9. if (lk = mk), (lk = 7x - 10), (kn=x + 3), (mn = 9x - 11), and (kj = 28), find (lj).
10. if (t) is the mid - point of (overline{su}), find (x).
11. if (g) is the mid - point of (overline{fh}), find (fg).
12. if (r) is the mid - point of (overline{qs}), find (qs).
13. if (b) is the mid - point of (overline{ac}), and (ac = 8x - 20), find (bc).
14. if (overline{ef}) bisects (overline{cd}), (cg = 5x - 1), (gd = 7x - 13), (ef = 6x - 4), and (gf = 13), find (eg).
15. if (r) is the mid - point of (overline{qs}), (rs = 2x - 4), (st = 4x - 1), and (rt = 8x - 43), find (qs).

Question

Accepted Answer

1. **Problem 9**:
   - Since $LK = MK$, we are not given enough information in the problem - statement to directly use this equality to find $LJ$. However, we know that $LJ=LK + KJ$. First, we need to find the value of $x$ using the other given line - segment relationships. But there is no clear way to do that with the information provided for this part. Let's assume we focus on the mid - point and line - segment equalities in other problems for a moment.
2. **Problem 10**:
   - # Explanation:
     - ## Step1: Set up the equation based on mid - point property
       - If $T$ is the mid - point of $SU$, then $ST = TU$. So, $8x + 11=12x−1$.
     - ## Step2: Solve for $x$
       - Subtract $8x$ from both sides: $11 = 12x−8x−1$, which simplifies to $11 = 4x−1$.
       - Add 1 to both sides: $11 + 1=4x$, so $12 = 4x$.
       - Divide both sides by 4: $x=\frac{12}{4}=3$.
   - # Answer:
     - $x = 3$
3. **Problem 11**:
   - # Explanation:
     - ## Step1: Set up the equation based on mid - point property
       - If $G$ is the mid - point of $FH$, then $FG = GH$. So, $11x−7 = 3x + 9$.
     - ## Step2: Solve for $x$
       - Subtract $3x$ from both sides: $11x−3x−7=9$, which simplifies to $8x−7 = 9$.
       - Add 7 to both sides: $8x=9 + 7$, so $8x = 16$.
       - Divide both sides by 8: $x = 2$.
     - ## Step3: Find $FG$
       - Substitute $x = 2$ into the expression for $FG$: $FG=11x−7=11\times2−7=22 - 7=15$.
   - # Answer:
     - $FG = 15$
4. **Problem 12**:
   - # Explanation:
     - ## Step1: Set up the equation based on mid - point property
       - If $R$ is the mid - point of $QS$, then $QR = RS$. So, $5x−3=21−x$.
     - ## Step2: Solve for $x$
       - Add $x$ to both sides: $5x+x−3=21$, which simplifies to $6x−3 = 21$.
       - Add 3 to both sides: $6x=21 + 3$, so $6x = 24$.
       - Divide both sides by 6: $x = 4$.
     - ## Step3: Find $QS$
       - $QS=QR + RS=(5x−3)+(21−x)$. Substitute $x = 4$: $QS=(5\times4−3)+(21−4)=(20 - 3)+17 = 17+17=34$.
   - # Answer:
     - $QS = 34$
5. **Problem 13**:
   - # Explanation:
     - ## Step1: Set up the equation based on mid - point property
       - If $B$ is the mid - point of $AC$, then $AB = BC$ and $AC = 2BC$. Also, $AB = 3x−1$ and $AC = 8x−20$. Since $AC = 2AB$, we have $8x−20 = 2(3x−1)$.
     - ## Step2: Expand and solve for $x$
       - Expand the right - hand side: $8x−20 = 6x−2$.
       - Subtract $6x$ from both sides: $8x−6x−20=-2$, which simplifies to $2x−20=-2$.
       - Add 20 to both sides: $2x=-2 + 20$, so $2x = 18$.
       - Divide both sides by 2: $x = 9$.
     - ## Step3: Find $BC$
       - Since $BC=\frac{AC}{2}$ and $AC = 8x−20$, substitute $x = 9$: $AC=8\times9−20=72 - 20 = 52$, so $BC=\frac{52}{2}=26$.
   - # Answer:
     - $BC = 26$
6. **Problem 14**:
   - Since $EF$ bisects $CD$, then $CG = GD$.
   - # Explanation:
     - ## Step1: Set up the equation
       - $5x−1=7x−13$.
     - ## Step2: Solve for $x$
       - Subtract $5x$ from both sides: $-1=7x−5x−13$, which simplifies to $-1 = 2x−13$.
       - Add 13 to both sides: $12 = 2x$.
       - Divide both sides by 2: $x = 6$.
     - ## Step3: Find $EF$
       - $EF = 6x−4=6\times6−4=36 - 4=32$.
     - ## Step4: Find $EG$
       - $EG=EF - GF$, and since $GF = 13$ and $EF = 32$, then $EG=32 - 13=19$.
   - # Answer:
     - $EG = 19$
7. **Problem 15**:
   - Since $R$ is the mid - point of $QS$, $QR = RS$. Also, $RT=RS + ST$.
   - # Explanation:
     - ## Step1: Set up the equation for $RT$
       - $RS + ST=RT$, so $(2x−4)+(4x−1)=8x−43$.
       - Combine like terms: $2x+4x−4−1=8x−43$, which simplifies to $6x−5=8x−43$.
     - ## Step2: Solve for $x$
       - Subtract $6x$ from both sides: $-5=8x−6x−43$, so $-5 = 2x−43$.
       - Add 43 to both sides: $38 = 2x$.
       - Divide both sides by 2: $x = 19$.
     - ## Step3: Find $QS$
       - $QS = 2RS$ (because $R$ is the mid - point of $QS$), and $RS=2x−4$. Substitute $x = 19$: $RS=2\times19−4=38 - 4 = 34$, so $QS = 2\times34=68$.
   - # Answer:
     - $QS = 68$

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