Find the equation of the line parallel to 4x-3y=7 and passing at a distance 4 from the point (1,-2).

The line to be found is parallel to 4x - 3y = 7. The slope
of 4x - 3y = 7 can be found by rewriting this as

3y = 4x -
7

=> y = (4/3)x -
7

Therefore the slope of the line is
(4/3).

Therefore the equation of the line is y = (4/3)x +
k

=> 4x - 3y + 3k =
0

The required line passes through a point which is at a
distance 4 from (1, -2).

We use the relation for
determining the distance d of a point (x1, y1) from the line ax+by +c = 0, which
is:

d = |ax1+by1+c|/ sqrt
(a^2+b^2)

4 = |4 + 6 + 3k|/ sqrt
(16+9)

20 = 10 + 3k

=>
3k = 10

Therefore the equation of the
required line is 4x -3y + 10 = 0.