Locus of Mid Point: X-Axis Cut Points

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In summary, the locus of all possible midpoints of lines drawn from (6, -8) to cut the x-axis is a horizontal line with equation y = -4. This can be proven by demonstrating that all points on this line have a common y-coordinate of -4, and the x-coordinate is determined by the value of x on the x-axis where the line is drawn through.
  • #1
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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  • #2
The expression you have gives you the locus. What do all of the midpoints have in common?
 
  • #3
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 :((
 
  • #4
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?
 
  • #5
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. :)
 
  • #6
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).
 

1. What is the Locus of Mid Point?

The Locus of Mid Point is a geometric concept that refers to the set of points that are equidistant from two given points. In other words, it is the path that a point takes when it is always located at the midpoint between two fixed points.

2. What is the X-Axis Cut Point of a Locus of Mid Point?

The X-Axis Cut Point of a Locus of Mid Point is the point on the X-axis where the Locus intersects. It is the point where the X-coordinate of the midpoint is equal to the X-coordinate of the two given points.

3. How is the Locus of Mid Point calculated?

The Locus of Mid Point can be calculated using the midpoint formula: ((x1 + x2)/2, (y1 + y2)/2), where (x1,y1) and (x2,y2) are the coordinates of the two given points.

4. What is the significance of the Locus of Mid Point in mathematics?

The Locus of Mid Point has many applications in mathematics, particularly in geometry and graphing. It can be used to find the center of a circle, the point of intersection between two lines, and to construct perpendicular bisectors.

5. Can the Locus of Mid Point have multiple solutions?

Yes, the Locus of Mid Point can have multiple solutions depending on the given points. For example, if the two given points are on opposite sides of the X-axis, there will be two points of intersection on the X-axis. However, if the two given points are on the same side of the X-axis, there will be no intersection on the X-axis.

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