6.8.8 Geometri Koordinat, SPM Praktis (Kertas 2)

Soalan 8:
Rajah di bawah menunjukkan sisi empat ABCD. Titik C terletak pada paksi-y.

Persamaan garis lurus AD ialah 2y = 5x – 21.
(a) Cari
(i) persamaan garis lurus AB,
(ii) koordinat A,
(b) Titik P bergerak dengan keadaan jaraknya dari titik D sentiasa 5 unit.
Cari persamaan lokus P.

$\begin{array}{l}2y=5x-21\\ y=\frac{5}{2}x-\frac{21}{2}\\ m_{AD}=\frac{5}{2}\\ m_{AB}\times m_{AD}=-1\\ m_{AB}\times \frac{5}{2}=-1\\ m_{AB}=-\frac{2}{5}\\ \\ \text{Persamaan }AB\\ y-y_1=m_{AB}\left(x-x_1\right)\\ y+1=-\frac{2}{5}\left(x+2\right)\\ 5y+5=-2x-4\\ 5y=-2x-9\end{array}$

(a)(ii)
$\begin{array}{l}2y=5x-21\mathrm{..........}\left(1\right)\\ 5y=-2x-9\mathrm{..........}\left(2\right)\\ \left(1\right)\times 5:10y=25x-105\mathrm{..........}\left(3\right)\\ \left(2\right)\times 2:10y=-4x-18\mathrm{..........}\left(4\right)\\ \left(2\right)-\left(4\right):0=29x-87\\ x=3\\ \text{Dari }\left(1\right),\\ 2y=15-21\\ 2y=-6\\ y=-3\\ A=\left(3\text{ , }-3\right)\end{array}$

(b)
$\begin{array}{l}y=2,\\ 4=5x-21\\ 5x=25\\ x=5\\ \text{Titik }D=\left(5,2\right)\\ PD=5\\ \sqrt{{\left(x-5\right)}^2+{\left(y-2\right)}^2}=5\\ {\left(x-5\right)}^2+{\left(y-2\right)}^2=25\\ x^2-10x+25+\left(y^2-4y+4\right)=25\\ x^2+y^2-10x-4y+4=0\end{array}$