SOLUTION: two cars leave an intersection, one traveling west and the other south. After some time the slower car is 7 mi nearer to the intersection than the faster car. At that time, two car

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Question 801179: two cars leave an intersection, one traveling west and the other south. After some time the slower car is 7 mi nearer to the intersection than the faster car. At that time, two cars are 17 mi apart. How far did each car travel?
Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
West and South are directions perpendicular to each other.
We assume that the trajectories of the two cars are perpendicular straight lines.
(Those must be long, straight North-South and East-West roads, like the ones I have seen in South Dakota).
We can call one car the slower car, and the other car the faster car, but there is no way to know which one was going West and which one was going South.


At a certain time after both cars have left the intersection, the distances traveled after leaving the intersection, and the distance between the two car, all measured in miles, are as follows:
= distance traveled by the faster car
= distance traveled by the slower car
= distance between the two cars
It's either or
Either way, we have a right triangle with leg lengths and and hypotenuse .
According to the Pythagorean theorem,

Solving that equation for :




Dividing both sides of the equation by 2, we simplify to get

Factoring , we transform the equation into
to solve it by factoring.
The solutions for that equation are
derived from (which does not make sense, because we expect a positive distance )
and from
Since , -->
The cars have traveled miles and miles away from the intersection.

OTHER OPTIONS:
can also be solved by applying the quadratic formula, or by "completing the square." For this particular equation, the quadratic formula may be more popular.
The quadratic formula says that the solutions to an equation of the form
are given by

For , , and , so

Since the negative solution does not make sense as a distance,
--> -->

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