Locus of Mid Point: X-Axis Cut Points

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Homework Help Overview

The problem involves determining the locus of midpoints of lines drawn from the point (6, -8) to various points on the x-axis. The subject area pertains to coordinate geometry and locus concepts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the expression for the midpoint and question the neglect of the x-term in determining the locus. There is exploration of the geometric interpretation involving triangles and midpoints.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about the locus and the role of the x-coordinate. Some guidance has been offered regarding the nature of the locus as a horizontal line, but there remains uncertainty about the reasoning behind neglecting the x-term.

Contextual Notes

There is a focus on understanding the implications of the midpoint formula and how it relates to the overall locus. Participants express confusion about the treatment of the x-coordinate in the context of the problem.

phymatter
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Homework Statement


From the point (6,-8) all possible lines are drawn to cut x-axis , find the locus of their middle ponts


Homework Equations



none

The Attempt at a Solution



i got the coordinates of the middle point as ((6+x)/2 ,-4 ) , but what will the locus be ?
 
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The expression you have gives you the locus. What do all of the midpoints have in common?
 
Mark44 said:
The expression you have gives you the locus. What do all of the midpoints have in common?

you mean that the locus is y=-4 neglecting the term containing x , but why are we neglecting x ? i mean if we put different values of x we get different mid points , so why are we neglecting it??
pls. help , its getting over my head :((
 
Draw a right angled triangle and join the midpoints of two sides. This new line will be parallel to the third side. Similarly, in any triangle, the line between the midpoints of two sides is parallel to the third side. You can prove this because the smaller triangle is similar to the larger, so the corresponding angles are equal.

If the apex is (6,-8) and the third side is the x-axis, the line between the midpoints is the locus. Does this convince you?
 
phymatter said:
you mean that the locus is y=-4 neglecting the term containing x , but why are we neglecting x ? i mean if we put different values of x we get different mid points , so why are we neglecting it??
pls. help , its getting over my head :((

You're neglecting the x, because the image of [tex]\frac{6 + x}{2}[/tex] spreads over the real number. What I mean is:

[tex]\forall x_0 \in \mathbb{R} , \exists x \in \mathbb{R} : \frac{6 + x}{2} = x_0[/tex].

So, take any real number x0, the point (x0; -4) is the midpoint of the line segment, of which 2 ends are (6; -8), and [tex]\left( 2x_0 - 6; 0 \right)[/tex].

Hope I'm being clear enough. :)
 
All points of the form ((6+x)/2 ,-4 ) satisfy y= -4 and that is the equation of the locus, a horizontal line. The "(6+x)/2" gives the x coordinate of a point on that line (for x being the x coordinate of the point (x, 0) that the line from (6, -8) is drawn through).
 

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