The point-slope form of the equation of line it's y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point the line passing through.
The slope-intercept form of the equation of line it's y = mx + b, where m is the slope and b is the y-intercept of the line.
1. Write the equation of the line that is parallel to y = 2x + 4 and passes through the point (-4, -1).
y=m₁x+b₁ ║ y=m₂x+b₂ ⇔ m₁ = m₂
{Two lines are parallel if their slopes are equal}
y = 2x + 4 ⇒ m₁ = 2 ⇒ m₂ = 2
(-4, -1) ⇒ x₁ = -4, y₁ = -1
point-slope form:
y - (-1) = 2(x - (-4))
y + 1 = 2(x + 4)
y + 1 = 2x + 8 {subtact 1 from both sides}
2. Write the equation of the line that is parallel to y = ⅓x - 3 and passes through the point (3, -1).
y = ⅓x - 3 ⇒ m₁ = ⅓ ⇒ m₂ = ⅓
(3, -1) ⇒ x₁ = 3, y₁ = -1
point-slope form:
y - (-1) = ⅓(x - 3)
y + 1 = ⅓x - 1 {subtact 1 from both sides}
3. Write the equation of the line that is perpendicular to y = ¾x - 1 and passes through the point (3, -3).
y=m₁x+b₁ ⊥ y=m₂x+b₂ ⇔ m₁×m₂ = -1
{Two lines are perpendicular if the product of theirs slopes is equal -1}
y = ¾x - 1 ⇒ m₁ = ¾
¾×m₂ = -1 ⇒ m₂ = -⁴/₃
(3, -3) ⇒ x₁ = 3, y₁ = -3
point-slope form:
y - (-3) = -⁴/₃(x - 3)
y + 3 = -⁴/₃x + 4 {subtact 3 from both sides}
4. Write the equation of the line that is perpendicular to y = -x - 5 and passes through the point (7, 3).
y = - x - 5 ⇒ m₁ = -1
-1×m₂ = -1 ⇒ m₂ = 1
(7, 3) ⇒ x₁ = 7, y₁ = 3
point-slope form:
y - 3 = -1(x - 7)
y - 3 = - x + 7 {add 3 to both sides}
