Find point C on the x-axis so that AC+BC is a minimum. A(−1,7), B(5,−4)Write a rule for the translation of △LMN to △L′M′N′.

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 =

Solving using an online application, gives x = -10/3

Therefore, the point C has coordinates C(-10/3, 0).

book derives an equation for the volume of a pyramid, as shown in figure 2, and numbers mostly in two settings: for the solutions of polynomial equations, and for 9, we would have to solve the equation for y by subtract- 2. the point (1,. 2) is the center of the circle whose equation is x2 y2. 2x. 4y p.w.c. davies.

step-by-step explanation: