Question

12 a) given: m is the mid - point of cd, draw a diagram. if cm = 6x - 1 and md = 10x - 13. find the value of x. 12 b) given the number line to the right, find ln and mp hint: just count! 12 c) given the number line to the right, determine the mid - point of mp 12 d) if lp = 3x - 13, find the value of x. solve for x using the segment addition postulate (\\(\overline{ab}+\overline{bc}=\overline{ac}\\)). 13) g •\\(\frac{x - 5}{h}\\)\\(\frac{2x - 8}{i}\\) 14) t •\\(\frac{9}{u}\\)\\(\frac{2x - 6}{v}\\) 15. given line m is perpendicular to n, complete the sentence below, then solve for x. show all work. *if two angles are complementary, then
m ← (8x - 1) (3x + 3)° n

Explanation:

12 a)

Step1: Apply mid - point property

Since M is the mid - point of CD, CM = MD. So, we set up the equation \(6x - 1=10x - 13\).

Step2: Isolate x terms

Subtract \(6x\) from both sides: \(-1 = 10x-6x - 13\), which simplifies to \(-1 = 4x-13\).

Step3: Isolate the constant

Add 13 to both sides: \(-1 + 13=4x\), so \(12 = 4x\).

Step4: Solve for x

Divide both sides by 4: \(x=\frac{12}{4}=3\).

Count the number of units between the points on the number - line.
For LN, starting from L at - 4 and ending at N at 1, the distance \(LN=|1-(-4)| = 5\).
For MP, starting from M at - 2 and ending at P at 4, the distance \(MP=|4 - (-2)|=6\).

The mid - point of a line segment with endpoints \(x_1\) and \(x_2\) on a number line is \(\frac{x_1 + x_2}{2}\). M is at - 2 and P is at 4.
The mid - point is \(\frac{-2 + 4}{2}=\frac{2}{2}=1\), which is the location of N.

Answer:

\(x = 3\)

12 b)