Straight lines passing through fixed point

[utkarshakash]

Homework Statement

If all the lines given by the equation $(3\sin \theta + 5\cos \theta )x+(7\sin \theta - 3\cos \theta )y+11(\sin \theta - \cos \theta)=0$ pass through a fixed point (a,b) forall theta in R then |a-b|=

Homework Equations

The Attempt at a Solution

Dividing both sides by (3sin theta + 5cos theta)
$x+ \dfrac{(7\sin \theta - 3\cos \theta)y+11(\sin \theta - \cos \theta)}{3\sin \theta + 5\cos \theta}=0$

This is of the form L1+λL2 and the fixed point is intersection of L1 and L2. Here x=0 and $y=\dfrac{-11(\sin \theta - \cos \theta)}{(7\sin \theta - 3\cos \theta )}$ However, the difference of the two depends on theta and is not constant. So how can it be fixed?

 

[ehild]

The line goes through the same point for all θ angles. So it goes through (a,b) when sinθ=cosθ.
Or sinθ=1, cosθ=0. ... Chose two angles and solve for x,y. Then check if that (x,y) is solution for the equation for any θ.

ehild

 

[utkarshakash]

The line goes through the same point for all θ angles. So it goes through (a,b) when sinθ=cosθ.
Or sinθ=1, cosθ=0. ... Chose two angles and solve for x,y. Then check if that (x,y) is solution for the equation for any θ.

ehild

Suppose I put theta=pi/2 then y=-11/7. Now what do you want me to do?

 

[ehild]

Suppose I put theta=pi/2 then y=-11/7. Now what do you want me to do?

You got y=-11/7 with the assumption that x=0, which is invalid.

I do not want anything from you. I just tried to help. Again, I suggest to choose two angles, and substitute their sine and cosine into the original equation, and solve the system of two equations for x and y. ehild

 

[utkarshakash]

You got y=-11/7 with the assumption that x=0, which is invalid.

I do not want anything from you. I just tried to help. Again, I suggest to choose two angles, and substitute their sine and cosine into the original equation, and solve the system of two equations for x and y.


ehild

OK I agree with you. But working backwards(since I know the answer), I get tan theta = -1/5. Now, this angle is very vague which is impossible to guess unless you really know this is the answer.

 

[Dick]

OK I agree with you. But working backwards(since I know the answer), I get tan theta = -1/5. Now, this angle is very vague which is impossible to guess unless you really know this is the answer.

I don't know what your are talking about. The idea is not to solve for theta. That's not the 'answer'. The answer would be the common intersection point. It's really easy, unlike some of your posts. As ehild suggested, intersect two lines corresponding to different values of theta and then look back and figure out why that intersection point is common for all values of theta. This isn't rocket science. It's simple.

 

[utkarshakash]

I don't know what your are talking about. The idea is not to solve for theta. That's not the 'answer'. The answer would be the common intersection point. It's really easy, unlike some of your posts. As ehild suggested, intersect two lines corresponding to different values of theta and then look back and figure out why that intersection point is common for all values of theta. This isn't rocket science. It's simple.

OMG! Now I realize how simple is this. I don't know how I overlooked that I can plug in different values of theta to get a number of eqns and solving any two of them to get the intersection point. At first I couldn't understand what ehild was trying to say. Thanks to both of you! :thumbs: