• 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?

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  • 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
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\(\large{distance}(ax+by+c=0, (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\)
  • DLS
How can we solve this using parametric form?
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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Distance between points (x1,y1) and (x2,y2) is \(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
  • DLS
\[\frac{x-x1}{\cos \Theta} = \frac{y-y1}{\sin \Theta} = r\]
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r is the distance
but,(2,1) does NOT lie on that line.
  • DLS
oh sorry its 2,-1
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3x+4y+6-6=2 3(x-2)=4(y+1) \[\frac{x-2}{-4/5}-\frac{y+1}{3/5}=r=+5 or -5\]
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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
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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....
  • DLS
i know that but why did we divide by sqrt of a^2+b^2 only?
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any special logic or anything?
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(....)
what else can u divide those numbers by, to get their sum of squares =1 ?
  • DLS
seems legit..thanks :D
welcome ^_^

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