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Coordinate GeometryQuestion and Answers: Page 1

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solve for x, y, z ∈R^+ x^2 +y^2 −2xy cos γ=c^2 y^2 +z^2 −2yz cos α=a^2 z^2 +x^2 −2zx cos β=b^2 with α+β+γ=360° example: a=12, b=8, c=10 α=120°, β=90°, γ=150°

$${solve}\:{for}\:{x},\:{y},\:{z}\:\in{R}^{+} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}\:\mathrm{cos}\:\gamma={c}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{yz}\:\mathrm{cos}\:\alpha={a}^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{zx}\:\mathrm{cos}\:\beta={b}^{\mathrm{2}} \\ $$$${with}\:\alpha+\beta+\gamma=\mathrm{360}° \\ $$$$ \\ $$$${example}:\: \\ $$$${a}=\mathrm{12},\:{b}=\mathrm{8},\:{c}=\mathrm{10} \\ $$$$\alpha=\mathrm{120}°,\:\beta=\mathrm{90}°,\:\gamma=\mathrm{150}° \\ $$

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I. A(−5, −1); B(3, −5); C(5, 2) ar(△ABC) = ? II. A(5, 3); B(2, 5); C(−5, 3); D(−4, −3) ar(□ABCD) = ? shortest solution

$$\mathrm{I}.\:\:\:\:\:\:\:\mathrm{A}\left(−\mathrm{5},\:−\mathrm{1}\right);\:\mathrm{B}\left(\mathrm{3},\:−\mathrm{5}\right);\:\mathrm{C}\left(\mathrm{5},\:\mathrm{2}\right)\:\:\:\:\:\:{ar}\left(\bigtriangleup\mathrm{ABC}\right)\:=\:? \\ $$$$\mathrm{II}.\:\:\:\:\:\mathrm{A}\left(\mathrm{5},\:\mathrm{3}\right);\:\mathrm{B}\left(\mathrm{2},\:\mathrm{5}\right);\:\mathrm{C}\left(−\mathrm{5},\:\mathrm{3}\right);\:\mathrm{D}\left(−\mathrm{4},\:−\mathrm{3}\right)\:\:\:\:\:\:\:{ar}\left(\Box\mathrm{ABCD}\right)\:=\:? \\ $$$$\mathrm{shortest}\:\mathrm{solution}\: \\ $$

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Focus and vertex of a parabola are at (3, 4) and (0,0). Find the equation of the directrix.

$$\mathrm{Focus}\:\mathrm{and}\:\mathrm{vertex}\:\mathrm{of}\:\mathrm{a}\:\mathrm{parabola}\:\mathrm{are}\:\mathrm{at}\:\left(\mathrm{3},\:\mathrm{4}\right)\:\mathrm{and}\:\left(\mathrm{0},\mathrm{0}\right). \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{directrix}. \\ $$

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Q202938 the value of x is 15 (Voir reponse develope )

$$\mathrm{Q202938} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{is}}\:\mathrm{15}\: \\ $$$$\left(\boldsymbol{{Voir}}\:\boldsymbol{{reponse}}\:\boldsymbol{{develope}}\:\right) \\ $$

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shortest distance from (−6,0)to x^2 −y^2 +16=0

$${shortest}\:{distance}\:{from}\:\left(−\mathrm{6},\mathrm{0}\right){to}\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{16}=\mathrm{0} \\ $$

Question Number 201660 Answers: 1 Comments: 0

An equilateral triangle inscribed in a parabola y^2 =4x. One of its vertices is at the vertex of the parabola. Find the length of each side of the triangle in units.

$${An}\:{equilateral}\:{triangle}\:{inscribed}\:{in}\:{a}\:{parabola} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x}.\:{One}\:{of}\:{its}\:{vertices}\:{is}\:{at}\:{the}\:{vertex}\:{of}\:\:{the}\:{parabola}. \\ $$$${Find}\:{the}\:{length}\:{of}\:{each}\:{side}\:{of}\:{the}\:{triangle}\:{in}\:{units}. \\ $$

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Find the shortest distance between point A(3,2) and curve y=(√x) (x>0).

$${Find}\:{the}\:{shortest}\:{distance}\:{between}\: \\ $$$${point}\:{A}\left(\mathrm{3},\mathrm{2}\right)\:{and}\:{curve}\:{y}=\sqrt{{x}}\:\left({x}>\mathrm{0}\right). \\ $$

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