Question

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

Explanation:

Step1: Recall mid - point property

If a point \(M\) is the mid - point of a line segment \(\overline{AB}\), then \(AM = MB\).

Step2: Solve problem 10

Since \(T\) is the mid - point of \(\overline{SU}\), we have \(ST=TU\). So, \(8x + 11=12x-1\).
Subtract \(8x\) from both sides: \(11 = 4x-1\).
Add 1 to both sides: \(12 = 4x\).
Divide both sides by 4: \(x = 3\).

Step3: Solve problem 11

Since \(G\) is the mid - point of \(\overline{FH}\), we have \(FG=GH\). So, \(11x-7=3x + 9\).
Subtract \(3x\) from both sides: \(8x-7=9\).
Add 7 to both sides: \(8x=16\).
Divide both sides by 8: \(x = 2\). Then \(FG=11x-7=11\times2-7=15\).

Step4: Solve problem 12

Since \(R\) is the mid - point of \(\overline{QS}\), we have \(QR = RS\). So, \(5x-3=21 - x\).
Add \(x\) to both sides: \(6x-3=21\).
Add 3 to both sides: \(6x=24\).
Divide both sides by 6: \(x = 4\). Then \(QS=QR + RS=(5x-3)+(21 - x)=4x + 18=4\times4+18=34\).

Step5: Solve problem 13

Since \(B\) is the mid - point of \(\overline{AC}\), \(BC=\frac{1}{2}AC\). Given \(AC = 8x-20\), then \(BC=\frac{8x-20}{2}=4x-10\). Also, since \(AB = BC\) and \(AB = 3x-1\), we have \(3x-1=4x-10\).
Subtract \(3x\) from both sides: \(-1=x - 10\).
Add 10 to both sides: \(x = 9\). Then \(BC=4x-10=4\times9-10=26\).

Step6: Solve problem 14

Since \(\overline{EF}\) bisects \(\overline{CD}\), \(CG=GD\). So, \(5x-1=7x-13\).
Subtract \(5x\) from both sides: \(-1=2x-13\).
Add 13 to both sides: \(12=2x\).
Divide both sides by 2: \(x = 6\). Then \(EF=6x-4=6\times6-4 = 32\), and since \(GF = 13\), \(EG=EF - GF=32-13=19\).

Step7: Solve problem 15

Since \(R\) is the mid - point of \(\overline{QS}\), \(QR=RS\). Also, \(RT=RS+ST\). Given \(RS = 2x-4\) and \(ST=4x-1\) and \(RT=8x-43\), we have \((2x-4)+(4x-1)=8x-43\).
Combine like - terms: \(6x-5=8x-43\).
Subtract \(6x\) from both sides: \(-5=2x-43\).
Add 43 to both sides: \(38=2x\).
Divide both sides by 2: \(x = 19\). Then \(QS=2RS=2(2x-4)=2(2\times19-4)=2\times34 = 68\).

Answer:

1. \(x = 3\)
2. \(FG = 15\)
3. \(QS = 34\)
4. \(BC = 26\)
5. \(EG = 19\)
6. \(QS = 68\)