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Line rotated by an angle ±θ from a given line whose slope is mb
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The slope of a given line is: mb = tan θb
or θb = tan-1 mb
The angles of the lines which are inclined by an angle θ from the original line are: θ1 = θb − θ
and θ2 = θb + θ
| Hence the slopes of the lines are: |
m1 = tan θ1 = tan (θb − θ) |
| m2 = tan θ2 = tan (θb + θ) |
And the inclined lines equations which are passing through the point
(x0 , y0) are:
| y − y0 = m1(x − x0) = tan(θb − θ)(x − x0) |
| y − y0 = m2(x − x0) = tan(θb + θ)(x − x0) |
Or expressed by mb
| y − y0 = tan(tan-1 mb ± θ)(x − x0) = tan(θb ± θ)(x − x0) |
the plus and minus signs stand for the two lines equations which lies either side of the original line.
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Example: find the equation of a line that is inclines by 15 degrees to the right of the line
2x −y + 3 = 0 and is passing through the point (1 , 5).
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| The slope of the given line is: |
mb = 2 |
| The angle of the given line is: |
θb = tan-1 2 = 63.43 degree |
| The angle of right incline line is: |
θ1 = 63.43 − 15 = 48.43 degree |
| The slope of right incline line is: |
mθ = tan 48.43 = 1.13 |
Hence the line equation is:
| (y − 5) = tan(63.43 − 15)(x − 1) |
| (y − 5) = 1.13(x − 1) |
| 1.13x − y + 3.87 = 0 |
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