC} = \overrightarrow{AC} - \overrightarrow{AD} = \begin{pmatrix} -5 \\ -1 \end{pmatrix} = \overrightarrow{AB}, so AB∥DCAB \parallel DC. Hence ABCDABCD is a parallelogram.

Worked solution (try it first)

(a)

  1. Write r=αp+βq\mathbf r = \alpha\mathbf p + \beta\mathbf q: (517)=(3α+2β5α−β)\begin{pmatrix} 5 \\ 17 \end{pmatrix} = \begin{pmatrix} 3\alpha + 2\beta \\ 5\alpha - \beta \end{pmatrix}.
  2. Compare components: 3α+2β=53\alpha + 2\beta = 5 and 5α−β=175\alpha - \beta = 17.
  3. Double the second and add: 13α=3913\alpha = 39, so α=3\alpha = 3 and β=5(3)−17=−2\beta = 5(3) - 17 = -2.
  4. So r=3p−2q\mathbf r = 3\mathbf p - 2\mathbf q.

(b)

  1. BC→=AC→−AB→\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB}
    =(−6+5−9+1)= \begin{pmatrix} -6 + 5 \\ -9 + 1 \end{pmatrix}
    =(−1−8)= \begin{pmatrix} -1 \\ -8 \end{pmatrix}.
  2. AD→=AB→+BD→\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD}
    =(−5+4−1−7)= \begin{pmatrix} -5 + 4 \\ -1 - 7 \end{pmatrix}
    =(−1−8)= \begin{pmatrix} -1 \\ -8 \end{pmatrix}
    =BC→= \overrightarrow{BC}, so ADAD is equal and parallel to BCBC.
  3. DC→=AC→−AD→\overrightarrow{DC} = \overrightarrow{AC} - \overrightarrow{AD}
    =(−6+1−9+8)= \begin{pmatrix} -6 + 1 \\ -9 + 8 \end{pmatrix}
    =(−5−1)= \begin{pmatrix} -5 \\ -1 \end{pmatrix}
    =AB→= \overrightarrow{AB}, so ABAB is equal and parallel to DCDC.
  4. Both pairs of opposite sides are parallel, so ABCDABCD is a parallelogram.

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