Answer
a.) $\frac{7}{4}x+y-\frac{23}{24}=0$
b.)$-\frac{4}{7}x+y+\frac{41}{42}=0$
Work Step by Step
The slope needs to be found first, which means solving the given equation for the variable 'y'.
$7x + 4y = 8$
$4y = -7x + 8$ Subtract 7x
$y = -\frac{7}{4}x + \frac{8}{4}$ Divide by 4, isolating y
This is in the form y=mx+b, so m, the slope, is equal to -7/4.
A line parallel to the line given will have slope -7/4, and will pass through the point $(\frac{5}{6},\frac{-1}{2})$, as given by the problem. There is enough information to plug this into the point slope equation, $y-y_1 =m(x-x_1)$. THis will then be put into the general form $Ax + By + C = 0$
a.) Parallel line equation: $ y-(-\frac{1}{2}) = -\frac{7}{4}(x-\frac{5}{6})$
Distribute 'm', and put it into the general form:
$\frac{7}{4}x+y-\frac{23}{24}=0$
This is the answer to (a).
(b) is solved in a similar manner, except the slope is different since the new line is perpendicular to the original. A perpendicular line has a slope of the negative reciprocal of the original slope.
$m_0=-\frac{7}{4}$
$m_{perp} = \frac{4}{7}$
Now plug this value into the same point slope equation, and solve for the general form.
Point slope form: $ y-(-\frac{1}{2}) = \frac{4}{7}(x-\frac{5}{6})$
Put in general form: $-\frac{4}{7}x+y+\frac{41}{42}=0$
This is the answer to (b).
