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

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### 2 Answers

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.**

Any line parallel to 4x-3y = 7 is of the form 4x-3y+k = 0.

We know that the distance d of a point (x1, y1) from the line ax+by +c = 0 is given by:

d = | ax1+by1+c)/(a^2+b^2)^(1/2).

Therefore the distance of the line 4x-3y+k = 0 from the point (1,-2) is :

4 = (4*1-3(-2)+k)/(4^2+5^2)^(1/2)

4 = |(4+3*2+k)|/5

20 = 10+k

So k = 20-10 = 10.

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