Matematik Tambahan SPM 2018, Kertas 2 (Soalan 3)

Soalan 3 (6 markah):
Penyelesaian secara lukisan berskala tidak diterima.
Rajah 1 menunjukkan segi tiga OCD.

Rajah 1

(a) Diberi luas segi tiga OCD ialah 30 unit², cari nilai h.

(b) Titik Q (2, 4) terletak pada garis lurus CD.
(i) Cari CQ : QD.
(ii) Titik P bergerak dengan keadaan PD = 2 PQ.
Cari persamaan lokus P.

Penyelesaian:
(a)

\begin{array}{l} Diberi luas △ O C D = 30 \\ \frac{1}{2} | \begin{array}{l} 0 h - 6 \\ 0     2      8 \end{array} \begin{array}{l} 0 \\ 0 \end{array} | = 30 \\ | (0) (2) + (h) (8) + (- 6) (0) - (0) (h) - (2) (- 6) - (8) (0) | = 60 \\ | 0 + 8 h + 0 - 0 + 12 - 0 | = 60 \\ | 8 h + 12 | = 60 \\ 8 h + 12 = 60 \\ 8 h = 48 \\ h = 6 \\ atau \\ 8 h + 12 = - 60 \\ 8 h = - 72 \\ h = - 9 (a b a i k a n) \end{array}

(b)(i)

\begin{array}{l} [\frac{6 (m) + (- 6) (n)}{m + n}, \frac{2 (m) + (8) (n)}{m + n}] = (2, 4) \\ \frac{6 m - 6 n}{m + n} = 2 \\ 6 m - 6 n = 2 m + 2 n \\ 4 m = 8 n \\ \frac{m}{n} = \frac{8}{4} \\ \frac{m}{n} = \frac{2}{1} \\ \frac{2 m + 8 n}{m + n} = 4 \\ 2 m + 8 n = 4 m + 4 n \\ 2 m = 4 n \\ \frac{m}{n} = \frac{4}{2} \\ \frac{m}{n} = \frac{2}{1} \\ Oleh itu, C Q = Q D = 2 : 1 \end{array}

(b)(ii)

\begin{array}{l} P D = 2 P Q \\ \sqrt{{(x - 6)}^{2} + {(y - 2)}^{2}} = 2 \sqrt{{(x - 2)}^{2} + {(y - 4)}^{2}} \\ {(x - 6)}^{2} + {(y - 2)}^{2} = 4 [{(x - 2)}^{2} + {(y - 4)}^{2}] \\ x^{2} - 12 x + 36 + y^{2} - 4 y + 4 = 4 [x^{2} - 4 x + 4 + y^{2} - 8 y + 16] \\ x^{2} - 12 x + 36 + y^{2} - 4 y + 4 = 4 x^{2} - 16 x + 16 + 4 y^{2} - 32 y + 64 \\ Persamaan lokus P : \\ 3 x^{2} + 3 y^{2} - 4 x - 28 y + 40 = 0 \end{array}