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What is the relationship between ax+by = c and bx -ay = c if a,b,c are not zero?
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High School Teacher
Given the equations:
ax + by = c
bx - ay = c
We notice that the above are equations of two lines.
We need to find the relation between the lines (i.e perpendicular, parallel, or neither).
To determine the relations, we will rewrite the equations into the slope form " y= mx + a) where m is the slope.
==> ax + by = c
==> by = -ax + c
==> y (-a/b) x + c/b...............(1)
Then slope for equation (1) is m1 = -a/b.
Now we will rewrite the second equation.
==> bx -ay = c
==> -ay = -bx + c
==> y= (b/a)x - c/a..............(2)
The slope for equation (2) is m2= b/a
Now we notice that m1 and m2 are NOT equal, then they are not parallel.
However, m1*m2 = -a/b * b/a = -1
Then, the relationship between the lines is that they are perpendicular.
Posted by hala718 on January 19, 2011 at 11:19 AM (Answer #1)
You can examine these equations more closely by putting them in standard form: y = mx + b
ax + by = c --> y = -a/bx + c/b
bx - ay = c --> y = b/a x - c/a
So the slope of the first equation is m1 = -a/b, and the slope of the second equation is m2 = b/a = -1/m1
When the slopes of two lines are the negative inverse of one another, they are perpendicular lines.
Therefore, the relationship between these two lines is that they are perpendicular.
Posted by kjcdb8er on January 19, 2011 at 11:24 AM (Answer #2)
High School Teacher
We know that the equations of the line are
ax+by = c...(1).
bx-ay = c....(2).
We rewrite both equations in the slope intercept form y = mx+k for, where m is the slope of the line and k = the y intercept.
The line (1): ax+by = c.
=> by = -ax+c
=> by/b = (-a/b)x+(c/b).
y = (-a/b)x+(a/b).........(i)
The line (2): bx-ay = c.
-ay = -ax+c.
-ay/-a = (-b/-a)x+(c/-a).
=> y = (b/a)x -c/a..........(ii)
So from (1) and (2)the slopes of the lines (1) and (2) are -a/b and b/a. The product of the slopes is -1. This indicates that the lines are perpendicular.
The lines ax-by = c and bx-ay = c are perpendicular.
Posted by neela on January 19, 2011 at 11:31 AM (Answer #4)
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