Question

1. if k is the mid - point of mj, mk = 7x - 12, kv=x + 3, mv = 9x - 11, and kj = 28, find lj.
2. if t is the mid - point of su, find x.
3. if g is the mid - point of fh, find fg.
4. if r is the mid - point of qs, find qs.
5. if b is the mid - point of ac, and ac = 8x - 20, find bc.
6. if ef bisects cd, cg = 5x - 1, gd = 7x - 13, ef = 6x - 4, and gf = 13, find eg.
7. if r is the mid - point of qs, rs = 2x - 4, st = 4x - 1, and rt = 8x - 43, find qs.

Explanation:

Step1: Use mid - point property

If a point is a mid - point, the two segments it divides are equal. For example, in problem 10, since \(T\) is the mid - point of \(SU\), \(8x + 11=12x−1\).

Step2: Solve for \(x\)

\(11 + 1=12x−8x\), \(4x = 12\), \(x = 3\).

Step3: Find the required length

In problem 11, since \(G\) is the mid - point of \(FH\), \(11x−7 = 3x + 9\), \(11x−3x=9 + 7\), \(8x = 16\), \(x = 2\), and \(FG=11x−7=11\times2−7 = 15\).

Answer:

1. \(x = 3\)
2. \(FG = 15\)
3. Solve \(5x−3=21−x\) for \(x\) (\(6x=24\), \(x = 4\)), \(QS=(5x−3)+(21−x)=28\)
4. Since \(B\) is mid - point of \(AC\), \(3x−1=\frac{8x−20}{2}\), \(3x−1 = 4x−10\), \(x = 9\), \(BC=3x−1=26\)
5. Since \(EF\) bisects \(CD\), \(5x−1=7x−13\), \(13 - 1=7x−5x\), \(2x = 12\), \(x = 6\), \(EF=6x−4 = 32\), \(EG=EF+GF=32 + 13=45\)
6. Since \(R\) is mid - point of \(QS\), and \(RS+ST = RT\), \((2x−4)+(4x−1)=8x−43\), \(6x−5=8x−43\), \(43 - 5=8x−6x\), \(2x = 38\), \(x = 19\), \(QS = 2RS=2(2x−4)=2\times(2\times19−4)=68\)