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-

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 .

Commented by mrW1 last updated on 29/Jun/17

Commented by mrW1 last updated on 29/Jun/17

in distance \frac{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 .

Commented by mrW1 last updated on 30/Jun/17

proof :

OB = OC

A_{Δ OBC} = 2 \times \frac{1}{2} \times OC \times \frac{k}{2} = \frac{OC}{2} \times k

\frac{1}{2} OC \times MP + \frac{1}{2} OB \times MQ = A_{Δ OBC}

\Rightarrow MP + MQ = \frac{2}{OC} A_{Δ OBC} = \frac{2}{OC} \times \frac{OC}{2} \times k = k

Commented by mrW1 last updated on 29/Jun/17

I used this knowledge for solving

Q 16066 .

Commented by mrW1 last updated on 29/Jun/17

Please show me the solution in your

book . I^{'} ll try .

Commented by mrW1 last updated on 30/Jun/17

that is mistake in editting with my

small smartphone .

it means certainly

[O B C] = \frac{O C . P M}{2} + \frac{O B . P Q}{2}

Commented by Tinkutara last updated on 30/Jun/17

Thanks Sir!

Commented by mrW1 last updated on 30/Jun/17

you are right . {it}^{'} s an other 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 O A = O B = O C = O D = A > 0 .

Suppose M lies in the interior of the

angle A O B .

If M_{A} and M_{B} are the projections of

M onto O A and O B, we have

{M M}_{A} + {M M}_{B} = k .

Multiplying by \frac{a}{2}, we obtain

[M O A] + [M O B] = \frac{k a}{2} .

But

[M O A] + [M O B] = [M A O B]

= [O A B] + [M A B] .

It follows that the area of M A B is

constant, and hence the locus of M in

the interior of ∠ A O B is a segment X Y

parallel to A B .

Considering the other three possible

locations, of M, we deduce that the

locus is a rectangle X Y Z T whose

diagonals are d and d^{'} .

Observation : It is worth mentioning

that if M lies, for instance, on the line

X Y but not in the interior of the

segment X Y, another equality occurs .

If M is on the half - line X Y as in Figure

5.15, we have

[M O A] - [M O B] = [O A B] + [M A B],

so we deduce that for those positions of

M, the difference of the distances to d

and d^{'} 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 .

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 Q 16066 . In Q 16066 if a = b^{'} = 2 b,

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 .

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 .

Commented by mrW1 last updated on 30/Jun/17

I think you can follow my answer better

than follow the answer in the 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!