Question

points a, b, and c are collinear. solve for x ab = 2x+1, bc = 15, and ac = 5x-5.
points a, b, and c are collinear. solve for ab ab = 4x-1, bc = 5x+11, and ac = 100.
solve for x (diagram: a---b---c, ab=5, bc=x²+6, ac=20)
solve for bc (diagram: a---b---c, ab=x+2, bc=3x+2, ac=5x-1)
solve for x (diagram: a---b---c, ab=x+12, bc=2(5-x), ac=32)

Explanation:

First Problem (Solve for \( x \) with \( AB = 2x + 1 \), \( BC = 15 \), \( AC = 5x - 5 \))

Step 1: Use collinearity ( \( AB + BC = AC \))

Since \( A \), \( B \), \( C \) are collinear, \( AB + BC = AC \). Substitute the given expressions:
\( (2x + 1) + 15 = 5x - 5 \)

Step 2: Simplify left side

Combine like terms: \( 2x + 16 = 5x - 5 \)

Step 3: Subtract \( 2x \) from both sides

\( 16 = 3x - 5 \)

Step 4: Add 5 to both sides

\( 21 = 3x \)

Step 5: Divide by 3

\( x = \frac{21}{3} = 7 \)

Step 1: Use collinearity ( \( AB + BC = AC \))

Substitute the expressions: \( (4x - 1) + (5x + 11) = 100 \)

Step 2: Simplify left side

Combine like terms: \( 9x + 10 = 100 \)

Step 3: Subtract 10 from both sides

\( 9x = 90 \)

Step 4: Divide by 9

\( x = 10 \)

Step 5: Find \( AB \) (substitute \( x = 10 \) into \( AB = 4x - 1 \))

\( AB = 4(10) - 1 = 40 - 1 = 39 \)

Step 1: Use collinearity ( \( AB + BC = AC \))

Substitute the expressions: \( 5 + (x^2 + 6) = 20 \)

Step 2: Simplify left side

Combine like terms: \( x^2 + 11 = 20 \)

Step 3: Subtract 11 from both sides

\( x^2 = 9 \)

Step 4: Take square root

\( x = \pm 3 \) (since length is positive, \( x = 3 \) or \( x = -3 \); but \( x^2 + 6 \) must be positive, both are valid, but typically \( x \) is real, so \( x = 3 \) or \( x = -3 \))

Answer:

\( x = 7 \)

Second Problem (Solve for \( AB \) with \( AB = 4x - 1 \), \( BC = 5x + 11 \), \( AC = 100 \))