Equation Of Locus

Geometry Level 2

A rod of length $l$ slides with its ends on the $x$ -axis and $y$ -axis.

Find the locus of its midpoint.

x^2 + y^2 = 4l^2

x^2 + 4y^2 = l^2

4x^2 + y^2 = l^2

4x^2 + 4y^2 = l^2

2 solutions

Maria Kozlowska
Mar 2, 2016

Let $M(x,y)$ be the midpoint of the ladder. Applying Pythagorean Theorem on a right triangle with hypotenuse ME we get:

$x^2+y^2=\left(\dfrac{l}{2} \right)^2 \Rightarrow 4x^2+4y^2=l^2$
Path of the ladder's midpoint is quarter of a circle.

This method is so easy, I didn't solve mine like this

Anthony Ezulike - 4 years, 8 months ago

this is not even a method. how do you simply assume the locus as circle. you cant

Christina Vincent - 1 year, 12 months ago

Anthony Ezulike
Oct 8, 2016

X=x/2 Y=y/2.......line one From pytagoras theory we have X^2 + Y^2 = l^2........line 2 Therefore mapping line 1 to fit line 2 we get 2X = x 2Y = y putting this in line 2 you get (2X)^2 + (2Y)^2 = l^2 4X^2 + 4Y^2 = l^2

Equation Of Locus

2 solutions

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