Answer:
The coordinates of the point C that minimizes AC + BC are (-20, 0) or (4, 0)
Step-by-step explanation:
The given coordinates of the points A and B are A(1, 6) and B(8, 8)
The location of the point C = The x-axis
Therefore;
The coordinates of the point C = (x, 0)
The length of the segment AC = √((1 - x)² + (6 - 0)²) = √((1 - x)² + 6²)
The length of the segment BC = √((8 - x)² + (8 - 0)²) = √((8 - x)² + 8²)
At minimum distance of AC + BC, we have;
d(√((1 - x)² + 6²) +√((8 - x)² + 8²))/dx = 0 = (1 - x) × 2 × (0.5 - 1)× (√((1 - x)² + 6²)^(0.5 - 1) + (8 - x) × 2 × (0.5 - 1)× √((8 - x)² + 8²)^(0.5 - 1)
∴ d(√((1 - x)² + 6²) +√((8 - x)² + 8²))/dx = 0 = -(1 - x)/√((1 - x)² + 6²) - (8 - x)/√((8 - x)² + 8²)
-(1 - x)/√((1 - x)² + 6²) = (8 - x)/√((8 - x)² + 8²)
(8 - x)·√((1 - x)² + 6²) = -(1 - x)·√((8 - x)² + 8²)
Squaring both sides gives;
(8 - x)²·((1 - x)² + 6²) = (1 - x)²·((8 - x)² + 8²)
Expanding, using an online tool, we get;
x⁴ - 18·x³ + 133·x² -720·x + 2368 = x⁴ - 18·x³ + 161·x² - 272·x + 128
Which gives;
(161 - 133)·x² - (272 - 720)·x + 128 - 2368 = 28·x² + 448·x - 2240 = 0
Dividing by 28 gives;
x² + 16·x - 80 = 0
(x + 20)·(x - 4) = 0
Therefore, x = -20 or x = 4
The coordinates of the point C that minimizes AC + BC are (-20, 0) or (4, 0)
