The image presents several problems. The first set of problems (5-8) asks us to find the equation of a line $L_2$ that passes through a given point $P$ and is parallel to a given line $L_1$. The second set of problems (1-6) asks us to find the equation of a line $L_2$ that passes through a given point $P$ and is perpendicular to a given line $L_1$. Let's solve the first problem, number 1.

1. Problem Description

The image presents several problems. The first set of problems (5-8) asks us to find the equation of a line $L_2$ that passes through a given point $P$ and is parallel to a given line $L_1$. The second set of problems (1-6) asks us to find the equation of a line $L_2$ that passes through a given point $P$ and is perpendicular to a given line $L_1$. Let's solve the first problem, number

1.

2. Solution Steps

Problem 1: Find the equation of the line $L_2$ that passes through $P(3,2)$ and is perpendicular to $L_1: 3x - y + 1 = 0$.

Step 1: Find the slope of line $L_1$. We can rewrite the equation of $L_1$ in slope-intercept form ($y = mx + b$) to find the slope.

$3x - y + 1 = 0$

$y = 3x + 1$

So, the slope of $L_1$ is $m_1 = 3$.

Step 2: Find the slope of line $L_2$. Since $L_2$ is perpendicular to $L_1$, the product of their slopes is $-1$. Therefore,

$m_1 * m_2 = -1$

$3 * m_2 = -1$

$m_2 = -\frac{1}{3}$

Step 3: Use the point-slope form of a line to find the equation of $L_2$. The point-slope form is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope. We have the point $P(3,2)$ and the slope $m_2 = -\frac{1}{3}$.

$y - 2 = -\frac{1}{3}(x - 3)$

Step 4: Convert the equation to slope-intercept form or standard form.

$y - 2 = -\frac{1}{3}x + 1$

$y = -\frac{1}{3}x + 3$

Multiplying by 3:

$3y = -x + 9$

$x + 3y = 9$

$x + 3y - 9 = 0$

3. Final Answer

The equation of the line $L_2$ is $x + 3y - 9 = 0$.