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DLS Group Title

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?

  • one year ago
  • one year ago

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

    • one year ago
  2. DLS Group Title
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    How can we solve this using parametric form?

    • one year ago
  3. hartnn Group Title
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    i didn't get your question actually, (2,1) does NOT lie on that line. can u draw a figure or something ?

    • one year ago
  4. hartnn Group Title
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    Distance between points (x1,y1) and (x2,y2) is \(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

    • one year ago
  5. DLS Group Title
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    \[\frac{x-x1}{\cos \Theta} = \frac{y-y1}{\sin \Theta} = r\]

    • one year ago
  6. DLS Group Title
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    r is the distance

    • one year ago
  7. hartnn Group Title
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    but,(2,1) does NOT lie on that line.

    • one year ago
  8. DLS Group Title
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    oh sorry its 2,-1

    • one year ago
  9. DLS Group Title
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    3x+4y+6-6=2 3(x-2)=4(y+1) \[\frac{x-2}{-4/5}-\frac{y+1}{3/5}=r=+5 or -5\]

    • one year ago
  10. DLS Group Title
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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

    • one year ago
  11. DLS Group Title
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    indirectly

    • one year ago
  12. hartnn Group Title
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    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

    • one year ago
  13. hartnn Group Title
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    so we can take one as -4/5 and other as 3/5

    • one year ago
  14. hartnn Group Title
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    like here u take cos t =-4/5, sin t = 3/5

    • one year ago
  15. hartnn Group Title
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    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

    • one year ago
  16. hartnn Group Title
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    its like normalization....

    • one year ago
  17. DLS Group Title
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    i know that but why did we divide by sqrt of a^2+b^2 only?

    • one year ago
  18. DLS Group Title
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    any special logic or anything?

    • one year ago
  19. hartnn Group Title
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    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(....)

    • one year ago
  20. hartnn Group Title
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    what else can u divide those numbers by, to get their sum of squares =1 ?

    • one year ago
  21. DLS Group Title
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    seems legit..thanks :D

    • one year ago
  22. hartnn Group Title
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    welcome ^_^

    • one year ago
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