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Question Number 2672 by Yozzi last updated on 24/Nov/15

I have 4 collinear points A(a,0), B(b,0), C(c,0) and D(d,0) where ∀a,b,c,d>0. Find a point E(x,y) such that the following expression is minimised: 2(AE+BE+CE+DE).

$${I}\:{have}\:\mathrm{4}\:{collinear}\:{points}\:{A}\left({a},\mathrm{0}\right), \\ $$ $${B}\left({b},\mathrm{0}\right),\:{C}\left({c},\mathrm{0}\right)\:{and}\:{D}\left({d},\mathrm{0}\right)\:{where}\: \\ $$ $$\forall{a},{b},{c},{d}>\mathrm{0}.\:{Find}\:{a}\:{point}\:{E}\left({x},{y}\right)\:{such} \\ $$ $${that}\:{the}\:{following}\:{expression}\:{is} \\ $$ $${minimised}: \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\left({AE}+{BE}+{CE}+{DE}\right). \\ $$

Commented byRasheed Soomro last updated on 24/Nov/15

∀a_(−) ,b,c,d>0 There is no a involved in the problem. Points are A(0,0),B(b,0), C(c,0) and E(x,y). Is there A(a,0) instead of A(0,0)?

$$\forall\underset{−} {{a}},{b},{c},{d}>\mathrm{0} \\ $$ $$\mathcal{T}{here}\:{is}\:{no}\:\:{a}\:\:{involved}\:{in}\:{the}\:{problem}. \\ $$ $${Points}\:{are}\:{A}\left(\mathrm{0},\mathrm{0}\right),{B}\left({b},\mathrm{0}\right),\:{C}\left({c},\mathrm{0}\right)\:{and}\:{E}\left({x},{y}\right). \\ $$ $${Is}\:{there}\:{A}\left({a},\mathrm{0}\right)\:{instead}\:{of}\:{A}\left(\mathrm{0},\mathrm{0}\right)? \\ $$

Commented byYozzi last updated on 24/Nov/15

Sorry. Error. I made it A(a,0) instead.

$${Sorry}.\:{Error}.\:{I}\:{made}\:{it}\:{A}\left({a},\mathrm{0}\right)\:{instead}. \\ $$

Answered by prakash jain last updated on 24/Nov/15

assume a<b<c<d Consider 3 points BCE. BE+CE≥BC. BE+CE=BC when E is on the same line as BC and in between them. b≤x≤d Similarly for points ADE. a≤x≤d I am assuming A′s coordinate is (a,0). If a<b<c<d Then coordinated for point E are (x,0) where b≤x≤c (assuming a<b<c<d).

$$\mathrm{assume}\:{a}<{b}<{c}<{d} \\ $$ $$\mathrm{Consider}\:\mathrm{3}\:\mathrm{points}\:\mathrm{BCE}. \\ $$ $$\mathrm{BE}+\mathrm{CE}\geqslant\mathrm{BC}. \\ $$ $$\mathrm{BE}+\mathrm{CE}=\mathrm{BC}\:\mathrm{when}\:\mathrm{E}\:\mathrm{is}\:\mathrm{on}\:\mathrm{the}\:\mathrm{same}\:\mathrm{line}\:\mathrm{as}\:\mathrm{BC} \\ $$ $$\mathrm{and}\:\mathrm{in}\:\mathrm{between}\:\mathrm{them}. \\ $$ $${b}\leqslant{x}\leqslant{d} \\ $$ $$\mathrm{Similarly}\:\mathrm{for}\:\mathrm{points}\:\mathrm{ADE}.\:{a}\leqslant{x}\leqslant{d} \\ $$ $$\mathrm{I}\:\mathrm{am}\:\mathrm{assuming}\:\mathrm{A}'{s}\:\mathrm{coordinate}\:\mathrm{is}\:\left({a},\mathrm{0}\right). \\ $$ $$\mathrm{If}\:{a}<{b}<{c}<{d} \\ $$ $$\mathrm{Then}\:\mathrm{coordinated}\:\mathrm{for}\:\mathrm{point}\:\mathrm{E}\:\mathrm{are} \\ $$ $$\left({x},\mathrm{0}\right)\:\mathrm{where}\:{b}\leqslant{x}\leqslant{c}\:\left({assuming}\:{a}<{b}<{c}<{d}\right). \\ $$

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