Reduce eq in normal form 3x-4y+4=0

Normal form of Line 3x-4y=-4 is , x cos k + y Sin k = p, where p is length of perpendicular from origin to that line and k is angle made by that perpendicular with positive x-axis. As both forms represents same line, Alternately we can consider them as two different coinciding lines. Hence we can write 3/(cos k) = -4/(sin k) = -4/p From this equation we can determine cos k and sin k and then value of p by using identity sqr(Sin k) + sqr(cos k)= 1. You can also determine tan k. As per my calculation tan k = -4/3 therefore angle k = tan inverse (-4/3) , angle k will be greater than 90 degree as tan k is negative. Also p = 4/5.. We can directly substitute cos k and sin k values in normal form x cos k + y Sin k = p, without knowing actual value of k.

x cos $ + y sin $ = p is normal form of same line 3x - 4y = - 4 , where p is perpendicular distance of line from origin and $ is angle made by that perpendicular with positive X axis. As both equation represents same line we can consider them as two different lines which are "coinciding" . Hence we can write (3/ cos $)=(-4/sin $)=(-4/p) , from these ratios we can find cos $, sin $ and hence tan $ and angle $. p can also determined using identity sqr(sin $) + sqr(cos $) = 1. After knowing value of p and $ , we can substitute in normal form equation..