Equation of a line. Line in coordinate geometry. Slopeintercept form ( 斜 截 式 ) Intercept form ( 截 距 式 ) Pointslope form ( 點 斜 式 )


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1 Chapter : Liear Equatios Chapter Liear Equatios Lie i coordiate geometr I Cartesia coordiate sstems ( 卡 笛 兒 坐 標 系 統 ), a lie ca be represeted b a liear equatio, i.e., a polomial with degree. But before we proceed, first itroduce some terms that should be kow. 0 4 Equatio of a lie The equatio of a lie ca be writte as equatio A + B + C 0 Where A, B ad C are real costats. This is called the geeral form ( 一 般 式 ) of a lie. We ca fid the equatio of a lie b give a two of the followig values: The slope The itercept The itercept Coordiate of a poit Coordiate of aother poit Slopeitercept form ( 斜 截 式 ) Whe the slope m ad the itercept (0,c) of a lie is kow, the the equatio of the lie is just: m + c Moreover, base o the above result, we ca kow the slope ad itercept of a lie if its geeral form A + B + C 0 is give: A C m, c B B Itercept form ( 截 距 式 ) The figure represets a lie. The lie crosses the ais at. The poit (0,) is called the itercept ( 截 距 ). Similarl, the poit (,0) is called the itercept ( 截 距 ). Moreover, there is a quatit called slope ( 斜 率 ) tellig us the obliquit of the lie. It ca be calculated b the followig formula: m Here, m is the smbol for slope, ad (, ), (, ) are two poits lig o the lie. I this lie, the slope is /. If a lie is parallel to the ais, its slope is 0. If a lie is parallel to the ais, its slope is udefied. Whe the  ad itercepts, (a, 0) ad (0,b) are give, the equatio of the lie is: + a b Poitslope form ( 點 斜 式 ) Whe the slope m ad a poit ( 0, 0 ) is give, the equatio is: m 0 0 9
2 Chapter : Liear Equatios Twopoit form ( 兩 點 式 ) Whe coordiates of two poits (, ), (, ) are give, the equatio is: Now take the lie above as eample. As said, the slope of that lie is /, the itercept is (0,), the itercept is (,0). We ca see that it also passes through the poits (4, ) ad (,). B slopeitercept form, we get the equatio: m+ c B itercept form, we get the equatio: + + a b + B poitslope form, we get the equatio: m ( ) 0 0 B twopoit form, we get the equatio: ( ) 4 ( ) 4 Simplifig all equatios above, ad we will all get our geeral formula for the lie: + 0 Itersectig Poit of Lies ( 線 之 交 點 ) The followig figure shows two lies itersectig at a poit:  0 Oe ma be iterested i fidig the coordiate of the itersectig poit. What should we do? Firstl, fid out the equatio of the two lies. Here, the two lies are + 0 ad + 0 respectivel. The, we set up a simultaeous equatios sstem ( 聯 立 方 程 系 統 ): Sice both equatios are liear, ad there are ukows, we call this a simultaeous liear equatios i ukows ( 聯 立 二 元 一 次 方 程 ), ad I ll abbreviate it as SLE. For SLE, we usuall write the equatios as: + Moreover, we would label the first equatio as (), the secod as (): + Now what? I order to solve SLE, we have methods: 0
3 Chapter : Liear Equatios Substitutio method ( 代 入 法 ) I (), we see that. Substitutig this ito (), we get: 5. That meas the itersectig poit is(, ). So, 5 Elimiatio method ( 消 元 法 ) Rather tha substitutio, we ma also solve the equatio b elimiatig a variable. How? I (), we see a . I (), we see a +. Does t it be good that the ca be cacelled b addig () ad () together? Surel! You ca just add () ad () together to form a ew equatio:  Not just, ca also be elimiated b doig () (), so:  5 Formula ( 公 式 ) The above two methods are eas to uderstad ad implemet, but would be iefficiet if the coefficiets are quite complicated, like, ? Actuall, there is a geeral formula solvig SLE. For a+ b c d+ e f Defie ae bd, bf ce, cd af ( is proouced as delta ). The,. To simplif memorizig, we ofte deote p q ps qr r s, so a b b c c a,,. d e e f f d Note that ot all SLE are solvable. Sometimes two lies are parallel, i.e., o itersectig poit. Usig substitutio or elimiatio method will ield a false statemet (e.g. 70), ad usig formula will give 0, but either or is ot zero. O the other had, a SLE ma also have ifiite solutios. Such a case is two lies overlap. At that time, usig substitutio or elimiatio method will give a idetit (e.g. ), ad usig formula will result i, ad all are zeroes. I additio, substitutio method ca also appl for other differet tpes of simultaeous equatios. For eample, cosider B substitutig ito (), we get, i.e.,. Ad ca easil be 4 derived from this. Etesio: If more tha lies itersect i oe poit ol, the are called cocurret ( 共 點 ) Absolute value ( 絕 對 值 ) Before we proceed, let s itroduce what is absolute value. The absolute value of is deoted b, where is defied as: if 0 if < 0 Or simpl: igore the egative sig. For eample, 5 5;  ; 56. The properties of absolute values are: 0 If a, ad a 0, the a or a. If a < 0, the equatio has o solutios. If, the or .
4 Chapter : Liear Equatios Distace Let us first review how to calculate the distace betwee two poits. B Pthagoras theorem ( 畢 氏 定 理 ), we kow the distace d betwee two poits (, ) ad (, ) is: d + How about the distace betwee a poit ad a lie? First we eed to defie the meaig the distace here, because there are so ma poits o a lie for ou to measure. We defie the distace betwee a poit ad a lie to be the shortest distace. (, /) / Amog all the lies passig though the poit ad the lie, we foud that the lie that both pass through the poit ad perpedicular to the lie make the distace shortest, as illustrated above. If the coordiates of a poit is (, ), ad the equatio of the lie is a + b + c 0, the the distace is: d a + b + c a + b For eample, the graph above tells us that the distace betwee poit (, /) ad lie + is: ( ) + ( ) + + d Properties of Slope Slope tells us how oblique a lie is. Actuall, b this, we ca give out some properties about slope. ) If the slopes of two lies are the same (i.e., m m ), the are parallel or the same. ) If the product of the slopes of two lies is (i.e. m m ), the are perpedicular to each other. Poit of Divisio of a Lie ( 線 之 分 點 ) A poit of divisio o a lie is the poit that divides a lie ito two parts. O lie AB, we defie a poit P such that it divides the lie ito ratio of r: s. If the coordiates of A is (, ), B is (, ), P is (, ), the s + r s + r (, ), s+ r s+ r This is called the sectio formula ( 分 點 公 式 ). Note that r ad s ma be egative, ad the poit is said to divide the lie eterall ( 外 分 ). B (,) A (0,) 0 4 For eample, if we wat to fid the poit that divides the lie segmet AB above i ratio :5, b sectio formula, the poit is,. 8
5 Chapter : Liear Equatios A special case of sectio formula is that r: s :. This ields the midpoit formula ( 中 點 + +,, 公 式 ), Area of Polgos Last but ot least, we tell ou how to calculate the area of a polgo with the coordiates of each verte is give. If the vertices of a polgo are (, ), (, ), (, ), (, ), the its area is: Here, otatio A meas ( ) ( ). Or, we ca memorize it like this: Note that the poits (, ), (, ), (, ), (, ) should be arraged i aticlockwise or clockwise order. Otherwise, the result ma be wrog. For eample, the area of the followig quadrilateral is: Revisio: A 0  I this chapter, we ve leart:. Lie i coordiate geometr. Equatio of a lie. Itersectig poit of lies 4. Absolute value 5. Distace betwee lie ad poit 6. Properties of slope 7. Poit of divisio 8. Area of polgo
6 Chapter : Liear Equatios Eercise I the followigs, if ot specified, k is a costat, ad is a iteger.. Fid out the slope,  ad itercept of the followigs: a) b) 9 c) + 4 d). Fid out the equatio of a lie which the product of its  ad itercept is 0, the itercept is o the right of the origi, ad it is parallel to the lie (HKCEE 990) I the figure, A (,0), B (0,5) ad C (0,) are three poits ad O (0,0) is the origi. D is a poit o AB such that the area of BCD equals half of the area of OAB. Fid the equatio of the lie CD. B C D 5. Prove the formula for solvig a SLE. 6. (HKMO 000 Heat) Fid the shortest distace betwee the lie 4 0 ad the poit (, ). 7. If + 5 4, fid all possible values of. [Hit: Cosider the coditios for <, < 5 ad 5] 8. (ISMC 000 Fial) Solve, ad z i: + z z z Cosider the lies L : ad L : + 0.Fid the equatio of the lie passig through the itersectio poit of L ad L, ad is perpedicular to L. 0. A lie L itersects the aes at A (a, 0) ad B (0, b). M (, 4) is the midpoit of AB. a) Fid a ad b. b) Fid the equatio of AB. c) C is a poit o the coordiate plat such that AC BC. The area of BCD is 5 square uits. Fid all possible coordiates of C.. Prove.. Let A (, ), B (, ) ad C (, ) be poits of vertices of a triagle. Let D, E, F be midpoits of BC, CA ad AB respectivel. a) Show that AD, BE ad CF are cocurret. b) Assume the three lies meet at G. Show that AG:GD BG:GE CG:GF : O A 4. I lie segmet AB, the coordiates of the two eds A ad B are (, 5) ad (7, ) respectivel. C (k, k) ad P ( 0, 0 ) is a poit o AB. D (, 4) is a poit. Fid: a) The equatio of AB. b) k. c) The ratio r:s that poit C divides AB. [Hit: r:s r/s] d) Legth of CD. e) The shortest legth of PD. f) The coordiates of P whe PD is the shortest. g) Legth of CP whe PD is the shortest. h) Area of PCD whe PC is the shortest. 4
7 Chapter : Liear Equatios Suggested Solutios for the Eercise a) Slope 5/6, itercept 7/6, itercept 7/5 b) Slope, itercept 9, itercept 9 c) Slope /4, itercept 6, itercept 8 d) Slope, itercept 0, itercept 0 ) ) a) b) 9 7 c) :4 d) e) f), g) 7 9 h) ) 0 7), 6 8) 6,, z6 9) a) a 6, b 8 b) c) (7,7), (,) 5
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