Locus Problem - Feet of Perpendicular

Locus Problem : A ladder of length L is sliding over the floor. Find the locus of point D such that D is the foot of perpendicular dropped from point C such that OACB is a rectangle.

Solution: Let A (a,0) be the feet of the ladder and B (0,b) be the top of the ladder rest against the wall . O is the intersection point of floor and wall. OACB is a rectangle. D is the feet of the perpendicular drawn from point C onto ladder AB . As the feet of the ladder slides, the point D travels on a path called asteroid whose equation is given by x^(2/3)+y^(2/3)=L^(2/3).

In the following applet, press the play button at the lower left corner to see the various positions of the ladder as it is dragged and the movement of point D of the ladder. You can also select the ‘Show Locus’ button to see the locus and its equation.

Locus of a General Point on Falling Ladder

In continuation to my previous post on locus of mid-point of a falling ladder , let us now explore the locus of a point lying anywhere on the falling ladder.

Example : A 6-foot ladder is placed vertically against a wall, and then the foot of the ladder is moved outward until the ladder lies flat on the floor with one end touching the wall. What is the locus of the point which divides the ladder in the ratio 2 : 3 as it slides?

Solution: Let P be the point which divides the slider AB (A (a,0) being the feet and B (0,b) rest on the wall) .O is the intersection point of floor and wall.As the feet of the ladder slides,the point P whose coordinates are(3a/5,2b/5) travels on an elliptical path given by x²/9+y²/4=(6/5)².

In the following applet, press the play button at the lower left corner to see the various positions of the ladder as it is dragged and the movement of point P of the ladder. You can also select the ‘Show Locus’ button to see the full locus and its equation.

Locus of Mid Point of Falling Ladder

Locus: A locus of points is the set of points, and only those points, that satisfies given conditions. The locus of points at a given distance from a given point is a circle whose center is the given point and whose radius is the given distance.

Example : A 6-foot ladder is placed vertically against a wall, and then the foot of the ladder is moved outward until the ladder lies flat on the floor with one end touching the wall. What is the locus of the midpoint of the ladder as it slides?

Solution: The midpoint is on the hypotenuse of the right triangle whose legs are on the wall and floor. Since a right triangle can be inscribed in a semicircle with the midpoint of the hypotenuse (diameter of the circle) as its center, we know the distance OC is a radius of this circle and therefore 3 feet. That is, no matter where the ladder is, OC will be 3 feet, and therefore the locus of midpoints is a quarter of a circle with center at the intersection of the floor and wall (point O) and radius 3 feet.

In the following applet, press the play button at the lower left corner to see the various positions of the ladder as it is dragged and the movement of midpoint C of the ladder. You can also select the ‘Show Locus’ button to see the full locus and its equation. The geometrical solution to this problem can be accessed by selecting the ‘Solution’ button and then dragging the point A.

Pedal Triangle - II

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com