Problem Breakdown and Approach

To determine the mutual arrangement of lines $L_1$, $L_2$, and $L_3$, and analyze their relationships, we proceed as follows:

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Line Representations

1. Line $L_1$:

   • Given in parametric form: $\vec{p} = (2, -1), \quad M_0 = (1, 0)$ Hence, its equation is: $\vec{r} = \vec{M}_0 + t\vec{p} \implies (x, y) = (1, 0) + t(2, -1)$

     Parametric equations: $x = 1 + 2t, \quad y = -t$

     Implicit equation can be derived: $y = -\frac{1}{2}(x - 1) \implies 2x + y - 2 = 0$

2. Line $L_2$:

   • Given by two points $N_0 = (-5, 3)$ and $N_1 = (-3, 2)$: Direction vector $\vec{d} = \vec{N}_1 - \vec{N}_0 = (-3 - (-5), 2 - 3) = (2, -1)$.

     Parametric equations: $x = -5 + 2t, \quad y = 3 - t$

     Implicit equation: $y = -\frac{1}{2}(x + 5) + 3 \implies 2x + y + 7 = 0$

3. Line $L_3$:

   • Given a normal vector $\vec{n} = (5, 1)$ and point $K_0 = (-2, 1)$: Implicit equation: $5(x + 2) + 1(y - 1) = 0 \implies 5x + y + 9 = 0$

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Mutual Arrangements

Step 1: Check if Lines are Parallel, Intersecting, or Coincident

For two lines $ax + by + c = 0$ and $a'x + b'y + c' = 0$:

1. Parallel: $\frac{a}{a'} = \frac{b}{b'}$ (equal direction ratios).
2. Coincident: Parallel and $\frac{c}{c'} = \frac{a}{a'} = \frac{b}{b'}$.
3. Intersecting: Not parallel.

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Step 2: Intersection Point (if Intersecting)

Solve the linear system of equations for the two lines.

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Step 3: Distance Between Parallel Lines

Distance $d$ between two parallel lines $ax + by + c = 0$ and $ax + by + c' = 0$: $d = \frac{|c - c'|}{\sqrt{a^2 + b^2}}$

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Step 4: Angle Between Intersecting Lines

Angle $\theta$ between two lines: $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|$ or, using normal vectors: $\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{\|\vec{n}_1\| \|\vec{n}_2\|}$

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Calculations

Parallel Check:

• $L_1: 2x + y - 2 = 0$
• $L_2: 2x + y + 7 = 0$
• $L_3: 5x + y + 9 = 0$

1. $L_1 \parallel L_2$ because $\frac{2}{2} = \frac{1}{1}$.
2. $L_1$ and $L_3$ are not parallel ($\frac{2}{5} \neq \frac{1}{1}$).
3. $L_2$ and $L_3$ are not parallel ($\frac{2}{5} \neq \frac{1}{1}$).

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Intersections:

1. $L_1 \cap L_3$: Solve: 2x + y - 2 = 0 \\ 5x + y + 9 = 0 \end{cases}$$ Subtract: $$3x = -11 \implies x = -\frac{11}{3}, \quad y = \frac{20}{3}$$ Intersection: $$(-\frac{11}{3}, \frac{20}{3})$$.
2. $L_2 \cap L_3$: Solve: 2x + y + 7 = 0 \\ 5x + y + 9 = 0 \end{cases}$$ Subtract: $$3x = -2 \implies x = -\frac{2}{3}, \quad y = -\frac{19}{3}$$ Intersection: $$(-\frac{2}{3}, -\frac{19}{3})$$.
3. $L_1 \cap L_2$: Solve: 2x + y - 2 = 0 \\ 2x + y + 7 = 0 \end{cases}$$ No solution (parallel).

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Distance Between $L_1$ and $L_2$:

$d = \frac{|(-2) - 7|}{\sqrt{2^2 + 1^2}} = \frac{9}{\sqrt{5}}$

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Angles:

• Angle between $L_1$ and $L_3$: $\cos \theta = \frac{|(2)(5) + (1)(1)|}{\sqrt{2^2 + 1^2} \sqrt{5^2 + 1^2}} = \frac{11}{\sqrt{5} \sqrt{26}}$ $\theta = \arccos\left(\frac{11}{\sqrt{5} \sqrt{26}}\right)$

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