SOLUTION: A is a point (-1,6) on a Cartesian graph, and B is point (14,9) on the same graph. Point C is on the x axis. The least value of line AC + line CB is...
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Question 1127347: A is a point (-1,6) on a Cartesian graph, and B is point (14,9) on the same graph. Point C is on the x axis. The least value of line AC + line CB is...
Found 2 solutions by greenestamps, solver91311:
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!
The phrase "least value of line AC + line CB" has no meaning. Lines don't have "values".
And lines don't have lengths, either. Mathematically, a line extends infinitely in both directions.
What you are clearly looking for in this problem is the minimum total length of line SEGMENTS AC and CB. Or more simply, the shortest total distance from A to C to B.
The standard method of solution for this problem is to define point D to be the reflection in the x-axis of A: D(-1,-6). Then the distance from A to C is the same as the distance from D to C. And since the shortest distance between two point is a straight line, the shortest distance from A to C to B is the length of segment DB.
Segment DB has horizontal and vertical components of 15 each; the Pythagorean Theorem gives the answer to the problem as 15*sqrt(2).
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The measure of the segment 
where 
is the point )
and 
is an arbitrary point 
-axis, )
is:
^2\ +\ (6\ -\ 0)^2})
Similarly, the measure of 
is:
^2\ +\ (9\ -\ 0)^2})
The total distance is the sum of these two expressions, so:
\ =\ \sqrt{x^2\ +\ 2x\ +\ 37}\ +\ \sqrt{x^2\ -\ 28x\ +\ 277})
\ =\ \frac{2x\ +\ 2}{2\sqrt{x^2\ +\ 2x\ +\ 37}}\ +\ \frac{2x\ -\ 28}{2\sqrt{x^2\ -\ 28x\ +\ 277}})
Set the first derivative equal to zero and solve for the critical values. Make sure to discard any extraneous roots. Then evaluate )
for that critical value.
John
My calculator said it, I believe it, that settles it
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