Find the equation of the locus of a point whose distance from the point (2, -2) is equal to its distance from the line x-y=0. I can't understand the question.
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This is related to Eccentricity.
Let the point be $P \equiv (x, y)$.
Distance of P from $(2, 2)$ is $\sqrt{(x-2)^2 + (y-2)^2} $ (1)
Distance of a point $(x_1, y_1)$ from a line $ax+by+c=0$ is $\left|\dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right|$
So distance of P from $x-y = 0$ is $\left|\dfrac{x-y}{\sqrt2}\right|$. (2)
Now, (1) and (2) are equal.
Therefore, $$\begin{align}\sqrt{(x-2)^2 + (y-2)^2} =\left|\dfrac{x-y}{\sqrt2}\right| \\(x-2)^2 + (y-2)^2 = \dfrac{(x-y)^2}{2}\end{align}$$
Can you simplify after this?
taninamdar
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i can but what did we do in the 3rd line ? after distance from P. – aki Sep 25 '14 at 17:58
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Distance between $(x_1, y_1), (x_2, y_2)$ is $\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$. Distance between a point $(x_1, y_1)$ and line $ax+by+c = 0$ is $\left|\dfrac{ax_1+by_1+c}{\sqrt{a^2 + b^2}}\right|$. Rest is just substitution and algebra. – taninamdar Sep 25 '14 at 18:00
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how did you substitute in $ ax_1 + by_1 +c $ ? – aki Sep 25 '14 at 19:09
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See this http://math.stackexchange.com/q/85761/66212 – taninamdar Sep 25 '14 at 19:10
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Ok I understand Thanks :) – aki Sep 25 '14 at 19:18
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The question is asking you to find something like this
If you need help solving this, just comment below, and I can work it out.
T.Woody
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