Questions: Find equations of the lines that pass through the given point and are parallel to and perpendicular to given line. 7x + 4y = 0, (1/4, 1/3) (a) parallel to the given line -7/4 x + 79/48 (b) perpendicular to the given line y = 4/7 x + 11/21

Solution Steps

To find the equations of the lines that pass through a given point and are parallel or perpendicular to a given line, follow these steps:

(a) Parallel Line:

1. Determine the slope of the given line by rewriting it in slope-intercept form \(y = mx + b\).
2. Use the same slope for the parallel line.
3. Use the point-slope form of a line equation \(y - y_1 = m(x - x_1)\) with the given point to find the equation of the parallel line.

(b) Perpendicular Line:

1. The slope of the perpendicular line is the negative reciprocal of the slope of the given line.
2. Use the point-slope form of a line equation with the given point to find the equation of the perpendicular line.

The given line is \(7x + 4y = 0\). To find its slope, we convert it to the slope-intercept form \(y = mx + b\).

\[ 4y = -7x \implies y = -\frac{7}{4}x \]

Thus, the slope of the given line is \(-\frac{7}{4}\).

A line parallel to the given line will have the same slope, \(-\frac{7}{4}\). Using the point \(\left(\frac{1}{4}, \frac{1}{3}\right)\) and the point-slope form \(y - y_1 = m(x - x_1)\), we have:

\[ y - \frac{1}{3} = -\frac{7}{4}\left(x - \frac{1}{4}\right) \]

Simplifying, we get:

\[ y = -\frac{7}{4}x + \frac{79}{48} \]

The slope of a line perpendicular to the given line is the negative reciprocal of \(-\frac{7}{4}\), which is \(\frac{4}{7}\). Using the same point \(\left(\frac{1}{4}, \frac{1}{3}\right)\), we apply the point-slope form:

\[ y - \frac{1}{3} = \frac{4}{7}\left(x - \frac{1}{4}\right) \]

Simplifying, we get:

\[ y = \frac{4}{7}x + \frac{11}{21} \]

Final Answer