Sliding Laddder PQ, w/ R between, RQ = 2*RP: find locus

Yuzu613

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5. A ladder PQ, 5 meters long, is resting against a wall. (P is against the wall; Q is on the floor, away from the wall.) Assume that the floor and the wall represent the x- and y-axis, respectively, and that R is a point on PQ such that RQ is twice RP. Find:

(a) the locus of point R when ladder PQ slides down the wall, and

(b) the coordinates of point R when it touches the floor.




As the picture shown, I got no clues how to solve it, can someone solve it with details step-by-step?

Here are my steps,

First I get the coordinate of R by using the formula and I get R(a/3 , 2b/3)


Since the ladder is 5m long, so x²+y²=25, following by x²=25-y² and y²=25-x², a²=25-y² b²=25-x²

Then I calculated the distance from the point R to the origin, O(0,0) and I get sqrt( (a/3-0)
²+(2b/3-0)² )

After that I don't know how to proceed anymore...
 

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5. A ladder PQ, 5 meters long, is resting against a wall. (P is against the wall; Q is on the floor, away from the wall.) Assume that the floor and the wall represent the x- and y-axis, respectively, and that R is a point on PQ such that RQ is twice RP. Find:

(a) the locus of point R when ladder PQ slides down the wall, and

(b) the coordinates of point R when it touches the floor.




As the picture shown, I got no clues how to solve it, can someone solve it with details step-by-step?
attachment.php


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Solved

Well somehow I solved the question,

First lets assume the coordinate for R is R(x,y)

By using the formula (nx1+mx2/n+m , ny1+my2/n+m), we get the coordinate of R which is R(a/3,2b/3)

Since R(x,y)=R(a/3,2b/3), so x=a/3 and we get 3x=a

The ladder is 5m long and divided to 3 segments, so each segment is 5/3 long, so the distance between R and Q is 10/3.

By using the distance formula, 10/3=sqrt(x-a/3)2+(y-0)2

and we will get the locus of the sliding ladder which is 36x2+9y2​-100=0
 
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