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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.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{a}\:\mathrm{point}\:\mathrm{which}\:\mathrm{moves}\:\mathrm{such}\:\mathrm{that}\:\mathrm{its} \\ $$$$\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{line}\:\:\:\mathrm{y}\:\:=\:\:\mathrm{4}\:\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}\:\:\:\mathrm{k}. \\ $$

Commented by TawaTawa1 last updated on 03/Mar/20

Please help.

$$\mathrm{Please}\:\mathrm{help}. \\ $$

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±k

$$\mathrm{this}\:\mathrm{is}\:\mathrm{easy}.\:\mathrm{scetch}\:\mathrm{it}.\:\mathrm{the}\:\mathrm{line}\:{y}=\mathrm{4}\:\mathrm{is}\:\mathrm{horizontal} \\ $$$$\mathrm{we}'\mathrm{re}\:\mathrm{searching}\:\mathrm{all}\:\mathrm{points}\:\mathrm{with}\:\mathrm{distance}\:{k}\:\mathrm{from} \\ $$$$\mathrm{this}\:\mathrm{line}.\:\mathrm{these}\:\mathrm{are}\:\mathrm{located}\:\mathrm{on}\:\mathrm{2}\:\mathrm{horizontal} \\ $$$$\mathrm{lines}:\:{y}=\mathrm{4}\pm{k} \\ $$

Commented by TawaTawa1 last updated on 04/Mar/20

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by mr W last updated on 04/Mar/20

the locus of a point which has a constant distance to a given line is a cylinder whose axis is the given line and whose radius is the given distance.

$${the}\:{locus}\:{of}\:{a}\:{point}\:{which}\:{has}\:{a} \\ $$$${constant}\:{distance}\:{to}\:{a}\:{given}\:{line} \\ $$$${is}\:{a}\:{cylinder}\:{whose}\:{axis}\:{is}\:{the}\:{given} \\ $$$${line}\:{and}\:{whose}\:{radius}\:{is}\:{the}\:{given} \\ $$$${distance}. \\ $$

Commented by TawaTawa1 last updated on 04/Mar/20

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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

$${let}\:{Y}={f}\left({x}\right)\:\:{bee}\:{equation}\:{of}\:{courbe}\:{werre}\:{tbis}\:{point}\:{moves} \\ $$$$\Rightarrow{D}\left(\left({x},{f}\left({x}\right)\right),{Y}=\mathrm{4}\right) \\ $$$${D}\left(\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right),{ax}+{by}+{c}=\mathrm{0}\right)=\frac{\mid{ax}_{\mathrm{1}} +{by}_{\mathrm{1}} +{c}\mid}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$=\frac{\mid{f}\left({x}\right)−\mathrm{4}\mid}{\mathrm{1}}=\mid{f}\left({x}\right)−\mathrm{4}\mid={k} \\ $$$$\Rightarrow\begin{cases}{{f}\left({x}\right)={k}+\mathrm{4}}\\{{f}\left({x}\right)=\mathrm{4}−{k}}\end{cases} \\ $$$$\Rightarrow{f}\left({x}\right)={c} \\ $$$$\Rightarrow{M}\in\left\{\left({x},{k}\right)\right\} \\ $$$${M}\in{Lign}\:{withe}\:{equation}\:{Y}={c}\:,{constante}\: \\ $$$$ \\ $$

Commented by jagoll last updated on 03/Mar/20

sir please q 83513

$$\mathrm{sir}\:\mathrm{please}\:\mathrm{q}\:\mathrm{83513} \\ $$

Commented by TawaTawa1 last updated on 04/Mar/20

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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