Find the 10

Find the value of t if 2tx+5y-6=0 and 5x-4y+8=0 are perpendicular and parallel. What angle does each of the lines make with the x-axis, and find the angle between the lines?

Correct answer:

t₁ = 2
A₁ = -38.6598 °
A₂ = -128.6598 °
t₂ = -3.125
B₁ = 51.3402 °
B₂ = -128.6598 °

Step-by-step explanation:

2 t_{x} + 5 y - 6 = 0 5 x - 4 y + 8 = 0 n_{1} = (2 t; 5) n_{2} = (5; - 4) n o r m a l n_{1} . n_{2} = 0 2 \cdot t_{1} \cdot 5 + 5 \cdot (- 4) = 0 10 t_{1} = 20 t_{1} = \frac{2 0}{1 0} = 2 t_{1} = 2

A_{1} = \frac{1 8 0 °}{π} \cdot arctan \frac{5}{2 \cdot t _{1}} - 90 = \frac{1 8 0 °}{π} \cdot arctan \frac{5}{2 \cdot 2} - 90 = - 38.659 8^{\circ} = - 38 ° 3 9^{'} 35 "

A_{2} = \frac{1 8 0 °}{π} \cdot arctan \frac{- 4}{5} - 90 = - 128.659 8^{\circ} = - 128 ° 3 9^{'} 35 "

p a r a l l e l n_{1} = k \cdot n_{2} k = 5 / (- 4) = - \frac{5}{4} = - 1 \frac{1}{4} = - 1.25 2 \cdot t_{2} = 5 / (- 4) \cdot 5 2 t_{2} = - 6.25 t_{2} = \frac{- 6 . 2 5}{2} = - 3.125 t_{2} = \frac{- 2 5}{8} = - 3.125

B_{1} = \frac{1 8 0 °}{π} \cdot arctan \frac{5}{2 \cdot t _{2}} - 90 = \frac{1 8 0 °}{π} \cdot arctan \frac{5}{2 \cdot ( - 3 . 1 2 5 )} - 90 = 51.340 2^{\circ} = 51 ° 2 0^{'} 25 "

B_{2} = \frac{1 8 0 °}{π} \cdot arctan \frac{- 4}{5} - 90 = - 128.659 8^{\circ} = - 128 ° 3 9^{'} 35 "

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