Question

find the angel between the straight line 2x+y=4 and x+3y=5..



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Answer 1

Answer:

2x−y=4 , so

y = 2x - 4 and the slope of this line is 2

so the angle of this line with the positive x-axis is tan−1(2)

3x+y=3, so

y = 3 - 3x, and the slope of this line is -3

so the angle of this line with the positive x-axis is tan−1(−3)

since tan−1(−3) is in the second quadrant and tan−1(2) is in the first quadrant

the acute angle between the two lines is therefore tan−1(−3)−tan−1(2)=45°

How will I find the angle between the two lines given by 2x-y=4 and 3x+y=3?

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Peter Chan

Answered 2 years ago

2x−y=4 , so

y = 2x - 4 and the slope of this line is 2

so the angle of this line with the positive x-axis is tan−1(2)

3x+y=3 , so

y = 3 - 3x, and the slope of this line is -3

so the angle of this line with the positive x-axis is tan−1(−3)

since tan−1(−3) is in the second quadrant and tan−1(2) is in the first quadrant

the acute angle between the two lines is therefore tan−1(−3)−tan−1(2)=45°

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You can verify this final result by using the following compound angle identity for tangent . . .

tan(A−B)≡tanA−tanB1+tanAtanB

thus we can let C=tan−1(−3)−tan−1(2)

then tanC=tan[tan−1(−3)−tan−1(2)]

tanC=tan[tan−1(−3)]−tan[tan−1(2)]1+tan[tan−1(−3)]tan[tan−1(2)]

tanC=−3−21+(−3)(2)=−51–6=−5−5=1

tanC=1

so C must be 45°

Explanation:

hope it helps

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Answer 2

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