7.8.4 Geometri Koordinat, SPM Praktis (Kertas 2)

Soalan 7:
Penyelesaian secara lukisan berskala tidak diterima.
Rajah di bawah menunjukkan segi tiga PRS. Sisi PR bersilang dengan paksi-y pada titik Q.


(a) Diberi PQ : QR = 2 : 3, cari
(i) koordinat P,
(ii) persamaan garis lurus PS,
(iii) luas, dalam unit², segi tiga PRS.

(b) Titik M bergerak dengan keadaan jaraknya dari titik R adalah sentiasa dua kali jaraknya dari titik S.
Cari persamaan lokus M.

$\begin{array}{l}P=\left(\frac{2\left(6\right)+3h}{2+3},\frac{2\left(12\right)+3k}{2+3}\right)\\ \left(0,6\right)=\left(\frac{12+3h}{5},\frac{24+3k}{5}\right)\\ \frac{12+3h}{5}=0\\ 3h=-12\\ h=-4\\ \\ \frac{24+3k}{5}=6\\ \text{3}k=30-24\\ k=2\\ \\ P=\left(-4,2\right)\end{array}$


(a)(ii)
$\begin{array}{l}{m}_{PS}=\frac{2-\left(-6\right)}{-4-2}\\ =-\frac{8}{6}\\ =-\frac{4}{3}\\ \\ \text{Persamaan }PS:\\ y-y_1=-\frac{4}{3}\left(x-2\right)\\ y-\left(-6\right)=-\frac{4}{3}x+\frac{8}{3}\\ 3y+18=-4x+8\\ 3y=-4x-10\end{array}$

(a)(iii)
$\begin{array}{l}\text{Luas }△PRS\\ =\frac{1}{2}|\begin{array}{l}-\text{4 2 6 }\\ \text{2 }-6\text{ 12 }\end{array}\begin{array}{l}-\text{4}\\ 2\end{array}|\\ =\frac{1}{2}|\left(24+24+12\right)-\left(4-36-48\right)|\\ =\frac{1}{2}|60-\left(-80\right)|\\ =70{\text{ unit}}^2\end{array}$

(b)
$\begin{array}{l}\text{Katakan }P=\left(x,y\right)\\ MR=2MS\\ \sqrt{{\left(x-6\right)}^2+{\left(y-12\right)}^2}=2\sqrt{{\left(x-2\right)}^2+{\left(y+6\right)}^2}\\ {\left(x-6\right)}^2+{\left(y-12\right)}^2=4\left[{\left(x-2\right)}^2+{\left(y+6\right)}^2\right]\\ x^2-12x+36+y^2-24y+144=4\left[x^2-4x+4+y^2+12y+36\right]\\ x^2-12x+y^2-24y+180=4x^2-16x+4y^2+48y+160\\ 3x^2+3y^2-4x+72y-20=0\end{array}$

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

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