Question

find the angle between the line 3x -4y= 420 & 4x + 3y = 420​

Answer 1

Answer:

θ = 90°

Step-by-step explanation:

property used:

If θ is the angle between two intersecting lines defined by y₁= m₁x₁+c₁ and y₂= m₂x₂+c₂, then, the angle θ is given by

tanθ=±(m₂-m₁) / (1+m₁m₂)

line1 => 3x-4y=420  => 4y = 3x-420  =>  y = $\frac{3x}{4}$ - $\frac{420}{4}$ => y = $\frac{3x}{4}$ - 105

line2 => 4x + 3y = 420​ => 3y = -4x + 420 => y = $\frac{-4x}{3}$ - $\frac{420}{3}$ => y = $\frac{-4x}{3}$ - 140

here,

m₁ = $\frac{3}{4}$ , m₂ = $\frac{-4}{3}$

=> tanθ = $\frac{| \frac{3}{4} - \frac{-4}{3} | }{1+(\frac{3}{4})*(\frac{-4}{3})}$

             = $\frac{| \frac{3}{4} + \frac{4}{3} | }{1-(\frac{3*4}{4*3})}$ = $\frac{| \frac{3}{4} + \frac{4}{3} | }{1-1}$ = $\frac{| \frac{3}{4} + \frac{4}{3} | }{0}$

             = ∞ (inifinity)

here, tanθ = ∞

but we know, tan 90° = ∞

=> θ = 90°