SOLUTION: Prove that for perpendicular lines, if the gradient of the two lines are m1 and m2, then m1 times m2 = -1 if m1 and m2 don't equal 0.

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Question 1182584: Prove that for perpendicular lines, if the gradient of the two lines are m1 and m2, then m1 times m2 = -1 if m1 and m2 don't equal 0.
Found 2 solutions by math_helper, math_tutor2020:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!





Draw the first line, slope m1, ending at point (a,b), r units out from the origin and making an angle alpha with respect to the x-axis:



   m1 = b/a = +%28r%2Asin%28alpha%29%29+%2F+%28r%2Acos%28alpha%29%29+



Now draw a 2nd line with slope m2, length r units, rotated 90 degrees with respect to the first line.  The endpoint at r units out is (c,d):



Make use of this identity:  +cos%28alpha%2Bbeta%29+=+cos%28alpha%29cos%28beta%29-sin%28alpha%29sin%28beta%29



   c = +r%2Acos%28alpha+%2B+90%29+=+-r%2Asin%28alpha%29+



Make use of this identity: +sin%28alpha%2Bbeta%29+=+sin%28alpha%29cos%28beta%29%2Bcos%28alpha%29sin%28beta%29+

   d = +r%2Asin%28alpha+%2B+90%29+=+r%2Acos%28alpha%29+ 





Noting m2 = d/c, you can now write:



   m1*m2 = (b/a)(d/c) =  = -1,  with the constraint neither slope is zero.

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Refer to the diagram below


Triangle ABC is a right triangle. The 90 degree is at angle A.

Sides AB and BC are p and q units long respectively. Where p & q are positive real numbers. We'll make p < q.

Notice how the slope of line BC is -q/p
rise = -q
run = p
The slope is negative since we're going downhill as we move from left to right (from C to B).

So we can consider m1 = -q/p.
We'll use this later.

----------------------------

Triangle CDE is also a right triangle with sides CD = p and DE = q

Notice how triangle CDE is a rotated version of triangle ABC. Triangle ABC and DCE are congruent triangles.

For either right triangle (ABC or DCE), the acute angles add to 90. Since we're dealing with identical triangles, this means that the acute angles DCE and ACB are also complementary.

Put another way:
angle ACB = angle DEC
(angle DCE) + (angle DEC) = 90
(angle DCE) + (angle ACB) = 90

Now since ACD is a straight angle, we can then say,
(angle DCE) + (angle ECB) + (angle ACB) = 180
[ (angle DCE) + (angle ACB) ] + (angle ECB) = 180
[ 90 ] + (angle ECB) = 180
angle ECB = 180 - 90
angle ECB = 90

This proves that angle ECB is a right angle, and it further shows segment BC is perpendicular to EC.

Note that the slope of line EC is rise/run = p/q
We'll call this m2, so,
m2 = p/q

----------------------------

Let's multiply the slopes m1 and m2

m1*m2 = (-q/p)*(p/q)
m1*m2 = (-q*p)/(p*q)
m1*m2 = -1
The p's and q's cancel when we divide them.

This concludes the proof.