# Slopes involving perpendicular lines

• Jul 28th 2012, 12:35 PM
astartleddeer
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 $y = 2x-2$ which passes through $x=1$ when $y=0$. 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:

$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x{1}} \cdot (x - x_{1})$

$y - 4 = \frac{8 - 4}{(-3) - (-1)} \cdot [x - (-1)] \Rightarrow y = -2x + 2$

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 $(-1)$

Therefore, $m_{1} \cdot m_{2} = (-1)$

$(2)\cdot(-2) \neq (-1)$

Delving deeper, one now remembers such that if you 'somehow' embrace the former line and rotate it through an angle of $90^{\circ}$ 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 $y = - \frac{1}{2}x + 1$

Therefore, $m_{1} \cdot m_{2} = (-1)$

$(2)\cdot(-\frac{1}{2}) = (-1)$

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 $y = -\frac{1}{2}x + 1$ using the former co-ordinate system I used to construct $y = 2x - 2$ they're not perpendicular.

What am I missing?

• Jul 28th 2012, 02:13 PM
HallsofIvy
Re: Slopes involving perpendicular lines
I don't know what to say. You assert that the graphs of $y= 2x- 2$ and $y= -\frac{1}{2}x+1$ are not perpendicular but they clearly are! The two lines intersect where $-\frac{1}{2}x+ 1= 2x- 2$ which is the same as $-\frac{5}{2}x= -3$ or $x= \frac{6}{5}$. That is, the two lines intersect at $\left(\frac{6}{5}, \frac{2}{5}\right)$. 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, $y= -\frac{1}{2}x+ 1$ gives y= 1 so (0, 1) is another point on $y= -\frac{1}{2}x+ 1$.

That is, $\left(\frac{6}{5}, \frac{2}{5}\right)$, (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 $a= \sqrt{(6/5- 2)^2+ (2/5- 2)^2}= \sqrt{(-4/5)^2+ (-8/5)^2}= \sqrt{80/25}= \frac{4\sqrt{5}}{5}$. The distance from (6/5, 2/5) to (0, 1) is $b= \sqrt{(6/5- 0)^2+ (2/5-1)^2}= \sqrt{36/25+ 9/25}= \frac{3}{5}\sqrt{5}$. The distance from (2, 2) to (0, 1) is $c= \sqrt{(2-0)^2+ (2-1)^2}= \sqrt{4+ 1}= \sqrt{5}$. But now we see that $a^2+ b^2= \frac{80}{5}+ \frac{45}{5}= \frac{125}{5}= 5= c^2$. 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.
• Jul 29th 2012, 04:51 AM
astartleddeer
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!