SOLUTION: Find the equation of the line parallel to the line x-3y=2 and passing at a distance 2√10 from the given line.

Algebra ->  Points-lines-and-rays -> SOLUTION: Find the equation of the line parallel to the line x-3y=2 and passing at a distance 2√10 from the given line.      Log On


   

Question 828236: Find the equation of the line parallel to the line x-3y=2 and passing at a distance 2√10 from the given line.
Found 2 solutions by josgarithmetic, josmiceli:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
The given line is y=%281%2F3%29x-2%2F3 so any general point of the line is (x,(1/3)(x-2)).

You want another line, y=%281%2F3%29x%2Bb, to be distance 2%2Asqrt%2810%29 from the given line. General point of the wanted line is (x,(1/3)x+b).

Distance Formula:
and what is needed is a formula for b in terms of x or no terms of x.
'
%280%2B%28%281%2F3%29%28x-2%29-%28%281%2F3%29x%2Bb%29%29%5E2%29=4%2A10
%28x%2F3-2%2F3-x%2F3-b%29%5E2=40
%28-2%2F3-b%29%5E2=40
%28b%2B2%2F3%29%5E2=40
b%2B2%2F3=0%2B-+2%2Asqrt%2810%29
b=-%282%2F3%29%2B-+2%2Asqrt%2810%29

Two possible equations then would be highlight%28y=%281%2F3%29x-%282%2F3%29-2%2Asqrt%2810%29%29 and highlight%28y=%281%2F3%29x-2%2F3%2B2%2Asqrt%2810%29%29

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+x+-+3y+=+2+
+3y+=+x+-+2+
+y+=+%281%2F3%29%2Ax+-+2%2F3+
--------------------------
The slope of the line I want will
have a slope +m+=+1%2F3+ also
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I need a point that this line will go through
to get the equation
-----------------------
Setting +y=0+,
+x+-+3%2A0+=+2+
+x+=+2+
So, the given line passes through (2,0)
----------------------------------------
The line through this point and perpendicular
to the given line will have slope = +-1%2Fm+=+-3+
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+%28+y-0+%29+%2F+%28+x-2+%29+=+-3+
+y+=+-3%2A%28+x-2+%29+
(1) +y+=+-3x+%2B+6+
-------------------
Now I need to find the point on this line that is
+d+=+2%2Asqrt%2810%29+ distance from (2,0)
----------------------------
+%28+2%2Asqrt%2810%29+%29%5E2+=+%28+y-0+%29%5E2+%2B+%28+x-2+%29%5E2+
+40+=+y%5E2+%2B+%28x-2%29%5E2+
Substituting from (1) above:
+40+=+%28-3x%2B6%29%5E2+%2B+%28x-2%29%5E2+
+40+=+9x%5E2+-36x+%2B+36+%2B+x%5E2+-+4x+%2B+4+
+10x%5E2+-+40x+=+0+
+x%5E2+-+4x+=+0+
+x%2A%28+x-4%29+=+0+
+x+=+4+
------------------
Plugging this into (1),
(1) +y+=+-3%2A4+%2B+6+
(1) +y+=+-12+%2B+6+
(1) +y+=+-6+
----------------------
Now I have +m+=+1%2F3+
and the point ( 4,-6)
Using the point-slope formula:
+%28y-%28-6%29+%29+%2F+%28+x-4%29+=+1%2F3+
+%28+y%2B6+%29+%2F+%28+x-4+%29+=+1%2F3+
+y+%2B+6+=+%281%2F3%29%2A%28x-4%29+
+y+%2B+6+=+%281%2F3%29%2Ax+-+4%2F3+
+y+=+%281%2F3%29%2Ax+-+18%2F3+-+4%2F3+
+y+=+%281%2F3%29%2Ax+-22%2F3+ answer
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Here's plots of the 3 lines: