0002000201000cVector-3.ACT0007020016(í°-í¡)î’(í°-í¢)=0.EAC010000000a3eÉ (í°-í¡)î’Š ¢)=0.EAC† Vector-3.ACT†*†-†7ˆ:†>Š Ž †NkE Ž'[ˆ4ŒQ Circle-2†† ¤…ŽŽ% \Œ6Geometry windowÎˆ0†ÎŒª§ŒˆŠŠ`ŠŒ¹ `™™™™™•†–šDvŽ†ßr† ŒìA† 'LŽ'‰Y‰D’ u˜x˜yŽ †qˆqˆ¹ˆB Š@‡¥Ð6–°†1mŠ«†Š‚„'GF†ˆ†Ä@! #†.ÈÈ `ˆ7B@†>Ð6–°†E’I–˜a††C †CŽU–Bgd†’Y™” Žˆs ™†3†³É’…A˜…’h`–…Š®˜…ˆÈŒÞ‰ŽYˆŸV„"ˆ¯Šöˆ!ˆãˆ%ˆˆó”ü[Š†When the point a=‰Ž3Œ'2Œ/ and Ž(1…0qy`ˆH#Y‡uYƒ†U ˜–0–šH®Æ$ÊHš©š¶´¨5°´þaþ ‰¼†Œ˜ °˜’®`–$ ™†OŠq[Ž;eAct020018(í°-í£)î’(í¡-í£)=r^2.EAC010000001cb0É¸Dkäûx[Vector Circle-3[ Tangent Line[ (í°-í£)î’(í¡-í£)=r^2[[ \Geometry windowÎÎ2%y`wP3G`` H8p–wA–F˜r@vrALCuxyˆB @Ð6–°fffffg ™E5Y’I™ ™u•Ffp ™Ä@ ÄH B@Ð6–°!2CVY™!2CVY™ÂÀ# @Ð6–°ÈÈ @Ð6–°ÆH! 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ŠˆªÐ6–°†$G!5•I™‡£ Y™”Bq™˜† ™•„ˆ`†Ä@ † ††ËÈ!`Š@†%Ð6–°†,$`Š9Œ†X ÈÈŽTr–C‰X)†z Y™WWc94ŠdPH†‘ ˜†C†›ÊÈ ¦…Qa)"Œ*XEŠºŠ6˜…ˆÇ¢ÝˆXˆŸV„"ˆ¯ˆˆ ‡ÉÈ ¤ý–ñˆ+®ýˆ(ˆG‹HŽ_<”E[‡When the line passesŠŽ! point Ž53†=5† and is Œparpendicular to the Ž vector í²=Ž6-2ŒO4ŒW, fi†Yˆ=Œilinear equation.[ŒŽ î~SoluˆRŠJŽ‘dotP(šnxŒ¼yŒ~-šŽ3ŒÜä,¼)R’r-2î’’lŽ-]*+4–'y '5ŠÆeans=0òeœf-†5Œ=0RŠŠ expand(ans/2)RŽ -x+2î’y-7=0[ŒB†ˆK Try your own.\Œ_Geomet†windowÎˆ4 †8Î†‡ˆŒˆˆŠŠŠX`Œ"–Šo–šDvŽ†ár† A† 'LŽ'‰C˜ u˜x˜yŽ ˆ ˆµŠv‡I Žõ CalculatorËˆÝ†ßŠ;Š´—l‹DÀSGraph2D,† †Š3Š@†ˆLISTSYSˆL4N†Modify†;€†($ˆ<STATCALC¤†Uˆ(ˆŒ<¬\ˆ<Sequence†<,ˆxSheet†Š4†|’ŠŒ°’olveEq†´ˆ´Lwr0†Í(Upˆ<’tupFLG1H†<(Š<†Lis†{p†dD‰ˆPic†´ÜViewWindØ†‰,äŠ±‡@ôxäŠ ŠŒä( ä<´äP$Èäd’ÜäŽx <†ä† † H’’T’’`’’l’’x’’„’’’’œ’’¨’ ’´’!’À’"’Ì’#’Ø’$’ä’%’ð’&’ü’'‘@’(‘T’)‘hˆ¾Œ*‘|,’+‘8’ˆ‹§D¸†ä-†P† †.’\’0’h’1’t’2’€’3’Œ’4’˜’5’¤’E’°’F’¼’H’È’I’Ô’J’à’K’ì’L’ø’M‘,’N‘@’O‘T’‰^‹k(’Q‘|‰ŒR‘@’S‘¤ˆ–ŒT‹¸X††ä]††d† ˆ^’h’_’l’`’p’a’t’b’x’”’|’Œ•’ˆ’Í’ˆ2ŒÎ’ ’Ð’¬Ü††ˆŒ!†” system‰list’‡]=’‡‰’†ü‰’†ø‰’†ô~[hÁŽŽž¾ ¤@hÁŽŽž¾ þ@þ€þÀÿÿ@ÿxhÁhŽŽž¾ þ@þ€þÀÿÿ@ÿx‡¸hÁŽŽž¾ þ@þ€þÀÿÿ@ÿx‡¸hÁŠŽž° ††a†F seq_histb†NewFoldeŠ]ˆ3 ’ systemä]l†0’^]=” –˜††î<ˆxö<þxìðÒ´ emhÁŠŒsystˆä]list’^]= ö<þxþ´<† ††††’’,† †0†@†@PSheet1úê¬^Š2ž3ž 4ž 5ž †©ŠX3D´|PSheet1èê¬^Š2ž3ž 4ž 5ž ŒH3D Šˆi™” ™††Š Ž ˆ8¸Dkäûx†††$’S’\¢œ†jD²<‡Žh†(žŽ `–PÝÆ$´$“$–p”x”,ˆˆ ™€ `”˜$’.ˆ>(1…0qy`ˆH#Y‡uYƒ†U ˜–0–šH®Æ$ÊHš©š¶´¨5°´þaþ ‹µ†Œ˜ °˜’®`–$ ™†OŠq[Ž;eAct020017Line í°=(1-t)í¡+tí¢.EAC01000000129fÉ Line í°=(1-t)í¡+tí¢.EAC† Vector-3.ACT†+†. ˆ;†?Š Ž †O†kE †(Œ'[ˆlŒQ and ˆwsŽa+tb†ŠlŽ \ŒGeometry windowÎˆ0†“ÎŒ¢ŒŸˆˆâ™™™™™’†`@É`Š”†9ˆŠ–DvŽ.†ór† ŒäA† 'LŽ'‰R‰<’ u˜x˜yŽ †qˆqˆ¹‰H ŠˆªÐ6–°†1–a5‡XY™ €Xui ™T• ug•†`††Ä@ † †ˆ'ÈÈ Š @†6Ð6–°†=$†<0†B †BŒTs–CIH‘Iw—†‹ Y™5PR4€Š%4%SW†¢ ˜†C†¬ÊÈ ¦…FTˆ5 ˜(“Š¹ ˆS˜…ˆî¢ÝˆXˆŸV„"ˆ¯ ‡Å˜* Ž‚§˜û•£r—J5TC %Y‹IB0P’Tž(Hw6› †Ä@ † ††† Žˆ"ˆˆ† †6Ž ˆ1Ž†V [ŠGiven a line passes through points Ž92ŒA3ŽJ&˜ -3œ), Žtfind the equation ofŠŒ”ˆŒ.[Œ¡Ž î~Soluˆ.RŠWŽÁ(1-t)îºœ+tœ"–žRŒ‘-å'î’t %1š73†î’Œ tŽ-Œ˜%Ž0†'ž%1’7+ŒS2¤\RŠlŠrexpand(ans)R–}Žš5’¢¢|”j ³¢#3[Œé Then we have:Ž x=-5î’t+2Ž&y=-t+3[Using substitution, ŒÇ eliminate t. 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