Let y = [tex] \frac{4}{3} [/tex] x be L₁ & the unknown be L₂
Now, if L₂ is perpendicular to L₁ then the product of their gradient (m) is -1. This implies that the gradient of L₂ is the negative reciprocal of L₁:
∴ if m of L₁ = [tex] \frac{4}{3} [/tex]
then m of L₂ = [tex] - \frac{3}{4} [/tex]
Now since L₂ passes through the line (4, 2),
then by using the point-slope form (y - y₁) = m (x - x₁)
⇒ (y - 2) = [tex] - \frac{3}{4} [/tex] (x - 4)
⇒ 4 (y - 2) = -3 (x - 4)
⇒ 4y - 8 = -3x + 12
⇒ 4y + 3x = 20
∴ the line that passes through the point (4,2) that is perpendicular to the line y = 4/3x is characterized by the equation 4y + 3x = 20
