Two cars leave an intersection at the same time. One travels north and the other travels east. When the eastbound car is 1 mile further than the northbound car from the intersection, the distance between them is 2 miles more than the northbound car's distance from the intersection. How far is the northbound car from the intersection?
use the pythagoras theorem :\[c^2 = x^2 + y^2\] where c = 2 miles x = 1 mile so plug this in the theorem and you'll get the following: \[(2)^2 - 1 = y^2\] \[y = \sqrt(4-1) \rightarrow y = \sqrt(3)\] I think so, correct me if I'm wrong ^_^
I think you made a mistake :)
I concur.
If the distance of the northbound car from the intersection point is x, then the distance of the eastbound car from the intersection is y+1, and the distance between the two cars is 2+y.
So, \[(2+y)^2=y^2+(y+1)^2\]
Solve the quadratic equation, you get y=-1 or y=3. Take only the positive value of y. Therefore the northbound car is 3 miles away from the intersection.
I should go prepare for my exam, or I am definitely getting a very bad grade.
why is it 2 + y?
oh, right, nvm lol, I again misread the question, sorry ^_^"
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