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
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.
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 .