Penjelasan dengan langkah-langkah:

Persamaan Garis Lurus

Materi 8 semester 1

ax + by + c = 0

sejajar : m = –a/b

tegak lurus : m = b/a

6) melalui (2, –1) sejajar y = 3x + 5

m = 3

x1 = 2

y1 = –1

y = m(x – x1) + y1

y = 3(x – 2) – 1

y = 3x – 7

7) melalui (–3, 2) sejajar 2x + 4y – 9

m = –2/4 = –½

x1 = –3

y1 = 2

y – y1 = m(x – x1)

y – 2 = –½(x + 3)

2(y – 2) = –(x + 3)

2y – 4 = – x – 3

x + 2y – 4 + 3 = 0

x + 2y – 1 = 0

8) melalui (1, 5) tegak lurus y = –3x + 4

m = ⅓

x1 = 1

y1 = 5

y – y1 = m(x – x1)

y – 5 = ⅓(x – 1)

y – 5 = ⅓x – ⅓

y = ⅓x + 5 – ⅓

y = ⅓x + 4⅔

9) melalui (–3, 4) bergradien –2

m = –2

x1 = –3

y1 = 4

y = m(x – x1) + y1

y = –2(x –(–3)) + 4

y = –2x – 6 + 4

y = –2x – 2

2x + y + 2 = 0

10) melalui pusat koordinat (0, 0) bergradien 2

m = 2

x1 = 0

y1 = 0

y – y1 = m(x – x1)

y – 0 = 2(x – 0)

y = 2x

2x – y = 0

11) melalui (–1, 2) tegak lurus 4y = -3x +5

m = 4/3

x1 = –1

y1 = 2

y – y1 = m(x – x1)

y – 2 = 4/3(x + 1)

3(y – 2) = 4(x + 1)

3y – 6 = 4x + 4

4x – 3y + 10 = 0

12) melalui (–2, 5) sejajar 2x + 3y + 6 = 0

m = –2/3

x1 = –2

y1 = 5

y – y1 = m(x – x1)

y – 5 = –⅔(x + 2)

3(y – 5) = –2(x + 2)

3y – 15 = –2x – 4

2x + 3y – 11 = 0