two lines are perpendicular only when...

[PG1995]

Hi

Please have a look on the attachment. You can find me query there. Please help me with it. Thank you.

Regards
PG

 

[misterT]

In your particular case (the slope is 45 degrees) -y/x is same as -x/y. In another words: x=y.

When you have a line 'a' with a slope y/x, line 'b' is perpendicular to it if its slope is -x/y.

 

[Ratchit]

PG1995,

Line l2 is 135°, not 90° as shown in your diagram. Why do you use a symmetrical diagram when an unsymmetrical diagram would immediately remove your confusion? Take 30° and 120°. tan 30° is y/x = 0.58.
tan 120° is x/-y = -1.73 . So, (y/x)*(x/-y)=-1 = 0.58*(-1.73) = -1 .

Ratch

 

[ARandomOWl]

From [LATEX]Tan 45=\frac{y}{x}[/LATEX] we get [LATEX]y_1=xtan45=x \therefore \frac{dy_1}{dx}=1[/LATEX]

And similarly [LATEX]y_2=-xtan45=-x \therefore \frac{dy_2}{dx}=-1[/LATEX]

[LATEX]\frac{dy_1}{dx}\cdot \frac{dy_2}{dx}=-1[/LATEX]

I can't quite figure which step is incorrect in your working.

I think Ratchit explained it better.

 

[misterT]

I can't quite figure which step is incorrect in your working.

There is no error, just a confusing example (as Ratch already pointed out). In the example case -y²/x² = -1

 

[PG1995]

Thank you, everyone.

Yes, The example I posted wasn't a good one. It didn't remove any confusion.

tan(33)*tan(57) = 1 = tan(33)*{1/tan(33)}

tan(57) = 1/tan(33)

You can use any other angles in place of 57 and 33 (they should sum up to 90).

tan(x) = 1/tan(y) [y+x = 90]

How do I reach this result? Please help me with it. Thanks.

Regards
PG

 

[misterT]

tan(x) = 1/tan(y) [y+x = 90]

How do I reach this result? Please help me with it. Thanks.

What result? do you mean: tan(x) = 1/tan(90-x)

 

[PG1995]

What result? do you mean: tan(x) = 1/tan(90-x)

Yes, Mr. T. Could you please help? Thanks.

 

[misterT]

I did not quite understand what the question was? Do you want to solve the equation for y? .. something like this: https://www.wolframalpha.com/input/?i=solve+for+y:+tan(x)+=+1/tan(y)

 

[PG1995]

Sorry for the confusion.

I'm saying that how do I prove that RHS equals LHS in: tan(x) = 1/tan(90-x)? Please let me know if you need further clarification. Thank you.

Regards
PG

 

[misterT]

With trigonometric identities:

tan(x) = sin(x)/cos(x)

so you get:

tan(x) = 1/tan(90-x) => sin(x)/cos(x) = cos(90-x)/sin(90-x)

and because sin(90-x) = cos(x) and cos(90-x) = sin(x), you'll get:

sin(x)/cos(x) = sin(x)/cos(x)


If you want a proof for the identities, do it geometrically using a right triangle:
https://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Ratio_identities

 

[PG1995]

With trigonometric identities:

tan(x) = sin(x)/cos(x)

so you get:

tan(x) = 1/tan(90-x) => sin(x)/cos(x) = cos(90-x)/sin(90-x)

and because sin(90-x) = cos(x) and cos(90-x) = sin(x), you'll get:

sin(x)/cos(x) = sin(x)/cos(x)


If you want a proof for the identities, do it geometrically using a right triangle:
https://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Ratio_identities

Thank you very much, Mr. T.

Best regards
PG