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.
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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 (a1, bm) and (a1, 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
(a1, bm) and (a1, 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
Hello, symmetry!
An interesting problem . . . I had to "invent" my own approach.
I'm certain others have shorter, more elegant proofs.
Quote:
Prove that if two nonvertical lines have slopes whose product is 1,
then the lines are perpendicular.
$\displaystyle \text{Let }m_1 = \frac{q}{p}\:\text{ and }\:m_2 = \frac{p}{q}$
Then: .$\displaystyle m_1\cdot m_2 \:=\:\left(\frac{1}{p}\right)\left(\frac{p}{q}\right) \:=\:1$
. . We have two slopes whose product is $\displaystyle 1$.
We know that: $\displaystyle slope \:=\:\frac{rise}{run}$
So $\displaystyle m_1 = \frac{q}{p}$ looks like this:Code:*
* q
* 
*      *
p
And $\displaystyle m_2 = \frac{p}{q}$ 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 $\displaystyle FOG$.
Let $\displaystyle a \:= \:OF \:=\:\sqrt{(p0)^2 + (q0)^2} \:=\:\sqrt{p^2+q^2} $
Let $\displaystyle b \:=\:OG\:=\:\sqrt{(q0)^2+(\text{}p0)^2} \:=\:\sqrt{p^2+q^2}$
Let $\displaystyle c \:=\:GF \:=\:\sqrt{(pq)^2 + (q+p)^2} \:=\:\sqrt{2p^2 + 2q^2}$
We note that: .$\displaystyle a^2 + b^2 \:=\:\left(\sqrt{p^2+q^2}\right)^2 + \left(\sqrt{p^2 + q^2}\right)^2 \:=\:2p^2+2q^2$
. . . and that: .$\displaystyle c^2\:=\:\left(\sqrt{2p^2+2q^2}\right)^2\:=\:2p^2+2 q^2$
Since $\displaystyle a^2 + b^2\:=\:c^2$, then $\displaystyle \Delta FOG$ is a right triangle with $\displaystyle \angle FOG = 90^o$.
I honor both replies but soroban did a fantastic job using the correct math symbols with LAtex.