1. ## [SOLVED] Locus

Hi guys

A ladder that is 6 m long rests with one end on the horizontal ground and the other end against a vertical wall. Considering the ground and the wall as the X- and Y-axis respectively, find the locus of the midpoint of the ladder.

For this question, I havn't been able to get far ...

I let the pt the ladder is touching the horizontal ground (and thus the X-axis) be x AND the pt the ladder is touching the wall (the Y-axis) be y. Thus, the midpt is (x/2,y/2).

I've tried to work it out the same method as the previous q.s in the txtbk e.g. saying this distance = that distance to get the locus.

2. Originally Posted by xwrathbringerx
Hi guys

A ladder that is 6 m long rests with one end on the horizontal ground and the other end against a vertical wall. Considering the ground and the wall as the X- and Y-axis respectively, find the locus of the midpoint of the ladder.

For this question, I havn't been able to get far ...

I let the pt the ladder is touching the horizontal ground (and thus the X-axis) be x AND the pt the ladder is touching the wall (the Y-axis) be y. Thus, the midpt is (x/2,y/2).

I've tried to work it out the same method as the previous q.s in the txtbk e.g. saying this distance = that distance to get the locus.

Let F denote the footpoint of the ladder. Then F(t, 0).

then the height of the ladder against the wall is

$h = \sqrt{36 - t^2}$

Thus the midpoint of the ladder has the coordinates $M\left(\dfrac t2\ ,\ \dfrac12 \sqrt{36-t^2} \right)$

Calculate the locus of all M by solving the system of equations:

$x = \dfrac t2~\implies~t = 2x$

$y = \dfrac12 \sqrt{36-t^2}~\implies~4y^2=36-4x^2~\implies~x^2+y^2=9$

That means the locus is a circle around the origin with a radius of 3.