Question 922397
3X+4Y-5=0  0r   {{{y = (-3/4)x + 5}}}
Line Perpendicular thru P(2,6) is {{{y = (4/3)x + 10/3}}}
Lines Intersect at P(-1,2)
(2,6)
(-1,2) D = {{{ sqrt ( (x[1]-x[2])^2+ (y[1]-y[2])^2 ) }}} = sqrt(9 + 16) = 5
{{{drawing(300,300,   -6, 6, -6, 8, grid(1), 

circle(2, 6,0.2),
circle(0,0,0.2),
circle(0,0,0.2),
graph( 300, 300, -6, 6, -6, 8,0,(-3/4)x + 5/4, (4/3)x + 10/3) )}}}