3x - 2y = 6 , 2x + 3= 6 are perpendicular lines but i need 3 ordered pairs for each. I used the m= y2-y1/x2-x1 method to establish that the slopes = -1. Any suggestions for ordered pairs for each line is much appreciated.
You must have written the second equation incorrectly. It needs a "y" term. It the productsa of the slopes, not the slopes themselves that must be -1.
Picking an ordered pair is just a matter of picking x and computing y, for any x that you want. Two ordered pairs define a straight line.
To find three ordered pairs for each line, we need to solve the equations and find the corresponding values of x and y. Let's start with the equation 3x - 2y = 6.
Step 1: Solve for y in terms of x:
3x - 2y = 6
-2y = -3x + 6
y = (3/2)x - 3
Now, let's substitute different values of x into this equation to find the corresponding y values.
For Line 1:
Ordered Pair 1: Let x = 0
y = (3/2)(0) - 3
y = -3
Ordered Pair 2: Let x = 1
y = (3/2)(1) - 3
y = -1.5 - 3
y = -4.5
Ordered Pair 3: Let x = 2
y = (3/2)(2) - 3
y = 3 - 3
y = 0
Therefore, the three ordered pairs for Line 1 are (0, -3), (1, -4.5), and (2, 0).
Now, let's move to the second equation: 2x + 3 = 6.
Step 2: Solve for x:
2x + 3 = 6
2x = 6 - 3
2x = 3
x = 3/2
x = 1.5
Substitute different values of x into the equation to find the corresponding y values:
For Line 2:
Ordered Pair 1: Let x = 0
2(0) + 3 = 6
3 = 6 (False)
Ordered Pair 2: Let x = 1
2(1) + 3 = 6
2 + 3 = 6 (False)
Ordered Pair 3: Let x = 2
2(2) + 3 = 6
4 + 3 = 6 (False)
It seems that the second equation does not define a line. Therefore, it is not possible to find three ordered pairs for Line 2.
In conclusion, the three ordered pairs for Line 1 are (0, -3), (1, -4.5), and (2, 0). However, we cannot find three ordered pairs for Line 2 as it does not define a line.