Question

write the equation of a line that is perpendicular to the given line and that passes through the given point. y-3=8/3(x+2);(-2,3)

Answer 1

`\large\boxed{y-3=-(3)/(8)(x+2)}`

Explanation:

Ths point-slope form of an equation of a line:

`y-y_1=m(x-x_1)`

m - slope

(x₁, y₁) - point

We have the equation:

`y-3=(8)/(3)(x+2)`

Therefore the slope is

`m=(8)/(3)`

Let k: y = m₁x + b₁ and l: y = m₂x + b₂.

l ⊥ k ⇔ m₁m₂ = -1 ⇒ m₂ = -1/m₁

Therefore

`m_1=(8)/(3)\Rightarrow m_2=-(1)/((8)/(3))=-(3)/(8)`

The line passes throguh the point (-2, 3).

We have the equation in point-slope form:

`y-3=-(3)/(8)(x-(-2))\\\\y-3=-(3)/(8)(x+2)`