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Question 1171826: Points B(b,0) and C(c,10) are to be placed on this graph so that the distance from A to B to C to D is a minimum. Find the value of b.
A) 3 4/15
B) 1 7/15
C) 2 2/15
D) 2 4/15
E) 2 7/15
https://ibb.co/PmXZ17b
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Note: If the link died or a student is not able to access said link, then it shows a diagram where points A and D are located at (0,4) and (8,9) respectively.
On the diagram given, draw the horizontal lines y = 0 and y = 10. The line y = 0 goes perfectly on top of the x axis. Point B is somewhere on this line. Point C is somewhere on the line y = 10.
Reflect A over the line y = 0, and you'll have A' = (0,-4)
Reflect D over the line y = 10, and you'll have D' = (8,11). Note how D and D' are the same distance away from the line y = 10
Now draw a straight line from A' to D'.
Plot point B at the intersection of y = 0 and line segment A'D'
Plot point C at the intersection of y = 10 and line segment A'D'
The distance from A' to D' is the shortest path possible such that we start at A, go to B, move to C, then finally arrive at D.
The proof is fairly straightforward (pun perhaps intended). This is what the diagram looks like
I used GeoGebra to make the diagram.
Note we have two pairs of congruent triangles.
One pair of congruent triangles is AEB and A'EB
The other pair of congruent triangles is CFD and CFD'
Since triangle AEB = triangle A'EB, we know that AB = A'B. Afterall we applied a reflection which preserves distance. Similarly, CD = CD'
The shortest path between any two points is always a straight line. Points B and C are on line segment A'D', which means that going from A' to D' will involve going through B and C as well.
In other words, the shortest path from A' to D' is: A' to B to C to D'
In terms of algebra, we know that A'B+BC+CD' is at its smallest or at its minimum.
But because AB = A'B and CD = CD', we know that the value of AB+BC+CD is the smallest possible as well. This concludes the proof and verifies that the diagram has the shortest path from A to B to C to D.
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From here we just need to figure out the location of B
We'll need the locations of A' and D' first
A' = (0,-4)
D' = (8,11)
The slope of the line going through these points is
m = (y2-y1)/(x2-x1)
m = (11-(-4))/(8-0)
m = (11+4)/(8-0)
m = (15)/(8-0)
m = 15/8
The y intercept of line A'D' is -4 since it's the y coordinate of A'(0,-4)
The equation y = mx+b turns into y = (15/8)x-4 which is the line going through points A' and D'.
Plug in y = 0 and solve for x to find the x intercept. This is the x coordinate of point B
y = (15/8)x-4
0 = (15/8)x-4
4 = (15/8)x
4*8 = 15x
32 = 15x
15x = 32
x = 32/15
x = (30+2)/15
x = 30/15 + 2/15
x = 2 + 2/15
x = 2 & 2/15
The location of B(b,0) is (2&2/15, 0)
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Answer: C) 2 & 2/15
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