Question

Write the equation of a line that is perpendicular to y=0.3x+6 and that passes through the point (3,-8)

Answer 1

First, find the perpendicular slope

y = 0.3x + 6   ⇒ m = 0.3  = $\frac{3}{10}$

perpendicular slope =  $-\frac{10}{3}$

Next, input the point (x₁, y₁) and slope (m) into the Point-Slope formula

  y - y₁ = m(x - x₁)

y - (-8) = $-\frac{10}{3}$(x - 3)

 y + 8 = $-\frac{10}{3}x$ + 10

      y = $-\frac{10}{3}$ + 2