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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hartnn
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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
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How can we solve this using parametric form?

hartnn
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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 ?

hartnn
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Distance between points (x1,y1) and (x2,y2) is
\(\huge d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\)

DLS
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\[\frac{xx1}{\cos \Theta} = \frac{yy1}{\sin \Theta} = r\]

DLS
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r is the distance

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

DLS
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oh sorry its 2,1

DLS
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3x+4y+66=2
3(x2)=4(y+1)
\[\frac{x2}{4/5}\frac{y+1}{3/5}=r=+5 or 5\]

DLS
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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

DLS
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indirectly

hartnn
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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

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

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

hartnn
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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

hartnn
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its like normalization....

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

DLS
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any special logic or anything?

hartnn
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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(....)

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

DLS
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seems legit..thanks :D

hartnn
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welcome ^_^