Question Number 61274 by alphaprime last updated on 31/May/19
$${If}\:{p}\:,\:{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,...{x}_{{i}} \:{and}\:{q},{y}_{\mathrm{1}} ,{y}_{\mathrm{2}} ,...{y}_{{i}} \:{form}\:{two}\: \\ $$$${infinite}\:{arithmetic}\:{sequences}\:{with}\:{common}\: \\ $$$${difference}\:\:{a}\:{and}\:{b}\:{respectively}\:, \\ $$$${then}\:{find}\:{the}\:{locus}\:{of}\:{the}\:{point}\:\left(\:\alpha\:,\:\beta\:\right)\: \\ $$$${where}\:\alpha\:=\:\frac{\mathrm{1}}{{n}}\:\sum_{{i}=\mathrm{1}} ^{{n}} {x}_{{i}} \:{and}\:\beta=\:\frac{\mathrm{1}}{{n}}\:\sum_{{i}=\mathrm{1}} ^{{n}} {y}_{{i}\:.} \\ $$
Answered by tanmay last updated on 31/May/19
![α=((x_1 +x_2 +x_3 +...+x_n )/n) α=(((p+a)+(p+2a)+...+(p+na))/n) α=((np+a[((n(n+1))/2)])/n) α=p+((a(n+1))/2) β=q+((b(n+1))/2) ((α−p)/(a/2))=((β−q)/(b/2)) so locus... b(x−p)=a(y−q) bx−ay+aq−bp=0 ean of straight line→locus st line...](Q61280.png)
$$\alpha=\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +...+{x}_{{n}} }{{n}} \\ $$$$\alpha=\frac{\left({p}+{a}\right)+\left({p}+\mathrm{2}{a}\right)+...+\left({p}+{na}\right)}{{n}} \\ $$$$\alpha=\frac{{np}+{a}\left[\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right]}{{n}} \\ $$$$\alpha={p}+\frac{{a}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\beta={q}+\frac{{b}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\frac{\alpha−{p}}{\frac{{a}}{\mathrm{2}}}=\frac{\beta−{q}}{\frac{{b}}{\mathrm{2}}} \\ $$$${so}\:{locus}... \\ $$$${b}\left({x}−{p}\right)={a}\left({y}−{q}\right) \\ $$$${bx}−{ay}+{aq}−{bp}=\mathrm{0} \\ $$$${ean}\:{of}\:{straight}\:{line}\rightarrow{locus}\:{st}\:{line}... \\ $$
Commented by bhanukumarb2@gmail.com last updated on 31/May/19
$${sir}\:{congrates}\:{for}\:{goiit}\:{approval}\:{pleasd} \\ $$$${share}\:{ur}\:{approch}\:{their}\:{also} \\ $$
Commented by bhanukumarb2@gmail.com last updated on 31/May/19
$${tanmay}\:{sir} \\ $$
Commented by tanmay last updated on 31/May/19
$${thank}\:{you}\:{sir}...{let}\:{me}\:{check}\:{in}\:{goiit}\: \\ $$