Slopes involving perpendicular lines

Hello,

I'm in minor confusion over slopes involving perpendicular lines.

Consider a rectangular co-ordinate system. The scales I'm using are such that ten small squares on my graph paper represent 1 unit on the x-axis and 2 units on the y-axis. I've drawn a straight line which passes through when . I've now drawn another straight line where both lines intersect at x=1 and are perpendicular (corner-to-corner for each square on the graph paper for both straight lines).

Now I define the equation for the second straight line choosing arbitrary points:

For the point of intersection we solve simultaneously, from which yields x=1. The solution for both equations holds.

Further, two straight lines are perpendicular - if and only if the product of their slopes equates to

Therefore,

Delving deeper, one now remembers such that if you 'somehow' embrace the former line and rotate it through an angle of then you have gone up x units and have subsequently moved towards the left by way of y units across.

We now define the equation for this perpendicular line from which is

Therefore,

I now explain my confusion. This is an absolute load of horse radish. A perpendicular line can only have a negative reciprocal for a slope if one inverts the axes. In other words, the x-axis becomes the y-axis and on the converse the y-axis becomes the x-axis.

What this tells me is that these lines cannot be constructed in an identical rectangular co-ordinate system in order to have the product of their slopes equate to (-1).

For example, if I plot a line by means of subbin' in x values for using the former co-ordinate system I used to construct they're not perpendicular.

What am I missing?

Thank you for your attention.

Re: Slopes involving perpendicular lines

I don't know what to say. You **assert** that the graphs of and are not perpendicular but they clearly are! The two lines intersect where which is the same as or . That is, the two lines intersect at . Now, when x= 2, y= 2x- 2 gives y= 4- 2= 2 so (2, 2) is a another point on the graph of y= 2x- 2.. When x= 0, gives y= 1 so (0, 1) is another point on .

That is, , (2, 2) and (0, 1) form a triangle with points on those two lines. The distance from (6/5, 2/5) to (2, 2) is . The distance from (6/5, 2/5) to (0, 1) is . The distance from (2, 2) to (0, 1) is . But now we see that . That is, the Pythagorean theorem holds- this is a right triangle with hypotenuse the line between (2, 2) and (0, 1). That proves that the intersection **is** a right triangle.

Perhaps your "squares" aren't true squares and you are "stretching" one way more than the other.

Rereading your post- "The scales I'm using are such that ten small squares on my graph paper represent 1 unit on the x-axis and 2 units on the y-axis", that is exactly what you are doing! If one your x axis is stretched more than your y axis, you certainly are not representing angles correctly. Redraw the graph using ten small squares representing 1 unit on both axes or representing 2 units on both axes.

Re: Slopes involving perpendicular lines

OOPS! Everything is working out fine now that I'm using the same scale for x and y on the axes.

False alarm!

Sorry for wastin' your time HallsofIvy. Thanks!