1. ## Lines

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.

2. 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

3. 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:
         q
* - - *
|
*   |
| -p
* |
|
*

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 $\angle FOG = 90^o$.

4. ## ok

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