The problem asks us to find the equations of lines $L_2$ that pass through a given point and are parallel or perpendicular to a given line $L_1$. The problems include finding the lines: 5) Parallel to $x - y + 5 = 0$ through $(-1, 1)$. 6) Parallel to $x + y + 1 = 0$ through $(0, -2)$. 7) Parallel to $x - 2y + 6 = 0$ through $(3, 0)$. 8) Parallel to $x + y - 3 = 0$ through $(-2, 2)$. 1) Perpendicular to $3x - y + 1 = 0$ through $(3, 2)$. 2) Perpendicular to $2x - 2y + 5 = 0$ through $(3, 0)$. 3) Perpendicular to $x - y + 5 = 0$ through $(-1, 1)$. 4) Perpendicular to $x + 3y - 3 = 0$ through $(-2, 2)$. 5) Perpendicular to $x + y + 1 = 0$ through $(0, 2)$. 6) Perpendicular to $x + 2y - 4 = 0$ through $(2, -1)$.

1. Problem Description

The problem asks us to find the equations of lines $L_2$ that pass through a given point and are parallel or perpendicular to a given line $L_1$. The problems include finding the lines:

5) Parallel to $x - y + 5 = 0$ through $(-1, 1)$.

6) Parallel to $x + y + 1 = 0$ through $(0, -2)$.

7) Parallel to $x - 2y + 6 = 0$ through $(3, 0)$.

8) Parallel to $x + y - 3 = 0$ through $(-2, 2)$.

1) Perpendicular to $3x - y + 1 = 0$ through $(3, 2)$.

2) Perpendicular to $2x - 2y + 5 = 0$ through $(3, 0)$.

3) Perpendicular to $x - y + 5 = 0$ through $(-1, 1)$.

4) Perpendicular to $x + 3y - 3 = 0$ through $(-2, 2)$.

5) Perpendicular to $x + y + 1 = 0$ through $(0, 2)$.

6) Perpendicular to $x + 2y - 4 = 0$ through $(2, -1)$.

2. Solution Steps

To find the equation of a line, we need a point and a slope. We are given the point. We need to find the slope.

For parallel lines, the slopes are equal: $m_2 = m_1$.

For perpendicular lines, the slopes are negative reciprocals: $m_2 = -1/m_1$.

Given a line $Ax + By + C = 0$, the slope is $m = -A/B$.

5) Parallel to $x - y + 5 = 0$ through $(-1, 1)$.

$m_1 = -1/(-1) = 1$. Therefore $m_2 = 1$.

$y - y_0 = m(x - x_0) \implies y - 1 = 1(x - (-1)) \implies y - 1 = x + 1 \implies y = x + 2$ or $x - y + 2 = 0$.

6) Parallel to $x + y + 1 = 0$ through $(0, -2)$.

$m_1 = -1/1 = -1$. Therefore $m_2 = -1$.

$y - (-2) = -1(x - 0) \implies y + 2 = -x \implies y = -x - 2$ or $x + y + 2 = 0$.

7) Parallel to $x - 2y + 6 = 0$ through $(3, 0)$.

$m_1 = -1/(-2) = 1/2$. Therefore $m_2 = 1/2$.

$y - 0 = (1/2)(x - 3) \implies y = (1/2)x - (3/2) \implies 2y = x - 3$ or $x - 2y - 3 = 0$.

8) Parallel to $x + y - 3 = 0$ through $(-2, 2)$.

$m_1 = -1/1 = -1$. Therefore $m_2 = -1$.

$y - 2 = -1(x - (-2)) \implies y - 2 = -x - 2 \implies y = -x$ or $x + y = 0$.

1) Perpendicular to $3x - y + 1 = 0$ through $(3, 2)$.

$m_1 = -3/(-1) = 3$. Therefore $m_2 = -1/3$.

$y - 2 = (-1/3)(x - 3) \implies 3y - 6 = -x + 3 \implies x + 3y - 9 = 0$.

2) Perpendicular to $2x - 2y + 5 = 0$ through $(3, 0)$.

$m_1 = -2/(-2) = 1$. Therefore $m_2 = -1$.

$y - 0 = -1(x - 3) \implies y = -x + 3 \implies x + y - 3 = 0$.

3) Perpendicular to $x - y + 5 = 0$ through $(-1, 1)$.

$m_1 = -1/(-1) = 1$. Therefore $m_2 = -1$.

$y - 1 = -1(x - (-1)) \implies y - 1 = -x - 1 \implies y = -x$ or $x + y = 0$.

4) Perpendicular to $x + 3y - 3 = 0$ through $(-2, 2)$.

$m_1 = -1/3$. Therefore $m_2 = 3$.

$y - 2 = 3(x - (-2)) \implies y - 2 = 3x + 6 \implies y = 3x + 8$ or $3x - y + 8 = 0$.

5) Perpendicular to $x + y + 1 = 0$ through $(0, 2)$.

$m_1 = -1/1 = -1$. Therefore $m_2 = 1$.

$y - 2 = 1(x - 0) \implies y - 2 = x \implies y = x + 2$ or $x - y + 2 = 0$.

6) Perpendicular to $x + 2y - 4 = 0$ through $(2, -1)$.

$m_1 = -1/2$. Therefore $m_2 = 2$.

$y - (-1) = 2(x - 2) \implies y + 1 = 2x - 4 \implies y = 2x - 5$ or $2x - y - 5 = 0$.

3. Final Answer

5) $x - y + 2 = 0$

6) $x + y + 2 = 0$

7) $x - 2y - 3 = 0$

8) $x + y = 0$

1) $x + 3y - 9 = 0$

2) $x + y - 3 = 0$

3) $x + y = 0$

4) $3x - y + 8 = 0$

5) $x - y + 2 = 0$

6) $2x - y - 5 = 0$