Question

I don't know how to do this please help. Thank you

Answer 1

Let y = $\frac{4}{3}$ 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 = $\frac{4}{3}$

then m of L₂ = $- \frac{3}{4}$

Now since L₂ passes through the line (4, 2),

then by using the point-slope form (y - y₁) = m (x - x₁)
     ⇒      (y - 2) =  $- \frac{3}{4}$ (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