Question 31

3/3 marks

Problem

31. If the points A(6, 1), B(p, 2), C(9, 4) and D(7, q) are the vertices of a parallelogram ABCD, then find the values of p and q. Hence, check whether ABCD is a rectangle or not.

Student work

given points ⇒A(6,1) B(p,2) C(9,4) D(7,q)\Rightarrow A(6,1)\ B(p,2)\ C(9,4)\ D(7,q)

A sketch of parallelogram ABCDABCD with diagonals crossing at O(x,y)O(x,y).

comparing yy coordinates, 52=q+22\dfrac{5}{2}=\dfrac{q+2}{2}, so q=3q=3.

A=(6,1)B=( ? 3,2)C=A=(6,1)\quad B=(\,?\,3,2)\quad C=
A=(6,1)B(8,2)C=(9,4)D=(7,3)A=(6,1)\quad B(8,2)\quad C=(9,4)\quad D=(7,3)

AC=(9−6)2+(4−1)2=32 unitsAC=\sqrt{(9-6)^2+(4-1)^2}=3\sqrt{2}\ \text{units}

BD=(8−7)2+(2−3)2=(1)2+(−1)=2 unitsBD=\sqrt{(8-7)^2+(2-3)^2}=\sqrt{(1)^2+(-1)}=\sqrt{2}\ \text{units}

As AC≠BDAC\ne BD, diagonals are not equal, hence parallelogram ABCDABCD is not a rectangle.

Diagonals of a parallelogram bisect each other, so OO is the midpoint of ACAC and BDBD.

considering ACAC, (152,52)\left(\dfrac{15}{2},\dfrac{5}{2}\right). considering BDBD, (7+p2,q+22)\left(\dfrac{7+p}{2},\dfrac{q+2}{2}\right).

comparing xx coordinates, 152=7+p2\dfrac{15}{2}=\dfrac{7+p}{2}, so p=8p=8.

Strengths

  • Used the midpoint property of diagonals for a parallelogram.
  • Solved the resulting equations correctly to get p=8p = 8 and q=3q = 3.
  • Computed diagonal lengths and used them correctly to conclude not a rectangle.
Marking rationale

All three method steps were shown and correctly executed: midpoint of ACAC found, midpoints equated to get pp and qq, and diagonals compared to test for a rectangle. The small notation slip in BDBD did not affect the numeric result.

What to improve

  • Algebra / notation Minor
    Wrote (−1)(-1) instead of (−1)2(-1)^2 when evaluating BD, a notation slip when squaring a negative number.

    From your work: BD=(8−7)2+(2−3)2=(1)2+(−1)=2 unitsBD=\sqrt{(8-7)^2+(2-3)^2}=\sqrt{(1)^2+(-1)}=\sqrt{2}\ \text{units}

    Try this: Practice writing and evaluating squares of negative numbers carefully; check each squared term is written as squared.

Solution — Question 31

For a parallelogram, the diagonals bisect each other, so the midpoint of diagonal ACAC equals the midpoint of diagonal BDBD:

(6+92,1+42)=(p+72,2+q2).\left(\frac{6+9}{2},\frac{1+4}{2}\right)=\left(\frac{p+7}{2},\frac{2+q}{2}\right).

Equating the corresponding coordinates gives 152=p+72\dfrac{15}{2}=\dfrac{p+7}{2}, so p=8p=8, and 52=2+q2\dfrac{5}{2}=\dfrac{2+q}{2}, so q=3q=3.

A rectangle has adjacent sides perpendicular, so the product of their slopes must be −1-1. For A(6,1)A(6,1), B(8,2)B(8,2), and C(9,4)C(9,4),

mAB=2−18−6=12,mBC=4−29−8=2,mABmBC=1≠−1.m_{AB}=\frac{2-1}{8-6}=\frac{1}{2},\qquad m_{BC}=\frac{4-2}{9-8}=2,\qquad m_{AB}m_{BC}=1\ne -1.

So, p=8p=8 and q=3q=3, and ABCDABCD is not a rectangle.

Final answer: p=8, q=3p=8,\ q=3; ABCDABCD is not a rectangle.

Key idea: Use equal diagonal midpoints to find the missing coordinates, then test whether adjacent sides are perpendicular.

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