The problem 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 $L_1$. We will solve problems 5 and 6. 5) $P(-1, 1)$ and $L_1: x - y + 5 = 0$ 6) $P(0, -2)$ and $L_1: x + y + 1 = 0$

1. Problem Description

The problem 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 $L_1$. We will solve problems 5 and

6.

5) $P(-1, 1)$ and $L_1: x - y + 5 = 0$

6) $P(0, -2)$ and $L_1: x + y + 1 = 0$

2. Solution Steps

5) Given point $P(-1, 1)$ and line $L_1: x - y + 5 = 0$.

We need to find a line $L_2$ that passes through $P$ and is parallel to $L_1$.

Since $L_2$ is parallel to $L_1$, their slopes are equal.

Rewrite $L_1$ in slope-intercept form: $y = x + 5$. The slope of $L_1$ is $m = 1$.

Therefore, the slope of $L_2$ is also $m = 1$.

Using the point-slope form of a line, we have:

$y - y_1 = m(x - x_1)$

Substituting $P(-1, 1)$ and $m = 1$, we get:

$y - 1 = 1(x - (-1))$

$y - 1 = x + 1$

$y = x + 2$

Therefore, $L_2: x - y + 2 = 0$.

6) Given point $P(0, -2)$ and line $L_1: x + y + 1 = 0$.

We need to find a line $L_2$ that passes through $P$ and is parallel to $L_1$.

Since $L_2$ is parallel to $L_1$, their slopes are equal.

Rewrite $L_1$ in slope-intercept form: $y = -x - 1$. The slope of $L_1$ is $m = -1$.

Therefore, the slope of $L_2$ is also $m = -1$.

Using the point-slope form of a line, we have:

$y - y_1 = m(x - x_1)$

Substituting $P(0, -2)$ and $m = -1$, we get:

$y - (-2) = -1(x - 0)$

$y + 2 = -x$

$y = -x - 2$

Therefore, $L_2: x + y + 2 = 0$.

3. Final Answer

5) $x - y + 2 = 0$

6) $x + y + 2 = 0$