Two parallel line segments move from Quadrant One to Quadrant Four. Their slopes does not change.
2 answers:
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To solve this, we must first put both lines in Slope Intercept Form (y=mx+b where m is the slope and b is the y-intercept).
y=3x-5 is already in SIF, so we only need to work on the other one.
x+3y=6
-x -x
3y=-x+6
/3 /3
y=-1/3x+2
Now we have both equations in slope intercept form, so we can start graphing from the y-intercepts and just follow the slopes.
When we do this, we will see that the lines meet at an exactly 90° angle. When a pair of lines does this, it means they are perpendicular.
Below I have attached an image that has both lines graphed so that you may visualize it. The green dots show the slopes, while the highlighted areas show the y-intercepts. Note that the lines intersect at a 90° angle, making them perpendicular.
y=3x-5 is already in SIF, so we only need to work on the other one.
x+3y=6
-x -x
3y=-x+6
/3 /3
y=-1/3x+2
Now we have both equations in slope intercept form, so we can start graphing from the y-intercepts and just follow the slopes.
When we do this, we will see that the lines meet at an exactly 90° angle. When a pair of lines does this, it means they are perpendicular.
Below I have attached an image that has both lines graphed so that you may visualize it. The green dots show the slopes, while the highlighted areas show the y-intercepts. Note that the lines intersect at a 90° angle, making them perpendicular.