Uji Minda 9.1b – Buku Teks Matematik Tingkatan 3 Bab 9 (Garis Lulus)

Soalan 1:
$$
\text { Tuliskan persamaan garis lurus berikut dalam bentuk }
$$ $$
\frac{x}{a}+\frac{y}{b}=1 \text { dan } y=m x+c
$$
$$
\text { (a) } 3 x-4 y=24
$$
$$
\text { (b) } 7 x+2 y=28
$$
$$
\text { (c) } 5 x-3 y=15
$$
$$
\text { (d) }-2 x+3 y=9
$$

Penyelesaian:
(a)(i)
$$
\begin{gathered}
3 x-4 y=24 \\
\frac{3 x}{24}-\frac{4 y}{24}=\frac{24}{24} \\
\frac{x}{8}-\frac{y}{6}=1
\end{gathered}
$$
(a)(ii)
$$
\begin{aligned}
3 x-4 y & =24 \\
-4 y & =-3 x+24 \\
\frac{-4 y}{(-4)} & =\frac{-3 x}{(-4)}+\frac{24}{(-4)} \\
y & =\frac{3}{4} x-6
\end{aligned}
$$

(b)(i)
$$
\begin{gathered}
7 x+2 y=28 \\
\frac{7 x}{28}+\frac{2 y}{28}=\frac{28}{28} \\
\frac{x}{4}+\frac{y}{14}=1
\end{gathered}
$$
(b)(ii)
$$
\begin{aligned}
7 x+2 y & =28 \\
2 y & =-7 x+28 \\
\frac{2 y}{2} & =\frac{-7 x}{2}+\frac{28}{2} \\
y & =-\frac{7}{2} x+14
\end{aligned}
$$

(c)(i)
$$
\begin{gathered}
5 x-3 y=15 \\
\frac{5 x}{15}-\frac{3 y}{15}=\frac{15}{15} \\
\frac{x}{3}-\frac{y}{5}=1
\end{gathered}
$$
(c)(ii)
$$
\begin{aligned}
5 x-3 y & =15 \\
-3 y & =-5 x+15 \\
\frac{-3 y}{(-3)} & =\frac{-5 x}{(-3)}+\frac{15}{(-3)} \\
y & =\frac{5}{3} x-5
\end{aligned}
$$

(d)(i)
$$
\begin{aligned}
& -2 x+3 y=9 \\
& \frac{-2 x}{9}+\frac{3 y}{9}=\frac{9}{9} \\
& -\frac{2 x}{9}+\frac{y}{3}=1
\end{aligned}
$$
(d)(ii)
$$
\begin{aligned}
-2 x+3 y & =9 \\
3 y & =2 x+9 \\
\frac{3 y}{3} & =\frac{2 x}{3}+\frac{9}{3} \\
y & =\frac{2}{3} x+3
\end{aligned}
$$

Soalan 2:
Tuliskan  persamaan  garis  lurus  berikut dalam bentuk ax + by = c  dan y = mx + c.
$$
\text { (a) } \frac{x}{4}+\frac{y}{3}=1
$$
$$
\text { (b) }-\frac{x}{3}+\frac{y}{6}=1
$$
$$
\text { (c) } \frac{3 x}{2}+\frac{y}{6}=1
$$
$$
\text { (d) } \frac{2 x}{3}-\frac{y}{4}=1
$$

Penyelesaian:
(a)(i)
$$
\begin{aligned}
\frac{x}{4}+\frac{y}{3} & =1 \\
\frac{3 x+4 y}{4(3)} & =1 \\
3 x+4 y & =12
\end{aligned}
$$
(a)(ii)
$$
\begin{aligned}
\frac{x}{4}+\frac{y}{3} & =1 \\
\frac{y}{3} & =-\frac{x}{4}+1 \\
\frac{(3) y}{3} & =-\frac{x(3)}{4}+1(3) \\
y & =-\frac{3}{4} x+3
\end{aligned}
$$

(b)(i)
$$
\begin{aligned}
-\frac{x}{3}+\frac{y}{6} & =1 \\
\frac{-6 x+3 y}{3(6)} & =1 \\
-6 x+3 y & =18
\end{aligned}
$$
(b)(ii)
$$
\begin{aligned}
-\frac{x}{3}+\frac{y}{6} & =1 \\
\frac{y}{6} & =\frac{x}{3}+1 \\
\frac{(6) y}{6} & =\frac{x(6)}{3}+1(6) \\
y & =2 x+6
\end{aligned}
$$

(c)(i)
$$
\begin{gathered}
\frac{3 x}{2}+\frac{y}{6}=1 \\
\frac{18 x+2 y}{2(6)}=1 \\
18 x+2 y=12 \\
9 x+y=6
\end{gathered}
$$
(c)(ii)
$$
\begin{aligned}
\frac{3 x}{2}+\frac{y}{6} & =1 \\
\frac{y}{6} & =-\frac{3 x}{2}+1 \\
\frac{(6) y}{6} & =-\frac{3 x(6)}{2}+1(6) \\
y & =-9 x+6
\end{aligned}
$$

(d)(i)
$$
\begin{gathered}
\frac{2 x}{3}-\frac{y}{4}=1 \\
\frac{8 x-3 y}{3(4)}=1 \\
8 x-3 y=12
\end{gathered}
$$
(d)(ii)
$$
\begin{aligned}
\frac{2 x}{3}-\frac{y}{4} & =1 \\
-\frac{y}{4} & =-\frac{2 x}{3}+1 \\
-\frac{(-4) y}{4} & =-\frac{2 x(-4)}{3}+1(-4) \\
y & =\frac{8}{3} x-4
\end{aligned}
$$

Soalan 3:
$$
\text { Tuliskan persamaan garis lurus berikut dalam bentuk } a x+b y=c \text { dan } \frac{x}{a}+\frac{y}{b}=1 \text {. }
$$
(a) y = 2x + 6
(b) y = 3x – 12
(c) y = –x + 5
(d) y = –2x – 4

Penyelesaian:
(a)(i)
$$
\begin{aligned}
y & =2 x+6 \\
-2 x+y & =6
\end{aligned}
$$
(a)(ii)
$$
\begin{aligned}
y & =2 x+6 \\
-2 x+y & =6 \\
\frac{-2 x}{6}+\frac{y}{6} & =\frac{6}{6} \\
-\frac{x}{3}+\frac{y}{6} & =1
\end{aligned}
$$

(b)(i)
$$
\begin{aligned}
y & =3 x-12 \\
-3 x+y & =-12 \\
3 x-y & =12
\end{aligned}
$$
(b)(ii)
$$
\begin{aligned}
y & =3 x-12 \\
-3 x+y & =-12 \\
\frac{-3 x}{(-12)}+\frac{y}{(-12)} & =\frac{-12}{(-12)} \\
\frac{x}{4}-\frac{y}{12} & =1
\end{aligned}
$$

(c)(i)
$$
\begin{aligned}
y & =-x+5 \\
x+y & =5
\end{aligned}
$$
(c)(ii)
$$
\begin{aligned}
y & =-x+5 \\
x+y & =5 \\
\frac{x}{5}+\frac{y}{5} & =\frac{5}{5} \\
\frac{x}{5}+\frac{y}{5} & =1
\end{aligned}
$$

(d)(i)
$$
\begin{aligned}
y & =-2 x-4 \\
2 x+y & =-4
\end{aligned}
$$
(d)(ii)
$$
\begin{aligned}
y & =-2 x-4 \\
2 x+y & =-4 \\
\frac{2 x}{(-4)}+\frac{y}{(-4)} & =\frac{-4}{(-4)} \\
-\frac{x}{2}-\frac{y}{4} & =1
\end{aligned}
$$