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.
Then: .
. . We have two slopes whose product is .
We know that:
So looks like this:Code:*
* q
* 
*      *
p
And 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 .
Let
Let
Let
We note that: .
. . . and that: .
Since , then is a right triangle with .
I honor both replies but soroban did a fantastic job using the correct math symbols with LAtex.