Question 927826
You could say there is a formula but you do not need it.  You could try to derive a formula, or you could just pick any point on either of the two given lines, find where a line intersects both of these and passes through you chosen point.


The two given lines have slope of {{{2/3}}}.  Any line perpendicular to them will have slope {{{-3/2}}}.  Try to put either equation into slope-intercept form if you not see this.


Pick an equation and find an axis intercept.
{{{2x=4+3y}}}
{{{x=4/2+3y/4}}}
{{{x=3y/4+2}}}
One point on this line is (2,0), which is x-intercept for 2x-3y=4.


The line containing (2,0) and with slope {{{-3/2}}}, using point-form to start, is {{{y-0=-(3/2)(x-2)}}}
{{{y=-3x/2+3}}}


You want to find what point is the intersection of this line with the second given line of {{{2x-3y=-4}}}.
Solve this second equation for y:
{{{-3y=-2x-4}}}
{{{y=(2/3)x+4/3}}}
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Equate the expressions of y for the perpendicular equation and the given second equation:
{{{highlight_green(-(3/2)x+3=(2/3)x+4/3)}}}
LCD is 6;
Multiply left and right by 6;
{{{-9x+18=4x+8}}}
{{{-9x-4x=8-18}}}
{{{-13x=-10}}}
{{{x=-10/13}}}----seems inconvenient.....
Use x to find y:
{{{y=(2/3)(-10/13)+4/3}}}
{{{4/3-20/39}}}
{{{52/39-20/39}}}
{{{(52-20)/39}}}
{{{32/39}}}
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Point on this second line is  (32/39, -10/13).


LAST, use the Distance Formula to find the distance between the points,  (32/39, -10/39)  and  (2,0).


(Process and final result not here shown...)