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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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.
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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.
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