Lines

**symmetry** · January 16th 2007, 01:56 PM

Prove that if two nonvertical lines have slopes whose product is -1, then the lines are perpendicular.

HINT GIVEN:

Use the converse of the Pythagorean Theorem.

**CaptainBlack** · January 16th 2007, 02:10 PM

Originally Posted by symmetry

Prove that if two nonvertical lines have slopes whose product is -1, then the lines are perpendicular.

HINT GIVEN:

Use the converse of the Pythagorean Theorem.

The converse of the Pythagorean theorem is that if triangle ABC has
sides a,b,c with the usual labelling, and a^2+b^2=c^2, then the angle
at C is right.

The lines meet at some point, call it (a,b), and let them have slopes m (!=0)
and 1/m.

Additional points on the two lines are (a-1, b-m) and (a-1, b+1/m).

Now we have three points, we have a triangle and can calculate the
lengths of the sides, and we will find the square of the side connecting
(a-1, b-m) and (a-1, b+1/m) is equal to the sum of the squares of the
other two sides.

The case where one of the slopes is 0 can be handled as a special case
and I hope you see that in that case the result is obvious.

RonL

**Soroban** · January 16th 2007, 03:37 PM

Hello, symmetry!

An interesting problem . . . I had to "invent" my own approach.
I'm certain others have shorter, more elegant proofs.

Prove that if two nonvertical lines have slopes whose product is -1,
then the lines are perpendicular.

$\text{Let }m_1 = \frac{q}{p}\:\text{ and }\:m_2 = -\frac{p}{q}$

Then: . $m_1\cdot m_2 \:=\:\left(\frac{1}{p}\right)\left(-\frac{p}{q}\right) \:=\:-1$
. . We have two slopes whose product is $-1$ .

We know that: $slope \:=\:\frac{rise}{run}$

So $m_1 = \frac{q}{p}$ looks like this:

Code:

                  *
              *   |q
          *       |
      * - - - - - *
              p

And $m_2 = -\frac{p}{q}$ looks like this;

Code:

Sketch them on the same graph.

Code:

      |           F
      |           * (p,q)
      |    a  *   |
      |   *       |
    O * - - + - - + -
 (0,0)|     |
      | *   |
      |     |
      | b * |
      |     |
      |     * (q,-p)
      |     G

Now consider the triangle $FOG$ .

Let $a \:= \:OF \:=\:\sqrt{(p-0)^2 + (q-0)^2} \:=\:\sqrt{p^2+q^2}$

Let $b \:=\:OG\:=\:\sqrt{(q-0)^2+(\text{-}p-0)^2} \:=\:\sqrt{p^2+q^2}$

Let $c \:=\:GF \:=\:\sqrt{(p-q)^2 + (q+p)^2} \:=\:\sqrt{2p^2 + 2q^2}$

We note that: . $a^2 + b^2 \:=\:\left(\sqrt{p^2+q^2}\right)^2 + \left(\sqrt{p^2 + q^2}\right)^2 \:=\:2p^2+2q^2$

. . . and that: . $c^2\:=\:\left(\sqrt{2p^2+2q^2}\right)^2\:=\:2p^2+2 q^2$

Since $a^2 + b^2\:=\:c^2$ , then $\Delta FOG$ is a right triangle with $\angle FOG = 90^o$ .

**symmetry** · January 16th 2007, 05:21 PM

I honor both replies but soroban did a fantastic job using the correct math symbols with LAtex.

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