DLS
  • DLS
Let (2,-1) be a point on line 3x+4y=2.Find the point at distance of 5 units on the line from the given point?
Mathematics
chestercat
  • chestercat
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hartnn
  • hartnn
\(\large{distance}(ax+by+c=0, (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\)
DLS
  • DLS
How can we solve this using parametric form?
hartnn
  • hartnn
i didn't get your question actually, (2,1) does NOT lie on that line. can u draw a figure or something ?

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hartnn
  • hartnn
Distance between points (x1,y1) and (x2,y2) is \(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
DLS
  • DLS
\[\frac{x-x1}{\cos \Theta} = \frac{y-y1}{\sin \Theta} = r\]
DLS
  • DLS
r is the distance
hartnn
  • hartnn
but,(2,1) does NOT lie on that line.
DLS
  • DLS
oh sorry its 2,-1
DLS
  • DLS
3x+4y+6-6=2 3(x-2)=4(y+1) \[\frac{x-2}{-4/5}-\frac{y+1}{3/5}=r=+5 or -5\]
DLS
  • DLS
why did we divide by sqrt of a^2+b^2 that is 5? They say by doing that we get sin Theta and cos Theta
DLS
  • DLS
indirectly
hartnn
  • hartnn
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
hartnn
  • hartnn
so we can take one as -4/5 and other as 3/5
hartnn
  • hartnn
like here u take cos t =-4/5, sin t = 3/5
hartnn
  • hartnn
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
hartnn
  • hartnn
its like normalization....
DLS
  • DLS
i know that but why did we divide by sqrt of a^2+b^2 only?
DLS
  • DLS
any special logic or anything?
hartnn
  • hartnn
for any 2 numbers \(\large(\frac{a}{\sqrt{a^2+b^2}})^2+(\frac{b}{\sqrt{a^2+b^2}})^2=1\) so there is a need to divide by sqrt(....)
hartnn
  • hartnn
what else can u divide those numbers by, to get their sum of squares =1 ?
DLS
  • DLS
seems legit..thanks :D
hartnn
  • hartnn
welcome ^_^

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