Wie löse ich sowas mit geogebra?
Ein Adler fliegt in gerader Richtung. Die Flugbahn lässt sich durch die Gerade f: -x + 9y = 2700 beschreiben.
Mit seinem neuen Smartphone kann Clemens Fotos von bis zu 200 Meter Entfernung in HD-Qualität machen. Wird Clemens mit seinem Handy ein qualitatives Foto machen können, wenn er sich am Punkt P = (500|100) befindet? Begründe deine Antwort! (1P)
If you are to justify the answer, geogebra will not be sufficient as proof. The aim is to determine the shortest distance from the observer to the possible flight path.
The trajectory is a plane -x + 9y = 2700 with the normal vector (-1,9,0)
The lottery line on the plane from the observer is therefore:
g(r) = (500,100,0) + r*(-1,9,0)
Use the lot straight into the plane to determine the lot point:
-1*(500-r) + 9*(100+9r) = 2700
Solution r = 1150/41
The point is:
g(r) = L = (500 – 1150/41, 100 + 9*1150/41, 0)
Distance Lot Point – Observer ~ 253.99 meters
So far away from the observer.
Does it have to be geogebra?
The distance between two points is the root (x1 -x2)2 + (y1-y2)
Select x1 and y1 as coordinates of P and x2 and y2 -x+9y=2700.
With geogebra, it would have to be done if you let the straight line draw, put a slider G on it and let the length of the distance PG indicate.
Depends on whether Clemens can handle the camera well
I don’t know myself with Geobra – I think myself –
but it’s probably about finding out if one
the points of the straight lines are at most 200m away from P.