WAEC 2016 · Paper 2 · Q13

  1. (a)

    Using a scale of 2 cm to 2 units on each axis, draw on a sheet of graph paper two perpendicular axes OxOx and OyOy for −10≤x≤10-10 \le x \le 10 and −12≤y≤12-12 \le y \le 12.

    Model answer
    −10−8−6−4−2246810−12−10−8−6−4−22468xyPQRSP1Q1R1S1P2Q2R2S2

    Draw both axes with 2 cm to 2 units, from −10-10 to 1010 across and −12-12 to 1212 up, and label them. Then plot and join each shape:

    • PQRSPQRS: P(−5,−4)P(-5, -4), Q(2,−1)Q(2, -1), R(0,3)R(0, 3), S(−8,4)S(-8, 4).
    • Translation by (3−8)\begin{pmatrix} 3 \\ -8 \end{pmatrix} (add 3 to xx, subtract 8 from yy): P1(−2,−12)P_1(-2, -12), Q1(5,−9)Q_1(5, -9), R1(3,−5)R_1(3, -5), S1(−5,−4)S_1(-5, -4).
    • Enlargement, scale factor −12-\frac12 from OO (halve each coordinate and change its sign): P2(2.5,2)P_2(2.5, 2), Q2(−1,0.5)Q_2(-1, 0.5), R2(0,−1.5)R_2(0, -1.5), S2(4,−2)S_2(4, -2). The image is upside down and on the other side of OO.

    Notice S1S_1 lands exactly on PP.

  2. (b)

    Draw on this graph, indicating the coordinates of all vertices: (i) the quadrilateral PQRSPQRS with vertices P(−5,−4)P(-5, -4), Q(2,−1)Q(2, -1), R(0,3)R(0, 3) and S(−8,4)S(-8, 4); (ii) the image P1Q1R1S1P_1Q_1R_1S_1 of PQRSPQRS under a translation by the vector (3−8)\begin{pmatrix} 3 \\ -8 \end{pmatrix}; (iii) the image P2Q2R2S2P_2Q_2R_2S_2 of PQRSPQRS under an enlargement from the origin with scale factor −12-\frac12.

    Model answer
    −10−8−6−4−2246810−12−10−8−6−4−22468xyPQRSP1Q1R1S1P2Q2R2S2

    Plot and join each shape, labelling every vertex with its coordinates: PQRSPQRS with P(−5,−4)P(-5, -4), Q(2,−1)Q(2, -1), R(0,3)R(0, 3), S(−8,4)S(-8, 4); the translation P1(−2,−12)P_1(-2, -12), Q1(5,−9)Q_1(5, -9), R1(3,−5)R_1(3, -5), S1(−5,−4)S_1(-5, -4) (dashed; S1S_1 lands on PP); and the enlargement with scale factor −12-\frac12 from OO, P2(2.5,2)P_2(2.5, 2), Q2(−1,0.5)Q_2(-1, 0.5), R2(0,−1.5)R_2(0, -1.5), S2(4,−2)S_2(4, -2) (highlighted), which is upside down on the other side of OO.

  3. (c)

    Find the equation of the line P1SP_1S.

    Show the answer

    8x+3y+52=08x + 3y + 52 = 0

Try it on a graph

PQRS (blue), the translated image (red), the enlarged image (green).

Worked solution (try it first)

(a)

  1. With 2 cm to 2 units, each 1 cm is 1 unit.
  2. Draw the xx-axis from −10-10 to 1010 and the yy-axis from −12-12 to 1212, crossing at OO, and number them.

(b)(i)

  1. Plot P(−5,−4)P(-5, -4), Q(2,−1)Q(2, -1), R(0,3)R(0, 3) and S(−8,4)S(-8, 4) and join them in order.

(ii)

  1. A translation by (3−8)\begin{pmatrix} 3 \\ -8 \end{pmatrix} adds 3 to each xx and takes 8 from each yy: P1(−2,−12)P_1(-2, -12), Q1(5,−9)Q_1(5, -9), R1(3,−5)R_1(3, -5), S1(−5,−4)S_1(-5, -4).
  2. Plot and join them.

(iii)

  1. An enlargement from the origin with scale factor −12-\frac12 multiplies each coordinate by −12-\frac12: P2(2.5,2)P_2(2.5, 2), Q2(−1,0.5)Q_2(-1, 0.5), R2(0,−1.5)R_2(0, -1.5), S2(4,−2)S_2(4, -2).
  2. Plot and join them.

(c)

  1. P1(−2,−12)P_1(-2, -12) and S(−8,4)S(-8, 4): gradient =4−(−12)−8−(−2)= \frac{4 - (-12)}{-8 - (-2)}
    =16−6= \frac{16}{-6}
    =−83= -\frac83.
  2. Through S(−8,4)S(-8, 4): y−4=−83(x+8)y - 4 = -\frac83(x + 8).
  3. Multiply by 3: 3y−12=−8x−643y - 12 = -8x - 64, so 8x+3y+52=08x + 3y + 52 = 0.
  4. (Check with P1P_1: −16−36+52=0-16 - 36 + 52 = 0 ✓.)

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