Let's tackle both parts of the problem step by step.

(a) Equation of the line through the point parallel to the given line

The equation of the given line is: $x + y = 5$ We can rewrite it in slope-intercept form $y = mx + b$, where $m$ is the slope. From $x + y = 5$, solving for $y$ gives: $y = -x + 5$ So, the slope of the line is $m = -1$.

Since the new line is parallel to the given line, it will have the same slope $m = -1$. The general equation of a line is given by: $y - y_1 = m(x - x_1)$ where $(x_1, y_1)$ is the given point $(-7, 6)$ and $m = -1$.

Substituting into the point-slope form: $y - 6 = -1(x + 7)$ Simplifying this: $y - 6 = -x - 7$ $y = -x - 1$ Thus, the equation of the line through the point $(-7, 6)$ that is parallel to the given line is: $y = -x - 1$

(b) Equation of the line through the point perpendicular to the given line

For perpendicular lines, the slope is the negative reciprocal of the original slope. Since the slope of the original line is $m = -1$, the slope of the perpendicular line will be: $m_{\perp} = \frac{1}{1} = 1$

Now, we use the point-slope form again with the point $(-7, 6)$ and the new slope $m_{\perp} = 1$: $y - 6 = 1(x + 7)$ Simplifying this: $y - 6 = x + 7$ $y = x + 13$ Thus, the equation of the line through the point $(-7, 6)$ that is perpendicular to the given line is: $y = x + 13$

Summary:

• The equation of the line parallel to $x + y = 5$ through $(-7, 6)$ is: $y = -x - 1$.
• The equation of the line perpendicular to $x + y = 5$ through $(-7, 6)$ is: $y = x + 13$.

Would you like more details or further explanation?

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Here are 5 related questions:

1. How do you find the slope of a line given its equation?
2. How do you convert from standard form to slope-intercept form of a line?
3. What is the point-slope form of a line, and when is it useful?
4. How do you determine if two lines are perpendicular?
5. How can you check if a point lies on a given line?

Tip: For parallel lines, the slopes are equal; for perpendicular lines, the slopes are negative reciprocals.