Mathematics, 19.12.2020 14:00 autumnguidry6867

Consider points A(1, 6) and B(8, 8). Find point C on the x-axis so AC +BC is a minimum. The point that minimizes AC +BC is C (_ , _)

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Answer from: lazybridplayer

The coordinates of a point is the position of the point, on the coordinate plane.

The point C that minimizes AC + BC are (-20, 0) and (4, 0)

The given parameters are:

$\mathbf{A = (1,6)}$

$\mathbf{B = (8,8)}$

From the question, we understand that point C is on the x-axis.

This means that:

$\mathbf{C = (x,0)}$

Calculate segments AC and BC using the following distance formula

$\mathbf{d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}}$

So, we have:

$\mathbf{AC = \sqrt{(1 - x)^2 + (6 - 0)^2}}$

$\mathbf{AC = \sqrt{(1 - x)^2 + 6^2}}$

$\mathbf{BC = \sqrt{(8 - x)^2 + (8 -0)^2}}$

$\mathbf{BC = \sqrt{(8 - x)^2 + 8 ^2}}$

Add both distance

$\mathbf{AB + BC = \sqrt{(1 - x)^2 + 6^2} + \sqrt{(8 - x)^2 + 8^2}}$

Differentiate, to calculate the minimum distance

$\mathbf{d' = (1 - x) \times 2 \times (0.5 - 1) \times ((1 - x)^2 + 6^2)^({0.5 - 1}) + (8 - x) \times 2 \times (0.5 - 1) \times ((8 - x)^2} + 8^2))}^{(0.5 - 1)} }$

Simplify

$\mathbf{d' = -\frac{1 - x}{\sqrt{(1 - x)^2 + 6^2}}- \frac{8 - x}{\sqrt{(8 - x)^2 + 8^2}}}$

Set to 0

$\mathbf{ -\frac{1 - x}{\sqrt{(1 - x)^2 + 6^2}} - \frac{8 - x}{\sqrt{(8 - x)^2 + 8^2}} = 0}$

Rewrite as:

$\mathbf{ -\frac{1 - x}{\sqrt{(1 - x)^2 + 6^2}} = \frac{8 - x}{\sqrt{(8 - x)^2 + 8^2}} }$

Cross multiply

$\mathbf{(8 - x) \times\sqrt{(1 - x)\² + 6\²} = -(1 - x) \times\sqrt{(8 - x)\² + 8\²}}$

Square both sides

$\mathbf{(8 - x)^2 \times (1 - x)\² + 6\² = (-(1 - x))^2 \times (8 - x)\² + 8\²}$

$\mathbf{(8 - x)^2 \times (1 - x)\² + 6\² = (1 - x)^2 \times (8 - x)\² + 8\²}$

Expand

$\mathbf{x^4 - 18x^3 + 133x^2 -720x + 2368 = x^4 - 18x^3 + 161x^2 - 272x + 128}$

Evaluate like terms

$\mathbf{133x^2 -720x + 2368 = 161x^2 - 272x + 128}$

Collect like terms

$\mahbf{161x^2 - 133x^2 - 272x + 720x + 128 - 2368 = 0}$

$\mathbf{28x^2 + 448x - 2240 = 0}$

Factor out 28

$\mathbf{28(x^2 + 16x - 80) = 0}$

Divide through by 28

$\mathbf{x^2 + 16x - 80 = 0}$

Expand

$\mathbf{x^2 + 20x - 4x - 80 = 0}$

Factorize

$\mathbf{x(x + 20) + 4(x + 20) = 0}$

Factor out x + 20

$\mathbf{(x + 20)(x - 4) = 0}$

Solve for x

$\mathbf{x =- 20\ or\ x = 4}$

Hence, the point C that minimizes AC + BC are (-20, 0) and (4, 0)

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