Question 1172331
<pre>
The left side of an equation of a line PARALLEL to a given line Ax+By=C has
the same left side Ax+By but a different number for C on the right.

The left side of an equation of a line PERPENDICULAR to a given line Ax+By=C
has the left side Bx-Ay but a different number for C on the right.

So a line perpendicular to 2x+3y=1 has left side 3x-2y. Let's say it's right
side is a number C.

Then we use the formula

Distance from the point (x<sub>1</sub>,y<sub>1</sub>)
to the line Ax+By+C=0 is

d = {{{abs(Ax[1]+By[1]+C)/sqrt(A^2+B^2)}}}

{{{abs(3(0)-2(0)+C)/sqrt((3)^2+(-2)^2)=7}}}

{{{abs(C/sqrt(9+4))=7}}}

{{{abs(C/sqrt(13))=7}}}

{{{abs(C)=7sqrt(14)}}}

{{{C= "" +- 7sqrt(14)}}}

So there are two solutions:

{{{3x-2y= "" +- 7sqrt(14)}}}

The given line is:

{{{drawing(400,400,-10,10,-10,10,graph(400,400,-10,10,-10,10),
line(-13,9,14,-9))}}}

Here is the first line {{{3x-2y= 7sqrt(14)}}} in red.
The green line to the origin is 7 units long:


{{{drawing(400,400,-10,10,-10,10,graph(400,400,-10,10,-10,10),
red(line(-11,-3.404199146,11,29.59580085)),
green(line(0,0,-6.044215778,4.029477186)),
line(-13,9,14,-9))}}}

Here is the second line {{{3x-2y= -7sqrt(14)}}}, also in red.
The green line to the origin is also 7 units long:

{{{drawing(400,400,-10,10,-10,10,graph(400,400,-10,10,-10,10),
red(line(-11,-3.404199146,11,29.59580085)),
green(line(0,0,-6.044215778,4.029477186)),

red(line(11,3.404199146,-11,-29.59580085)),
green(line(0,0,6.044215778,-4.029477186)),


line(-13,9,14,-9))}}}

Edwin</pre>