Anna's method is described below
The line Ax+By+C=0 will always be perpendicular to the line Bx-Ay+C=0
Is Anna's method reasonable?? Explain why or why not
Anna claims she has discovered a method for determining the general form equation of a line that is perpendicular to a give line?
2015-11-22 5:07 am
回答 (3)
2015-11-22 5:59 am
A perpendicular line has a negative inverse slope.
Slope is m in the equation y = mx +b
Ax+By+C=0
by = -ax
y = -a/b x
⊥ would be +b/a
Looking at the 2nd equation:
Bx-Ay+C=0
-ay = -bx
y = -b/-a x
y = b/a x
Yes, they are ⊥
Slope is m in the equation y = mx +b
Ax+By+C=0
by = -ax
y = -a/b x
⊥ would be +b/a
Looking at the 2nd equation:
Bx-Ay+C=0
-ay = -bx
y = -b/-a x
y = b/a x
Yes, they are ⊥
2015-11-22 5:09 am
Anna sounds like a real khunt
2015-11-22 5:26 am
Ax + By +C=0 this gives
y = -A x/D - C/D.
slope
-A/D = tan theta. New line should be perpendicular to it
-A/D = tan (theta+pi/2)= sin(theta+pi/2)/cos( theta+pi/2)= cos theta/(-sin theta)= - 1/ tan theta. The new slope is D/A. We write
y = D/A x + f, This gives
Ay - Dx -fA=0. This is the new line.
y = -A x/D - C/D.
slope
-A/D = tan theta. New line should be perpendicular to it
-A/D = tan (theta+pi/2)= sin(theta+pi/2)/cos( theta+pi/2)= cos theta/(-sin theta)= - 1/ tan theta. The new slope is D/A. We write
y = D/A x + f, This gives
Ay - Dx -fA=0. This is the new line.
收錄日期: 2021-04-21 15:24:36
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