bentuklah persamaan linear yang garisnya melalui pasangan titik-titik berikut
(4 : –2) & (0 : 6)
(2 : 2) & (10 : 12)
Jawab
A (4 : –2) & B (0 :6)
Dik : X1 = 4
X2 = 0
Y1 = –2
Y2 = 6
Penyelesaian:
X – X1 Y – Y1
————— = —————
X2 – X1 Y2 – Y1
X – 4 Y – (–2)
————— = —————
0 – 4 6 – (–2)
X – 4 Y + 2
————— = —————
0 – 4 6 + 2
X + 4 Y + 2
————— = —————
–4 8
8 (X –4) = –4 (Y + 2)
8X – 32 = –4Y –8
8X = –4Y + 32 + 8
8X = –4Y + 24
X = –4Y + 24
—————
8
X = –4Y 24
—— + ——
8 8
X = –4
—— + 3
8Y
(2 : 2) & (10 : 12)
Dik : X1 = 4
X2 = 0
Y1 = –2
Y2 = 6
Penyelesaian:
X – X1 Y – Y1
————— = —————
X2 – X1 Y2 – Y1
X – 2 Y – 2
————— = —————
20 – 2 12 – 2
X + 2 Y – 2
————— = —————
8 10
10 (X – 2) = 8 (Y – 2)
10X – 20 = 8Y – 16
10X = 8Y + 4
X = 8Y + 4
—————
10
X = 8Y 4
—— + ——
10 10
untuk setiap tittik koordinat (X : Y) & koofisien kemiringan berikut ini. Carilah persamaan garis lurus!
(–1 : 3) & b= 5
(–1 : 3) & b= 2
Jawab
(–1 : 3) & b= 5
Y – Y1 = b (X – X1)
Dik: X1 = –1
Y1 = 3
b = 5
penyelesaian:
Y –Y1 = b (X –X1)
Y – 3 = 5 (X – (–1)
Y – 3 = 5 (X + 1)
Y – 3 = 5 X + 5
Y = 5 X + 5 + 3
Y = 5X + 8
b.
carilah kemiringan / lereng / koofisien arah berikut ini, dengan diketahui titik-titik koordinat berikut ini!
A (–5 : 2) & B (5 : 6)
A (3 : –5) & B (8 : 9)
Jawab
A (–5 : 2) & B (5 : 6)
∆Y Y2 – Y1 6 – 2 4 4 2
b = —— = ————— = ————— = ————— = —— = —
∆X X2 – X1 5 – (– 5) 5 + 5 10 5
A (3 : –5) & B (8 : 9)
∆Y Y2 – Y1 9 – (–5) 9 + 5 14 4
b = —— = ————— = ————— = ————— = —— = 2 —
∆X X2 – X1 8 – 3 5 5 5
fungsi permintaan sebuah barang di tunjukan oleh perusahaan Q = 75 – 3P
-
Q = 75 – 3P
If, Q = 10
Q = 75 – 3 (10)
Q = 75 – 30
Q = 45
Q = 75 – 3P
If, P = 0
Q = 75 – 3 (0)
Q = 75 – 0
Q = 75
Q = 75 – 3P
If, Q = 15
Q = 75 – 3P
15 = 75 – 3P
15 – 75 = – 3P
– 60 = – 3
60
P = — = 20
3
Q = 75 – 3P
If, Q = 0
Q = 75 – 3P
0 = 75 – 3P
–75 = –3P
P = 25
dik : P1 = 5000
Q1 = 30 orang
P2 = 8000
Q2 = 10 orang
Q1 = a – bp
Q2 = a – bp
Penyelesaian
Q1 = a – bp
30 = a – 5000 b
Q2 = a – bp
10 = a – 8000 b
30 = a – 5000 b
10 = a – 8000 b
———————— –
20 = 3000 b
20 1
b = —— = 0,007 / —
3000 150
Jadi, Q1 = a – bp
1
30 = a – 0,007 (5000) / ( — x 5000 )
50
30 = a – 33,3
30 + 33,3 = a
63,3 = a
Fungsinya: Q = a – bp
1
Q = 63,3 – 0,007p / 63,3 – — p
50
Untuk Q2 & P2
Q2 = a – bp2
10 = a – 8000 b
b. kurvanya
jika Q = 0
P = 0 → Q = 63,3 – 1/ 50 (0)
Q = 63,3
1
Q = 63,3 – — p
50
1
0 = 63,3 – — p
50
1
— p = 63,3 – 0
50
p = 63,3
——
1
— p
50
p = 63,3 X 150
P = 9.495
dik : P1 = 300
Q1 = 1000
P2 = 300 + 100 = 400
Q2 = 1000 + 400 = 1400
Penyelesaian:
Q 1= –a + bp1
1000 = –a + 300p
Q2 = –a + bp2
1400 = –a + 400 b
1000 = –a + 300 b
1400 = –a + 400 b
———————— –
–400 = –100 b
b = –400
—— b
–100
jadi : Q1 = –a + 300 b
1000 = –a + 4 (300)
1000 = –a + 1200
a = 1200 – 1000
a = 200
fungsinya : Q1 = –a + bp1
Q = –200 + 4P
untuk Q2 & P2
Q2 = –a + bp2
1400 = –a + 4b
jika Q = 0
Q = –a + bp
0 = –200 + 4p
4p = –200 – 0
200
p = —— = 50
4
P = 0
Q = –a + bp
Q = –200 + 4 (0)
Q = –200 + 0
Q = –200
P =15 – Q (permintaan)
P = 3 + 0,5 Q (penawaran)
D= 5
15 – Q = 3 + 0,5 Q
–Q – 0,5Q = –15 + 3
–1,5Q = –12
Q = 8
P = 15 – Q / P = 3 + 0,5 Q
P = 15 – 8 / P = 3 + 0,5 (8)
p = 7 / P = 3 + 4 = 7
jadi E (7 : 7)
kurva
P = 15 – Q (permintaan)
if, P = 0 → Q
P = 15 – Q
0 = 15 – Q
Q = 15
if, Q = 0 → P
P = 15 – Q
P = 15 – 0
P = 15
P = 3 + 0,5 Q (penawaran)
if, Q = 0 → Q?
P = 3 + 0,5 (Q)
0 = 3 + 0,5 Q
–0,5Q = 3
Q = 3 1
— = 3 X – —
5/10 5
= –6
if, q = 0 → p
p = 3 + 0,5 (Q)
P = 3 + 0,5 (0)
P = 3
