Hi can you help me in solving this from coordinate geometry?

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Homework Help Overview

The original poster attempts to find the equation of a line that passes through the point A (1, 2) and has the maximum perpendicular distance from the origin. This problem falls within the subject area of coordinate geometry.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants suggest drawing sketches to visualize the problem and propose determining the perpendicular distance from an arbitrary line to the origin. Others discuss the implications of the line's degrees of freedom and suggest using trigonometric identities to express the line's equation.

Discussion Status

The discussion is actively exploring different mathematical approaches to derive the line's equation. Participants are engaging with concepts of maximizing functions and vector products, indicating a productive direction without reaching a consensus or final solution.

Contextual Notes

There is an emphasis on the constraints of the problem, particularly regarding the degrees of freedom in the line's equation and the requirement for the line to pass through a specific point.

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Find the equation of a line passing through point A (1, 2) and whose perpendicular distance from origin is maximum.
 
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Start by drawing a sketch showing the coordinate axes and point A.

Draw an arbitrary line which passes thru point A.

Can you figure out how to determine the perpendicular distance from this arbitrary line to the origin?
 
SteamKing said:
Start by drawing a sketch showing the coordinate axes and point A.

Draw an arbitrary line which passes thru point A.

Can you figure out how to determine the perpendicular distance from this arbitrary line to the origin?
Yes. If that eqn is Ax + By + C = 0
then the distance from origin is |C|/√(A^2 + B^2)!
 
Ax + By + C = 0 isn't the whole story.

A straight line in the plane has two degrees of freedom only, not three like your equation suggests. And this line has to go through point (1,2), so there can't be more than one degree of freedom (e.g. the slope).

You can eliminate one degree of freedom by requiring A2 + B2 = 1 (effectively dividing by ##\sqrt{A^2+B^2}## -- and no fear of dividing by zero; why not ?)

That means the new A and B can be written as ##\cos\phi## and ##\sin\phi## for some angle ##\phi## (well, not just 'some' angle...)

Congrats! you have just derived the so-called http://doubleroot.in/straight-line-normal-form/equation for a straight line !Not there yet: the line has to go through (1,2), so you now eliminate C by substituting the coordinates of point A.

According to your equation the new C ( let's call it C' ) is then the distance to the origin, so you have that in terms of xA and yA.

If you know about maximizing a function, you can differentiate that expression and thus find ##\phi##

If you know about vector products, you can also see that the left hand side is a dot product of two vectors: ##|C'| = (1,2)\;\cdot\; (\cos\phi, \sin\phi)## and the absolute value of that is $$|C'| = |(1,2)| \; |(\cos\phi, \sin\phi)| \; cos\alpha$$ where ##\alpha## is the angle between the two vectors.

And then |C'| is clearly maximum if ##\alpha = 0## (the two vectors are collinear).
 

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