Slopes product = -1 <===> lines perpendicular

[Dank2]

Homework Statement

Proof that if the slopes of two lines a1, a2 (that are not vertical), m1,m2 satisfy:
m1*m2 = -1, then the lines are perpendicular.

Homework Equations

The Attempt at a Solution

I tried to use the tan function, so that m1 = tanΘ where Θ₁ is the angle of the line formed from x axis.
and m2 = tanΘ₂.
now tanΘ₂*Θ₁ = -1,
I tried using the tan(Θ1+Θ2) = but after simplifing i got (tanΘ1 - tankΘ2)/2
not sure if I am on the right way, rather solve it using trig.

ok found the answer, i had to use the points (1,m1), (1, m2), form a triangle and check for Pythagorean theorem.

 

[fresh_42]

Hint: You can use the addition theorem of cosine for $\Theta_1 - \Theta_2$.

 

[SammyS]

Homework Statement

Proof that if the slopes of two lines a1, a2 (that are not vertical), m1,m2 satisfy:
m1*m2 = -1, then the lines are perpendicular.

Homework Equations

The Attempt at a Solution

I tried to use the tan function, so that m1 = tanΘ where Θ₁ is the angle of the line formed from x axis.
and m2 = tanΘ₂.
now tanΘ₂*Θ₁ = -1,
I tried using the tan(Θ1+Θ2) = but after simplifing i got (tanΘ1 - tankΘ2)/2
not sure if I'm on the right way, rather solve it using trig.

ok found the answer, i had to use the points (1,m₁), (1, m₂), form a triangle and check for Pythagorean theorem.

if you use the points, (1,m₁), and (1, m₂), then what's the third point you use to form the triangle?

How are these points related to the lines a₁ and a₂ and the slopes of these lines?

 

[Dank2]

Hint: You can use the addition theorem of cosine for $\Theta_1 - \Theta_2$.

Yes, cos(Θ1-Θ2) = cosΘ1*cosΘ2 + sinΘ1*sinΘ2 = 1/sqrt(1+m1²)* -1/sqrt(1 + (-1/m)²) + m1/sqrt(1+m1²)*(-1/m1)/sqrt(1 + (-1/m)²) = 1 / * + -1 / * = 0 ==> Θ1-Θ2 = right angle, thanks. i see i must use the triangle to proof it in both ways.

 

[Dank2]

if you use the points, (1,m₁), and (1, m₂), then what's the third point you use to form the triangle?

How are these points related to the lines a₁ and a₂ and the slopes of these lines?

hey sammy, I've already seen a solution for it , and i marked the thread as solved. thanks.

 

[SammyS]

hey sammy, I've already seen a solution for it , and i marked the thread as solved. thanks.

Well, it's true that I'm not grading your work, but if I were grading it, I would expect those issues to be addressed.

The points (1, m₁) and (1, m₂) are not likely to be on the lines involved .

 

[Dank2]

The points (1, m1) and (1, m2) are not likely to be on the lines involved .

Why not? If m1 = tanΘ =y/x
Then (x, m1x) is on the line and so (1, m1)

 

[Dank2]

Well, it's true that I'm not grading your work, but if I were grading it, I would expect those issues to be addressed.

The points (1, m₁) and (1, m₂) are not likely to be on the lines involved .

Maybe I should have added needed to be copied so that they will intersect with (0,0), and then we can always have the mentioned dots and form the triangle?

 

[SammyS]

Maybe I should have added needed to be copied so that they will intersect with (0,0), and then we can always have the mentioned dots and form the triangle?

... and perhaps:

The points (0, 0) and (1, m₁) either lie on line a₁ or lie on a line parallel to a₁.

Similarly for (0, 0) and (1, m₂) and line a₂ ...

 

[fresh_42]

Maybe I should have added needed to be copied so that they will intersect with (0,0), and then we can always have the mentioned dots and form the triangle?

You don't need this as requirement. None of your lines is vertical. Thus you can always draw a coordinate system, in which the origin is the intersection point and, say the $x-$axis a third line (also intersecting at the origin of course). With that you have defined $\Theta_1$ and $\Theta_2$.

 

[mizanshoeb1]

most probably you can use addition theorem of cosine for $\theta_1$ and $\theta_ 2$... i hope this hint will help you..