Answer:
The coordinate of the point C is (-10/3, 0)
Step-by-step explanation:
The coordinates of the given points are A(-1, 7), B(5, -4)
Let the coordinate of the point C on the x-axis = C(x, 0)
Therefore, we have;
The square of the distance, d², between two points having coordinates, (x₁, y₁), and (x₂, y₂), is given by the following formula;
d² = (x₂ - x₁)² + (y₂ - y₁)²
The square of the distance from the point C to A, AC², is given as follows;
AC² = (x - (-1))² + (0 - 7)² = (x + 1)² + 7² = x² + 2·x + 1 + 49 = x² + 2·x + 50
Similarly, the square of the distance from the point C to B, AB², is given as follows;
BC² = (x - (-4))² + (0 - 5)² = (x + 4)² + 5² = x² + 10·x + 25 + 25= x² + 10·x + 50
Therefore;
AC² + BC² = x² + 2·x + 50 + x² + 10·x + 50 = 2·x² + 12·x + 100
(AC + BC)² = AC² + BC² + 2·AB·BC
2·AB·BC = √(4·AB²·BC²) =√(4 × (x² + 2·x + 50) × (x² + 10·x + 50))
2·AB·BC = √(4·x⁴ + 48·x³ + 480·x² + 2400·x + 10,000)
(AC + BC)² = 2·x² + 12·x + 100 + √(4·x⁴ + 48·x³ + 480·x² + 2400·x + 10,000)
AB + BC will be smallest when (AC + BC)² is smallest
Differentiating (AC + BC)² and equating to zero, to find the minimum point, gives;
d(2·x² + 12·x + 100 + √(4·x⁴ + 48·x³ + 480·x² + 2400·x + 10,000))/dx = [tex]\dfrac{16\cdot x^{3} + 144\cdot x^{2} + 960\cdot x + 2400}{2\cdot \sqrt{4\cdot x^{4} + 48\cdot x^{3} + 480\cdot x^{2} + 2400\cdot x + 10000}} + 4\cdot x +12 = 0[/tex]
Solving using an online application, gives x = -10/3
Therefore, the point C has coordinates C(-10/3, 0).
