Find the locus of a point which moves such that its distance from the line y = 4 is a constant k.


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Question Number 83570 by TawaTawa1 last updated on 03/Mar/20

$Find the locus of a point which moves such that its$ $distance from the line y = 4 is a constant k .$
Commented by TawaTawa1 last updated on 03/Mar/20

$Please help .$
Commented by MJS last updated on 03/Mar/20

$this is easy . scetch it . the line y = 4 is horizontal$ ${we}^{'} re searching all points with distance k from$ $this line . these are located on 2 horizontal$ $lines : y = 4 \pm k$
Commented by TawaTawa1 last updated on 04/Mar/20

$God bless you sir .$
	Answered by mr W last updated on 04/Mar/20

	$t h e l o c u s o f a p o i n t w h i c h h a s a$ $c o n s t a n t d i s t a n c e t o a g i v e n l i n e$ $i s a c y l i n d e r w h o s e a x i s i s t h e g i v e n$ $l i n e a n d w h o s e r a d i u s i s t h e g i v e n$ $d i s t a n c e .$
	Commented by TawaTawa1 last updated on 04/Mar/20

	$God bless you sir .$
	Answered by mind is power last updated on 03/Mar/20
	$let Y=f(x) bee equation of courbe werre tbis point moves ⇒D((x,f(x)),Y=4) D((x_1 ,y_1 ),ax+by+c=0)=((∣ax_1 +by_1 +c∣)/(√(a^2 +b^2 ))) =((∣f(x)−4∣)/1)=∣f(x)−4∣=k ⇒ { ((f(x)=k+4)),((f(x)=4−k)) :} ⇒f(x)=c ⇒M∈{(x,k)} M∈Lign withe equation Y=c ,constante$
	$l e t Y = f (x) b e e e q u a t i o n o f c o u r b e w e r r e t b i s p o i n t m o v e s$ $\Rightarrow D ((x, f (x)), Y = 4)$ $D ((x_{1}, y_{1}), a x + b y + c = 0) = \frac{∣ {a x}_{1} + {b y}_{1} + c ∣}{\sqrt{a^{2} + b^{2}}}$ $= \frac{∣ f (x) - 4 ∣}{1} =∣ f (x) - 4 ∣= k$ $\Rightarrow {\begin{cases} f (x) = k + 4 \\ f (x) = 4 - k \end{cases}$ $\Rightarrow f (x) = c$ $\Rightarrow M \in {(x, k)}$ $M \in L i g n w i t h e e q u a t i o n Y = c, c o n s t a n t e$
	Commented by jagoll last updated on 03/Mar/20

	$sir please q 83513$
	Commented by TawaTawa1 last updated on 04/Mar/20

	$God bless you sir$
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