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

Answer:

$\large\boxed{y-3=-\dfrac{3}{8}(x+2)}$

Step-by-step 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:

$y-3=\dfrac{8}{3}(x+2)$

Therefore the slope is

$m=\dfrac{8}{3}$

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

l ⊥ k ⇔ m₁m₂ = -1 ⇒ m₂ = -1/m₁

Therefore

$m_1=\dfrac{8}{3}\Rightarrow m_2=-\dfrac{1}{\frac{8}{3}}=-\dfrac{3}{8}$

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

We have the equation in point-slope form:

$y-3=-\dfrac{3}{8}(x-(-2))\\\\y-3=-\dfrac{3}{8}(x+2)$