Menu Close

Let-d-d-be-two-nonparallel-lines-in-the-plane-and-let-k-gt-0-Find-the-locus-of-points-the-sum-of-whose-distances-to-d-and-d-is-equal-to-k-

Tinku Tara 0 Comments
Pin




Question Number 16067 by Tinkutara last updated on 21/Jun/17
Let d, d′ be two nonparallel lines in the plane and let k > 0. Find the locus of points, the sum of whose distances to d and d′ is equal to k.
$$\mathrm{Let}\:{d},\:{d}'\:\mathrm{be}\:\mathrm{two}\:\mathrm{nonparallel}\:\mathrm{lines}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{plane}\:\mathrm{and}\:\mathrm{let}\:{k}\:>\:\mathrm{0}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of} \\ $$$$\mathrm{points},\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{whose}\:\mathrm{distances}\:\mathrm{to} \\ $$$${d}\:\mathrm{and}\:{d}'\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:{k}. \\ $$
Commented by mrW1 last updated on 29/Jun/17
Commented by mrW1 last updated on 29/Jun/17
in distance (k/2) to line d_1 and to d_2 we draw e_1 // d_1 and e_2 // d_2 . e_1 and e_2 meet at A. OA is the bisector of angle between d_1 and d_2 . we draw a line at A perpendicularly to OA which intersects d_1 and d_2 at B and C. BC is the locus of M whose sum of distances to d_1 and d_2 is k.
$$\mathrm{in}\:\mathrm{distance}\:\frac{\mathrm{k}}{\mathrm{2}}\:\mathrm{to}\:\mathrm{line}\:\mathrm{d}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{to}\:\mathrm{d}_{\mathrm{2}} \:\mathrm{we} \\ $$$$\mathrm{draw}\:\mathrm{e}_{\mathrm{1}} \://\:\mathrm{d}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{e}_{\mathrm{2}} \://\:\mathrm{d}_{\mathrm{2}} . \\ $$$$\mathrm{e}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{e}_{\mathrm{2}} \:\mathrm{meet}\:\mathrm{at}\:\mathrm{A}. \\ $$$$\mathrm{OA}\:\mathrm{is}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{d}_{\mathrm{1}} \\ $$$$\mathrm{and}\:\mathrm{d}_{\mathrm{2}} . \\ $$$$\mathrm{we}\:\mathrm{draw}\:\mathrm{a}\:\mathrm{line}\:\mathrm{at}\:\mathrm{A}\:\mathrm{perpendicularly} \\ $$$$\mathrm{to}\:\mathrm{OA}\:\mathrm{which}\:\mathrm{intersects}\:\mathrm{d}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{d}_{\mathrm{2}} \:\mathrm{at} \\ $$$$\mathrm{B}\:\mathrm{and}\:\mathrm{C}. \\ $$$$\mathrm{BC}\:\mathrm{is}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{M}\:\mathrm{whose}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{distances}\:\mathrm{to}\:\mathrm{d}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{d}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{k}. \\ $$
Commented by mrW1 last updated on 30/Jun/17
proof: OB=OC A_(ΔOBC) =2×(1/2)×OC×(k/2)=((OC)/2)×k (1/2)OC×MP+(1/2)OB×MQ=A_(ΔOBC) ⇒MP+MQ=(2/(OC))A_(ΔOBC) =(2/(OC))×((OC)/2)×k=k
$$\mathrm{proof}: \\ $$$$\mathrm{OB}=\mathrm{OC} \\ $$$$\mathrm{A}_{\Delta\mathrm{OBC}} =\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{OC}×\frac{\mathrm{k}}{\mathrm{2}}=\frac{\mathrm{OC}}{\mathrm{2}}×\mathrm{k} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{OC}×\mathrm{MP}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{OB}×\mathrm{MQ}=\mathrm{A}_{\Delta\mathrm{OBC}} \\ $$$$\Rightarrow\mathrm{MP}+\mathrm{MQ}=\frac{\mathrm{2}}{\mathrm{OC}}\mathrm{A}_{\Delta\mathrm{OBC}} =\frac{\mathrm{2}}{\mathrm{OC}}×\frac{\mathrm{OC}}{\mathrm{2}}×\mathrm{k}=\mathrm{k} \\ $$
Commented by mrW1 last updated on 29/Jun/17
I used this knowledge for solving Q16066.
$$\mathrm{I}\:\mathrm{used}\:\mathrm{this}\:\mathrm{knowledge}\:\mathrm{for}\:\mathrm{solving}\: \\ $$$$\mathrm{Q16066}. \\ $$
Commented by mrW1 last updated on 29/Jun/17
Please show me the solution in your book. I′ll try.
$$\mathrm{Please}\:\mathrm{show}\:\mathrm{me}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{in}\:\mathrm{your} \\ $$$$\mathrm{book}.\:\mathrm{I}'\mathrm{ll}\:\mathrm{try}. \\ $$
Commented by mrW1 last updated on 30/Jun/17
that is mistake in editting with my small smartphone. it means certainly [OBC] = ((OC.PM)/2) + ((OB.PQ)/2)
$$\mathrm{that}\:\mathrm{is}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{editting}\:\mathrm{with}\:\mathrm{my} \\ $$$$\mathrm{small}\:\mathrm{smartphone}. \\ $$$$\mathrm{it}\:\mathrm{means}\:\mathrm{certainly} \\ $$$$\left[{OBC}\right]\:=\:\frac{{OC}.{PM}}{\mathrm{2}}\:+\:\frac{{OB}.{PQ}}{\mathrm{2}}\: \\ $$
Commented by Tinkutara last updated on 30/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Commented by mrW1 last updated on 30/Jun/17
you are right. it′s an other typo.
$$\mathrm{you}\:\mathrm{are}\:\mathrm{right}.\:\mathrm{it}'\mathrm{s}\:\mathrm{an}\:\mathrm{other}\:\mathrm{typo}. \\ $$
Answered by Tinkutara last updated on 30/Jun/17
Let O be the point of intersection between d and d′. We consider the points A, B, C, D, as in Figure, such that OA = OB = OC = OD = A > 0. Suppose M lies in the interior of the angle AOB. If M_A and M_B are the projections of M onto OA and OB, we have MM_A + MM_B = k. Multiplying by (a/2) , we obtain [MOA] + [MOB] = ((ka)/2) . But [MOA] + [MOB] = [MAOB] = [OAB] + [MAB]. It follows that the area of MAB is constant, and hence the locus of M in the interior of ∠AOB is a segment XY parallel to AB. Considering the other three possible locations, of M, we deduce that the locus is a rectangle XYZT whose diagonals are d and d′. Observation: It is worth mentioning that if M lies, for instance, on the line XY but not in the interior of the segment XY, another equality occurs. If M is on the half-line XY as in Figure 5.15, we have [MOA] − [MOB] = [OAB] + [MAB], so we deduce that for those positions of M, the difference of the distances to d and d′ equals k.
$$\mathrm{Let}\:{O}\:\mathrm{be}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{intersection} \\ $$$$\mathrm{between}\:{d}\:\mathrm{and}\:{d}'.\:\mathrm{We}\:\mathrm{consider}\:\mathrm{the} \\ $$$$\mathrm{points}\:{A},\:{B},\:{C},\:{D},\:\mathrm{as}\:\mathrm{in}\:\mathrm{Figure},\:\mathrm{such} \\ $$$$\mathrm{that}\:{OA}\:=\:{OB}\:=\:{OC}\:=\:{OD}\:=\:{A}\:>\:\mathrm{0}. \\ $$$$\mathrm{Suppose}\:{M}\:\mathrm{lies}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{angle}\:{AOB}. \\ $$$$\mathrm{If}\:{M}_{{A}} \:\mathrm{and}\:{M}_{{B}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{projections}\:\mathrm{of} \\ $$$${M}\:\mathrm{onto}\:{OA}\:\mathrm{and}\:{OB},\:\mathrm{we}\:\mathrm{have} \\ $$$${MM}_{{A}} \:+\:{MM}_{{B}} \:=\:{k}. \\ $$$$\mathrm{Multiplying}\:\mathrm{by}\:\frac{{a}}{\mathrm{2}}\:,\:\mathrm{we}\:\mathrm{obtain} \\ $$$$\left[{MOA}\right]\:+\:\left[{MOB}\right]\:=\:\frac{{ka}}{\mathrm{2}}\:. \\ $$$$\mathrm{But} \\ $$$$\left[{MOA}\right]\:+\:\left[{MOB}\right]\:=\:\left[{MAOB}\right] \\ $$$$=\:\left[{OAB}\right]\:+\:\left[{MAB}\right]. \\ $$$$\mathrm{It}\:\mathrm{follows}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:{MAB}\:\mathrm{is} \\ $$$$\mathrm{constant},\:\mathrm{and}\:\mathrm{hence}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{M}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{interior}\:\mathrm{of}\:\angle{AOB}\:\mathrm{is}\:\mathrm{a}\:\mathrm{segment}\:{XY} \\ $$$$\mathrm{parallel}\:\mathrm{to}\:{AB}. \\ $$$$\mathrm{Considering}\:\mathrm{the}\:\mathrm{other}\:\mathrm{three}\:\mathrm{possible} \\ $$$$\mathrm{locations},\:\mathrm{of}\:\mathrm{M},\:\mathrm{we}\:\mathrm{deduce}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{locus}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rectangle}\:{XYZT}\:\mathrm{whose} \\ $$$$\mathrm{diagonals}\:\mathrm{are}\:\mathrm{d}\:\mathrm{and}\:{d}'. \\ $$$$\boldsymbol{\mathrm{Observation}}:\:\mathrm{It}\:\mathrm{is}\:\mathrm{worth}\:\mathrm{mentioning} \\ $$$$\mathrm{that}\:\mathrm{if}\:{M}\:\mathrm{lies},\:\mathrm{for}\:\mathrm{instance},\:\mathrm{on}\:\mathrm{the}\:\mathrm{line} \\ $$$${XY}\:\mathrm{but}\:\mathrm{not}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{segment}\:{XY},\:\mathrm{another}\:\mathrm{equality}\:\mathrm{occurs}. \\ $$$$\mathrm{If}\:{M}\:\mathrm{is}\:\mathrm{on}\:\mathrm{the}\:\mathrm{half}-\mathrm{line}\:{XY}\:\mathrm{as}\:\mathrm{in}\:\mathrm{Figure} \\ $$$$\mathrm{5}.\mathrm{15},\:\mathrm{we}\:\mathrm{have} \\ $$$$\left[{MOA}\right]\:−\:\left[{MOB}\right]\:=\:\left[{OAB}\right]\:+\:\left[{MAB}\right], \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{deduce}\:\mathrm{that}\:\mathrm{for}\:\mathrm{those}\:\mathrm{positions}\:\mathrm{of} \\ $$$${M},\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{distances}\:\mathrm{to}\:\mathrm{d} \\ $$$$\mathrm{and}\:{d}'\:\mathrm{equals}\:{k}. \\ $$
Commented by Tinkutara last updated on 30/Jun/17
Commented by Tinkutara last updated on 30/Jun/17
Commented by Tinkutara last updated on 30/Jun/17
Figure 5.15. Maybe something is not printed correctly.
$$\mathrm{Figure}\:\mathrm{5}.\mathrm{15}.\:\mathrm{Maybe}\:\mathrm{something}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{printed}\:\mathrm{correctly}. \\ $$
Commented by mrW1 last updated on 30/Jun/17
The answer in book is not so well explained. The figures don′t match the question. But I can catch its statement and it′s in principle the same as my answer. In fact this question is in core the same as in Q16066. In Q16066 if a=b′=2b, we get the symmetry and the locus of M (line EF) is perdendicular to the bisector of angle α and OE=OF. I think my answer to this question is again much easier and clear than the answer in book.
$$\mathrm{The}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{book}\:\mathrm{is}\:\mathrm{not}\:\mathrm{so}\:\mathrm{well}\: \\ $$$$\mathrm{explained}.\:\mathrm{The}\:\mathrm{figures}\:\mathrm{don}'\mathrm{t}\:\mathrm{match} \\ $$$$\mathrm{the}\:\mathrm{question}.\:\mathrm{But}\:\mathrm{I}\:\mathrm{can}\:\mathrm{catch}\:\mathrm{its} \\ $$$$\mathrm{statement}\:\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{in}\:\mathrm{principle}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{as}\:\mathrm{my}\:\mathrm{answer}. \\ $$$$\mathrm{In}\:\mathrm{fact}\:\mathrm{this}\:\mathrm{question}\:\mathrm{is}\:\mathrm{in}\:\mathrm{core}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{as}\:\mathrm{in}\:\mathrm{Q16066}.\:\mathrm{In}\:\mathrm{Q16066}\:\mathrm{if}\:\mathrm{a}=\mathrm{b}'=\mathrm{2b}, \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{symmetry}\:\mathrm{and}\:\mathrm{the}\:\mathrm{locus} \\ $$$$\mathrm{of}\:\mathrm{M}\:\left(\mathrm{line}\:\mathrm{EF}\right)\:\mathrm{is}\:\mathrm{perdendicular}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{angle}\:\alpha\:\mathrm{and}\:\mathrm{OE}=\mathrm{OF}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{this}\:\mathrm{question}\:\mathrm{is} \\ $$$$\mathrm{again}\:\mathrm{much}\:\mathrm{easier}\:\mathrm{and}\:\mathrm{clear}\:\mathrm{than}\:\mathrm{the} \\ $$$$\mathrm{answer}\:\mathrm{in}\:\mathrm{book}. \\ $$
Commented by mrW1 last updated on 30/Jun/17
I think both answers are correct. none is better.But I think my answer trys to explain more clearly and naturally.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{both}\:\mathrm{answers}\:\mathrm{are}\:\mathrm{correct}.\:\mathrm{none} \\ $$$$\mathrm{is}\:\mathrm{better}.\mathrm{But}\:\mathrm{I}\:\mathrm{think}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{trys} \\ $$$$\mathrm{to}\:\mathrm{explain}\:\mathrm{more}\:\mathrm{clearly}\:\mathrm{and}\:\mathrm{naturally}. \\ $$
Commented by mrW1 last updated on 30/Jun/17
I think you can follow my answer better than follow the answer in the book.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{can}\:\mathrm{follow}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{better} \\ $$$$\mathrm{than}\:\mathrm{follow}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{the}\:\mathrm{book}. \\ $$
Commented by Tinkutara last updated on 30/Jun/17
Yes Sir! I do it. I posted my book′s solution thinking it would be easier or you will get a new method. But actually your answers are excellent!
$$\mathrm{Yes}\:\mathrm{Sir}!\:\mathrm{I}\:\mathrm{do}\:\mathrm{it}.\:\mathrm{I}\:\mathrm{posted}\:\mathrm{my}\:\mathrm{book}'\mathrm{s} \\ $$$$\mathrm{solution}\:\mathrm{thinking}\:\mathrm{it}\:\mathrm{would}\:\mathrm{be}\:\mathrm{easier}\:\mathrm{or} \\ $$$$\mathrm{you}\:\mathrm{will}\:\mathrm{get}\:\mathrm{a}\:\mathrm{new}\:\mathrm{method}.\:\mathrm{But} \\ $$$$\mathrm{actually}\:\mathrm{your}\:\mathrm{answers}\:\mathrm{are}\:\mathrm{excellent}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *