Question #134446
Find the equation of the line that is parallel to 4x-5y = 2 and whose distance from the
given line is 5.
1
Expert's answer
2020-10-05T18:22:40-0400

we know that distance between the parallel lines Ax+By+C1=0 and Ax+By+C2=0Ax+By+C_1=0 \space and \space Ax+By+C_2=0

is given by

d=C1C2A2+B2d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}

The rough diagram of the given question is given below




Equation of any line parallel to 4x5y2=04x-5y-2=0 is of the form 4x5y+k=04x-5y+k=0, where k is constant and its value can be found out by given other geometrical condition.


Now Applying the formula of distance between the parallel lines, we get

On comparing we get A=4,B=5,C1=2,and C2=k and distance(d)=5 unitsA=4,B=-5 ,C_1=-2 ,and \space C_2=k \ and \space distance (d) = 5 \space units


d=C1C2A2+B2or,5=2k42+(5)2or,5=±(k2)16+25Taking ve sign,we getor,5=k+241or,k+2=541or,k=5412\therefore d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}} \\ or,5=\frac{|-2-k|}{\sqrt{4^2+(-5)^2}}\\or, 5= \pm\frac{(-k-2)}{\sqrt{16+25}}\\Taking \space '-ve' \space sign ,we \space get\\or, 5=\frac{k+2}{\sqrt{41}}\\or, k+2 = 5\sqrt{41}\\or,k= 5\sqrt{41}-2


Taking +ve sign,we get5=(k+2)41or,k+2=541or,k=5412or,k=(541+2)Taking \space '+ve' \space sign ,we \space get\\ 5=\frac{-(k+2)}{\sqrt{41}}\\or, k+2 = -5\sqrt{41}\\or,k= -5\sqrt{41}-2\\or, k = -(5\sqrt{41} +2)


So the equations of a required line is

4x5y+(5412)=0.or4x5y(541+2)=0.4x-5y+(5\sqrt{41}-2)=0.\\or\\4x-5y-(5\sqrt{41}+2)=0.



ANSWER




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Comments

Assignment Expert
15.12.20, 21:35

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MAYA
14.12.20, 15:40

thanks

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