GCSE Mathematics (9-1)

Chapter 22: Review

The chapter problem, solved

Take AA as the origin. Let AB→=u\overrightarrow{AB}=\mathbf{u} and AD→=v\overrightarrow{AD}=\mathbf{v}. Since opposite sides of a parallelogram have the same directed vectors, CC has position vector u+v\mathbf{u}+\mathbf{v}.

The midpoint of ACAC has position vector 12(u+v)\frac12(\mathbf{u}+\mathbf{v}). The midpoint of BDBD has position vector 12(u+v)\frac12(\mathbf{u}+\mathbf{v}) too. The midpoints are the same point, so the diagonals bisect each other. The prediction from the chapter opener was correct.

Mixed practice — choose the method first

For each item, name the method before you work it out. Where asked, explain why that method applies.

  1. Point P=(−2,4)P=(-2,4) is translated by (5−3)\begin{pmatrix}5\\-3\end{pmatrix}. Give its image coordinates.
  2. Find the vector from A=(3,−1)A=(3,-1) to B=(−2,5)B=(-2,5).
  3. Work out (2−4)+(−73)\begin{pmatrix}2\\-4\end{pmatrix}+\begin{pmatrix}-7\\3\end{pmatrix}.
  4. Work out 2(−35)−(4−2)2\begin{pmatrix}-3\\5\end{pmatrix}-\begin{pmatrix}4\\-2\end{pmatrix}.
  5. In triangle PQRPQR, PQ→=a\overrightarrow{PQ}=\mathbf{a} and QR→=b\overrightarrow{QR}=\mathbf{b}. Find PR→\overrightarrow{PR}.
  6. In parallelogram ABCDABCD, AB→=u\overrightarrow{AB}=\mathbf{u} and AD→=v\overrightarrow{AD}=\mathbf{v}. Find AC→\overrightarrow{AC} and DB→\overrightarrow{DB}.
  7. Explain how to prove that the diagonals of a parallelogram bisect each other using vectors.
  8. Solve 3x+4=193x+4=19 from earlier algebra. Which operation undoes the addition first?
  9. A rectangle has width xx and length x+3x+3. Its perimeter is 30 cm. Form an equation and find its dimensions.
Answers and concise working
  1. (3,1)(3,1); add the vector components to the coordinates.
  2. (−2−35−(−1))=(−56)\begin{pmatrix}-2-3\\5-(-1)\end{pmatrix}=\begin{pmatrix}-5\\6\end{pmatrix}.
  3. (−5−1)\begin{pmatrix}-5\\-1\end{pmatrix}; add matching components.
  4. (−1012)\begin{pmatrix}-10\\12\end{pmatrix}; multiply first, then subtract.
  5. a+b\mathbf{a}+\mathbf{b}; add the vectors along the route.
  6. AC→=u+v\overrightarrow{AC}=\mathbf{u}+\mathbf{v} and DB→=u−v\overrightarrow{DB}=\mathbf{u}-\mathbf{v}.
  7. Use position vectors A=0A=\mathbf{0}, B=uB=\mathbf{u}, D=vD=\mathbf{v} and C=u+vC=\mathbf{u}+\mathbf{v}. Both diagonal midpoints are 12(u+v)\frac12(\mathbf{u}+\mathbf{v}).
  8. Subtract 4 first: 3x=153x=15, then divide by 3, so x=5x=5.
  9. 2x+2(x+3)=302x+2(x+3)=30, so 4x=244x=24 and x=6x=6. The dimensions are 6 cm by 9 cm.

What you can now do

Your relevance reflection

Return to the example you wrote at the start. Explain where vectors could describe movement in that situation. If the connection has changed, write the new example and say what changed.

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