To solve this problem, we need to write the equations of lines that are parallel and perpendicular to the given line [tex]\( k \)[/tex], which passes through the point [tex]\( (24, -15) \)[/tex].
Step-by-Step Solution:
### Part (a): Equation of the Parallel Line
A line that is parallel to another line has the same slope.
1. Identify the slope of the given line [tex]\( k \)[/tex]:
The equation of line [tex]\( k \)[/tex] is [tex]\( y = \frac{3}{4} x + 42 \)[/tex]. The slope (m) of this line is [tex]\( \frac{3}{4} \)[/tex].
2. Use the Point-Slope Form to find the equation of the parallel line:
The point-slope form for the equation of a line is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
Here, [tex]\( m = \frac{3}{4} \)[/tex], [tex]\( x_1 = 24 \)[/tex], and [tex]\( y_1 = -15 \)[/tex].
3. Substitute the values into the point-slope form:
[tex]\[
y - (-15) = \frac{3}{4} (x - 24)
\][/tex]
Simplifying this:
[tex]\[
y + 15 = \frac{3}{4}x - \frac{3}{4} \cdot 24
\][/tex]
[tex]\[
y + 15 = \frac{3}{4}x - 18
\][/tex]
4. Rewrite it into slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[
y = \frac{3}{4} x - 18 - 15
\][/tex]
[tex]\[
y = \frac{3}{4} x - 33
\][/tex]
So the equation of the line parallel to line [tex]\( k \)[/tex] that passes through [tex]\( (24, -15) \)[/tex] is:
[tex]\[ y = \frac{3}{4} x - 33 \][/tex]
### Part (b): Equation of the Perpendicular Line
A line that is perpendicular to another line has a slope that is the negative reciprocal of the original line's slope.
1. Identify the slope of the given line [tex]\( k \)[/tex]:
The slope of line [tex]\( k \)[/tex] is [tex]\( \frac{3}{4} \)[/tex].
2. Find the negative reciprocal of the slope:
The negative reciprocal of [tex]\( \frac{3}{4} \)[/tex] is [tex]\( -\frac{4}{3} \)[/tex].
3. Use the Point-Slope Form to find the equation of the perpendicular line:
Again, use the point-slope form with:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
Here, [tex]\( m = -\frac{4}{3} \)[/tex], [tex]\( x_1 = 24 \)[/tex], and [tex]\( y_1 = -15 \)[/tex].
4. Substitute the values into the point-slope form:
[tex]\[
y - (-15) = -\frac{4}{3} (x - 24)
\][/tex]
Simplifying this:
[tex]\[
y + 15 = -\frac{4}{3}x + \frac{4}{3} \cdot 24
\][/tex]
[tex]\[
y + 15 = -\frac{4}{3}x + 32
\][/tex]
5. Rewrite it into slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[
y = -\frac{4}{3} x + 32 - 15
\][/tex]
[tex]\[
y = -\frac{4}{3} x + 17
\][/tex]
So the equation of the line perpendicular to line [tex]\( k \)[/tex] that passes through [tex]\( (24, -15) \)[/tex] is:
[tex]\[ y = -\frac{4}{3} x + 17 \][/tex]
Therefore, the final results are:
- The parallel line equation: [tex]\( y = \frac{3}{4} x - 33 \)[/tex]
- The perpendicular line equation: [tex]\( y = -\frac{4}{3} x + 17 \)[/tex]
