**Soalan 8:**

Rajah di bawah menunjukkan sisi empat

*ABCD*. Titik

*C*terletak pada paksi-

*y*.

Persamaan garis lurus

*AD*ialah 2

*y*= 5

*x*– 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*.

*Penyelesaian:***(a)(i)**

**(a)(ii)**

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

**(b)**

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