Find the 10

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

Result

t1 =  2
A1 =  -38.66 °
A2 =  -128.66 °
t2 =  -3.125
B1 =  51.34 °
B2 =  -128.66 °

Solution:

2tx+5y−6=0 5x−4y+8=0  n1=(2t;5) n2=(5;−4)   normal n1.n2=0   2⋅ t1⋅ 5+5⋅ (−4)=0  10t1=20  t1=2 =22tx+5y-6=0 \ \\ 5x-4y+8=0 \ \\ \ \\ n_{1}=(2t; 5) \ \\ n_{2}=(5; -4) \ \\ \ \\ \ \\ normal \ \\ n_{1}.n_{2}=0 \ \\ \ \\ \ \\ 2 \cdot \ t_{1} \cdot \ 5+5 \cdot \ (-4)=0 \ \\ \ \\ 10t_{1}=20 \ \\ \ \\ t_{1}=2 \ \\ =2
A1=180∘π⋅arctan⁡52⋅ t1−90=180∘π⋅arctan⁡52⋅ 2−90≐−38.6598≐−38.66∘≐−38∘39′35"A_{1}=\dfrac{ 180^\circ }{ \pi } \cdot \arctan \dfrac{ 5 }{ 2 \cdot \ t_{1} } -90=\dfrac{ 180^\circ }{ \pi } \cdot \arctan \dfrac{ 5 }{ 2 \cdot \ 2 } -90 \doteq -38.6598 \doteq -38.66 ^\circ \doteq -38^\circ 39'35"
A2=180∘π⋅arctan⁡−45−90≐−128.6598=−128.66∘≐−128∘39′35"A_{2}=\dfrac{ 180^\circ }{ \pi } \cdot \arctan \dfrac{ -4 }{ 5 } -90 \doteq -128.6598=-128.66 ^\circ \doteq -128^\circ 39'35"
parallel n1=k⋅ n2 k=5/(−4)=−54=−1.25   2⋅ t2=5/(−4)⋅ 5  2t2=−6.25  t2=−258=−3.125 =−258=−3.125parallel \ \\ n_{1}=k \cdot \ n_{2} \ \\ k=5/(-4)=- \dfrac{ 5 }{ 4 }=-1.25 \ \\ \ \\ \ \\ 2 \cdot \ t_{2}=5/(-4) \cdot \ 5 \ \\ \ \\ 2t_{2}=-6.25 \ \\ \ \\ t_{2}=\dfrac{ -25 }{ 8 }=-3.125 \ \\ =- \dfrac{ 25 }{ 8 }=-3.125
B1=180∘π⋅arctan⁡52⋅ t2−90=180∘π⋅arctan⁡52⋅ (−3.125)−90≐51.3402≐51.34∘≐51∘20′25"B_{1}=\dfrac{ 180^\circ }{ \pi } \cdot \arctan \dfrac{ 5 }{ 2 \cdot \ t_{2} } -90=\dfrac{ 180^\circ }{ \pi } \cdot \arctan \dfrac{ 5 }{ 2 \cdot \ (-3.125) } -90 \doteq 51.3402 \doteq 51.34 ^\circ \doteq 51^\circ 20'25"
B2=180∘π⋅arctan⁡−45−90≐−128.6598=−128.66∘≐−128∘39′35"B_{2}=\dfrac{ 180^\circ }{ \pi } \cdot \arctan \dfrac{ -4 }{ 5 } -90 \doteq -128.6598=-128.66 ^\circ \doteq -128^\circ 39'35"

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