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 \(AB\), then \(AM = MB\).

Step2: Solve problem 10

Since \(T\) is the mid - point of \(\overline{SU}\), we have \(8x + 11=12x - 1\).
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 \(11x-7 = 3x + 9\).
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 \(5x-3=21 - x\).
Add \(x\) to both sides: \(6x-3 = 21\).
Add 3 to both sides: \(6x=24\).
Divide both sides by 6: \(x = 4\).
\(QS=(5x - 3)+(21 - x)=4x + 18\). Substitute \(x = 4\), \(QS=4\times4+18=34\).

Step5: Solve problem 13

Since \(B\) is the mid - point of \(\overline{AC}\) and \(AC = 8x-20\), then \(BC=\frac{1}{2}AC\).
Also, since \(AB = 3x - 1\) and \(BC=\frac{1}{2}(8x - 20)=4x-10\), and \(AC=AB + BC\), \(8x-20=(3x - 1)+(4x-10)\).
\(8x-20=7x-11\).
Subtract \(7x\) from both sides: \(x - 20=-11\).
Add 20 to both sides: \(x = 9\).
\(BC=4x-10=4\times9-10 = 26\).

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: \(x = 6\).
\(EF=6x - 4=6\times6-4 = 32\).
Since \(EF=EG + GF\) and \(GF = 13\), then \(EG=EF - GF=32-13 = 19\).

Step7: Solve problem 15

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

Answer:

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