Question

1. if s is the midpoint of rt, rs = 5x + 17, and st = 8x - 31, find rs. 12. if line y bisects ac, ab = 4 - 5x, and bc = 2x + 25, find ac. 13. if b is the midpoint of ac, ac = cd, ab = 3x + 4, ac = 11x - 17, and ce = 49, find de.

Explanation:

Step1: Use mid - point property for problem 11

Since \(S\) is the mid - point of \(\overline{RT}\), then \(RS = ST\). Set up the equation \(5x + 17=8x - 31\).
\[

$$\begin{align*} 5x+17&=8x - 31\\ 17 + 31&=8x-5x\\ 48&=3x\\ x& = 16 \end{align*}$$

\]
Then find \(RS\) by substituting \(x = 16\) into the expression for \(RS\): \(RS=5x + 17=5\times16+17=80 + 17=97\).

Step2: Use mid - point property for problem 12

Since line \(y\) bisects \(\overline{AC}\), then \(AB = BC\). Set up the equation \(4-5x=2x + 25\).
\[

$$\begin{align*} 4-25&=2x + 5x\\ -21&=7x\\ x&=- 3 \end{align*}$$

\]
Then find \(AC=AB + BC\). Substitute \(x=-3\) into the expressions for \(AB\) and \(BC\): \(AB=4-5\times(-3)=4 + 15 = 19\), \(BC=2\times(-3)+25=-6 + 25 = 19\), so \(AC=19+19 = 38\).

Step3: Use mid - point property for problem 13

Since \(B\) is the mid - point of \(\overline{AC}\), then \(AB=\frac{1}{2}AC\). Given \(AB = 3x + 4\) and \(AC=11x-17\), we have \(3x + 4=\frac{1}{2}(11x - 17)\).
\[

$$\begin{align*} 2(3x + 4)&=11x-17\\ 6x+8&=11x - 17\\ 8 + 17&=11x-6x\\ 25&=5x\\ x&=5 \end{align*}$$

\]
Since \(AC = CD\) and \(AC=11x-17\), when \(x = 5\), \(AC=11\times5-17=55 - 17 = 38\). Given \(CE = 49\), then \(DE=CE - CD=49-38 = 11\).

Answer:

1. \(RS = 97\)
2. \(AC = 38\)
3. \(DE = 11\)