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Find the locus of points, the distance between them and the point $(2,1)$ is equal to the distance between them and the straight line $4x = 3y$

I know that it is the definition of a parabola But I do not know how to find Solution

Could one help me ?

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Given a line $\ell: \ ax+by+c=0$ in the $(x,y)$-plane and a point $P=(x,y)$, the distance from $P$ to $\ell$ computes according to $$d(P,\ell)=\left|{ax+by+c\over\sqrt{a^2+b^2}}\right|\ .$$ In your case this amounts to $$d(P,\ell)=\left|{4\over5}x-{3\over5}y\right|\ .$$ The geometric condition in your problem now reads $$ \left|{4\over5}x-{3\over5}y\right|=\sqrt{(x-2)^2+(y-1)^2}\ .$$ After squaring and collecting terms you have the solution. The resulting equation will be of second degree in $x$ and in $y$, but one cannot verify by inspection that it is the equation of a parabola.

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enter image description here

  1. Re-write 4x = 3y into general form and get 4x - 3y = 0.

  2. The square of the distance between (2, 1) and P(x, y) [a point on the locus] is $(x-2)^2 + (y-1)^2$

  3. The square of the distance between P(x, y) and the line is $\frac {(4(x) - 3(y))^2} {(4)^2+ (-3)^2}$

  4. Equate the equations obtained in step-2 and step-3 will give you the equation of the locus.

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The line $\frac{3 x}{4}-0.5$ is parallel to $4x=3y$ and pass through point $(2,1)$. Line $\frac{3 x}{4}-0.25$ between those lines (yellow line) is the locus between line and point. You can see this in the illustrative figure.

Lines

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  • $\begingroup$ I want the distance between them and the point$ (2,1)$ is equal to the distance between them and the straight line $4x=3y$ $\endgroup$ – Hamada Al Apr 30 '14 at 23:40
  • $\begingroup$ Ok, excuse me I misunderstood the question. $\endgroup$ – Toni E May 1 '14 at 10:54

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