Question

@marco26
come here

Answer 1

yes

Answer 2

line : y+Ax+B=0
distance from axes origin=1 so we have
\[\frac{\left| B \right|}{\sqrt{1+A^2}}=1\\then\\ B^2=1+A^2\]
am i clear?

Answer 3

where did the first equation come from? the one with absolute value

Answer 4

ok take a look
http://www.intmath.com/plane-analytic-geometry/perpendicular-distance-point-line.php

Answer 5

ok

Answer 6

well then u have
y+Ax+B=0
diff wrt x --> y'+A=0 --> y'=-A --> y'^2=A^2
put y'=-A in the original equation
u have
y-xy'+B=0
y-xy'=-B
(y-xy')^2=B^2=1+A^2=1+y'^2

Answer 7

is that right?

Answer 8

thanks!

Answer 9

your welcome