Answer:
[tex]\large\boxed{y-3=-\dfrac{3}{8}(x+2)}[/tex]
Step-by-step explanation:
Ths point-slope form of an equation of a line:
[tex]y-y_1=m(x-x_1)[/tex]
m - slope
(x₁, y₁) - point
We have the equation:
[tex]y-3=\dfrac{8}{3}(x+2)[/tex]
Therefore the slope is
[tex]m=\dfrac{8}{3}[/tex]
Let k: y = m₁x + b₁ and l: y = m₂x + b₂.
l ⊥ k ⇔ m₁m₂ = -1 ⇒ m₂ = -1/m₁
Therefore
[tex]m_1=\dfrac{8}{3}\Rightarrow m_2=-\dfrac{1}{\frac{8}{3}}=-\dfrac{3}{8}[/tex]
The line passes throguh the point (-2, 3).
We have the equation in point-slope form:
[tex]y-3=-\dfrac{3}{8}(x-(-2))\\\\y-3=-\dfrac{3}{8}(x+2)[/tex]
