Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 175176 by nadovic last updated on 21/Aug/22

Find the equation of the locus of   points equidistant from the point  A(4, −1) and the line x−y+2=0.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\: \\ $$$$\mathrm{points}\:\mathrm{equidistant}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point} \\ $$$${A}\left(\mathrm{4},\:−\mathrm{1}\right)\:\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:{x}−{y}+\mathrm{2}=\mathrm{0}. \\ $$

Answered by som(math1967) last updated on 22/Aug/22

let movale point(h,k)  from the condition  (√((h−4)^2 +(k+1)^2 ))=((∣h−k+2∣)/( (√((1)^2 +(−1)^2 ))))  2(h−4)^2 +2(k+1)^2 =h^2 +k^2 +4−2hk                                                +4h−4k    [squaring both side]  h^2 +k^2 +12h−8k−2hk−30=0  ∴ equation of the locus  x^2 +y^2 +12x−8y−2xy−30=0  locus path is parabola

$${let}\:{movale}\:{point}\left({h},{k}\right) \\ $$$${from}\:{the}\:{condition} \\ $$$$\sqrt{\left({h}−\mathrm{4}\right)^{\mathrm{2}} +\left({k}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mid{h}−{k}+\mathrm{2}\mid}{\:\sqrt{\left(\mathrm{1}\right)^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{2}\left({h}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{2}\left({k}+\mathrm{1}\right)^{\mathrm{2}} ={h}^{\mathrm{2}} +{k}^{\mathrm{2}} +\mathrm{4}−\mathrm{2}{hk} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}{h}−\mathrm{4}{k} \\ $$$$\:\:\left[{squaring}\:{both}\:{side}\right] \\ $$$${h}^{\mathrm{2}} +{k}^{\mathrm{2}} +\mathrm{12}{h}−\mathrm{8}{k}−\mathrm{2}{hk}−\mathrm{30}=\mathrm{0} \\ $$$$\therefore\:{equation}\:{of}\:{the}\:{locus} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\mathrm{12}\boldsymbol{{x}}−\mathrm{8}\boldsymbol{{y}}−\mathrm{2}\boldsymbol{{xy}}−\mathrm{30}=\mathrm{0} \\ $$$$\boldsymbol{{locus}}\:\boldsymbol{{path}}\:\boldsymbol{{is}}\:\boldsymbol{{parabola}} \\ $$

Commented by nadovic last updated on 22/Aug/22

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

Answered by a.lgnaoui last updated on 22/Aug/22

  l ensemble des points situes a la meme distance de A a la droite (D)  y=x+2 au pont M(4,−1) sont les points de la droite parallele a (D)note (D′) qui s  qui passe par M    y=x+m  x=4      y=−1  4+m=−1     m=−5  y=x−5

$$ \\ $$$${l}\:{ensemble}\:{des}\:{points}\:{situes}\:{a}\:{la}\:{meme}\:{distance}\:{de}\:{A}\:{a}\:{la}\:{droite}\:\left({D}\right) \\ $$$${y}={x}+\mathrm{2}\:{au}\:{pont}\:{M}\left(\mathrm{4},−\mathrm{1}\right)\:{sont}\:{les}\:{points}\:{de}\:{la}\:{droite}\:{parallele}\:{a}\:\left({D}\right){note}\:\left({D}'\right)\:{qui}\:{s} \\ $$$${qui}\:{passe}\:{par}\:{M}\:\:\:\:{y}={x}+{m} \\ $$$${x}=\mathrm{4}\:\:\:\:\:\:{y}=−\mathrm{1} \\ $$$$\mathrm{4}+{m}=−\mathrm{1}\:\:\:\:\:{m}=−\mathrm{5} \\ $$$${y}={x}−\mathrm{5}\:\:\: \\ $$

Commented by som(math1967) last updated on 22/Aug/22

(4,−1) not equidistace to y=x+2  and y=x−5

$$\left(\mathrm{4},−\mathrm{1}\right)\:{not}\:{equidistace}\:{to}\:{y}={x}+\mathrm{2} \\ $$$${and}\:{y}={x}−\mathrm{5} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com