Question

1. points a, b, c, and d are on a line such that b is between a and c, and c is between b and d. the distance from a to b is 6 units. the distance from b to c is twice the distance from a to b, and the distance from c to d is twice the distance from b to c. what is the distance, in units, from the midpoint of bc to the midpoint of cd? a. 18 b. 14 c. 12 d. 9 d. 6

Explanation:

Step1: Define the given distances

Let \(AB = 6\) units. Since \(BC\) is between \(B\) and \(D\) and \(B\) is between \(A\) and \(C\), and the distance from \(B\) to \(C\) is twice the distance from \(C\) to \(D\), let \(CD=x\), then \(BC = 2x\). Also, the distance from \(A\) to \(B\) is \(6\) units.

Step2: Find the mid - points

Let \(M_1\) be the mid - point of \(BC\) and \(M_2\) be the mid - point of \(CD\). The length of \(BC = 2x\), so the mid - point \(M_1\) of \(BC\) divides \(BC\) into two equal parts, each of length \(x\). The length of \(CD=x\), so the mid - point \(M_2\) of \(CD\) divides \(CD\) into two equal parts, each of length \(\frac{x}{2}\).

Step3: Calculate the distance between the mid - points

The distance from the mid - point of \(BC\) to the mid - point of \(CD\) is \(x+\frac{x}{2}=\frac{2x + x}{2}=\frac{3x}{2}\).
We know that \(AB = 6\), \(BC = 2x\), and \(CD=x\). Since the points are on a line, we don't need to consider the overall line structure for this part of the problem. The distance between the mid - point of \(BC\) and the mid - point of \(CD\) is related to the lengths of \(BC\) and \(CD\). Since \(BC = 2x\) and \(CD=x\), the distance between the mid - point of \(BC\) and the mid - point of \(CD\) is \(\frac{3}{2}\times CD\).
If we assume \(CD = 6\) (because if we consider the relationships and the fact that we are looking for a distance value), then the distance from the mid - point of \(BC\) to the mid - point of \(CD\) is \(\frac{3}{2}\times6=9\) units.

Answer:

D. 9