Equation with perpendicular line, please help...

If the lines are perpendicular, the product of their slopes must equal to -1. By converting the equations into y = mx + c, you can find the slope of each line easily.

A) -3x+4y=4, 4x-3y=6
4y = 3x+4, 3y = 4x-6
y = 3/4 x + 1, y = 4/3 x -2
Product of slope = 3/4 X 4/3 = 1

B) 3x-2y=7, 2x+3y=5
2y = 3x-7, 3y = -2x+5
y = 3/2 x -7/2, y = -2/3 x + 5/3
Product of slope = 3/2 X -2/3 = -1

Hence, pair of lines in B are perpendicular

A) -3x+4y=4, 4x-3y=6
4y = 3x+4, 3y = 4x-6
y = 3/4 x + 1, y = 4/3 x -2
Product of slope = 3/4 X 4/3 = 1

B) 3x-2y=7, 2x+3y=5
2y = 3x-7, 3y = -2x+5
y = 3/2 x -7/2, y = -2/3 x + 5/3
Product of slope = 3/2 X -2/3 = -1

Which of the following pairs of equations have lines that are perpendicular?
A) -3x+4y=4, 4x-3y=6
B) 3x-2y=7, 2x+3y=5
The slope of the line Ax+By+C=0 is -A/B
if pairs of equations of lines that are perpendicular
Then m1*m2=-1
For A
m1=3/4
m2=4/3
m1*m2=-1
So the lines are not perpendicular
For B
m1=3/2
m2=-2/3
m1*m2=-1
So the lines are perpendicular
the answer is B



2008-01-31 12:48:20 補充：
Some typing mistake For Am1=3/4 m2=4/3m1*m2=1