Question 828236
{{{ x - 3y = 2 }}}
{{{ 3y = x - 2 }}}
{{{ y = (1/3)*x - 2/3 }}}
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The slope of the line I want will
have a slope {{{ m = 1/3 }}} also
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I need a point that this line will go through
to get the equation
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Setting {{{ y=0 }}},
{{{ x - 3*0 = 2 }}}
{{{ x = 2 }}}
So, the given line passes through (2,0)
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The line through this point and perpendicular 
to the given line will have slope = {{{ -1/m = -3 }}}
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{{{ ( y-0 ) / ( x-2 ) = -3 }}}
{{{ y = -3*( x-2 ) }}}
(1) {{{ y = -3x + 6 }}}
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Now I need to find the point on this line that is 
{{{ d = 2*sqrt(10) }}} distance from (2,0)
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{{{ ( 2*sqrt(10) )^2 = ( y-0 )^2 + ( x-2 )^2 }}}
{{{ 40 = y^2 + (x-2)^2 }}}
Substituting from (1) above:
{{{ 40 = (-3x+6)^2 + (x-2)^2 }}}
{{{ 40 = 9x^2 -36x + 36 + x^2 - 4x + 4 }}}
{{{ 10x^2 - 40x = 0 }}}
{{{ x^2 - 4x = 0 }}}
{{{ x*( x-4) = 0 }}}
{{{ x = 4 }}}
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Plugging this into (1),
(1) {{{ y = -3*4 + 6 }}}
(1) {{{ y = -12 + 6 }}}
(1) {{{ y = -6 }}}
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Now I have {{{ m = 1/3 }}}
and the point ( 4,-6)
Using the point-slope formula:
{{{ (y-(-6) ) / ( x-4) = 1/3 }}}
{{{ ( y+6 ) / ( x-4 ) = 1/3 }}}
{{{ y + 6 = (1/3)*(x-4) }}}
{{{ y + 6 = (1/3)*x - 4/3 }}}
{{{ y = (1/3)*x - 18/3 - 4/3 }}}
{{{ y = (1/3)*x -22/3 }}} answer
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Here's plots of the 3 lines:
{{{ graph( 400, 400, -10, 10, -10, 10,(1/3)*x-2/3,-3x+6,(1/3)*x-22/3 ) }}}