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\(\large{distance}(ax+by+c=0, (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\)

How can we solve this using parametric form?

Distance between points (x1,y1) and (x2,y2) is
\(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

\[\frac{x-x1}{\cos \Theta} = \frac{y-y1}{\sin \Theta} = r\]

r is the distance

but,(2,1) does NOT lie on that line.

oh sorry its 2,-1

3x+4y+6-6=2
3(x-2)=4(y+1)
\[\frac{x-2}{-4/5}-\frac{y+1}{3/5}=r=+5 or -5\]

indirectly

so that the two numbers(-4/5 and 3/5) here satisfy, a^2+b^2=1
which is the property of sin and cos

so we can take one as -4/5 and other as 3/5

like here u take cos t =-4/5, sin t = 3/5

if you did not divide by 5,
then 4^2+3^2 does not equal 1
so u cannot take cos t=4, and sin t =3

its like normalization....

i know that but why did we divide by sqrt of a^2+b^2 only?

any special logic or anything?

what else can u divide those numbers by, to get their sum of squares =1 ?

seems legit..thanks :D

welcome ^_^