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= 2</b> </p> <p><img width="26" height="2" src="file:///C:/Users/jp%20computer/AppData/Local/Packages/oice_16_974fa576_32c1d314_f69/AC/Temp/msohtmlclip1/01/clip_imaXXXXX.XXX" v:XXXXXX="_x0000_s1026"> </p> <p> </p> <p><b>m=2</b> </p> <p> </p> <p><b>XXXXXXXX in XXXXXXX is y=mx+b where b XX XXXXXXXXX</b> </p> <p> </p> <p><b>XXX calculating it XXX we XXX XXXXX X(6,XX)</b> </p> <p> </p> <p><b>XX we XXXX here y=XX XXX x=X XXXX put in the XXXXXXXX</b> </p> <p> </p> <p><b>X=XX+b</b> </p> <p> </p> <p><b>11=X×6+b</b> </p> <p> </p> <p><b>b=-X</b> </p> <p> </p> <p> </p> <p><b>so equation XX</b> <b>= 2 − X</b> </p> <p> </p> <ul><li><b>XXXX XXX XXXXXXXX of XXX XXXXXXXXXXXXX XXXXXXXX of line AB (XXXX)</b> </li></XX> <p> </p> <p> </p> <p><b>XXX doing so first calculate XXXXX XX the XXXXXXXXXXXXX bisector</b> </p> <p> </p> <table> <tbody><tr> <td> <p><b>Midpoint XX XX XXX (</b> </p></td><td> <p> </p></td><td> <p><b>,</b> </p></td><td> <p> </p></td><td> <p><b>)=(5,X)</b> </p></td><td> <p> </p></td><td> <p> </p></td></tr> <tr> <td> <p> </p></td><td> <p> </p></td><td> <p> </p></td><td> <p> </p></td></tr> <tr> <td> <p><b>Slope of XXX XX can XXXXXXXXX from XXXXX points as</b> </p></td><td> <p> </p></td><td> <p><b>= = 2</b> </p></td></tr> <tr> <td> <p> </p></td></tr> <tr> <td> <p><b>m=2</b> </p></td><td> <p> </p></td><td> <p> </p></td><td> <p> </p></td><td> <p> </p></td><td> <p> </p></td><td> <p> </p></td></tr> </tbody></table> <p><XXX width="X" XXXXXX="3" XXX="file:///C:/XXXXX/XX%20computer/XXXXXXX/XXXXX/XXXXXXXX/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/AC/XXXX/msohtmlclip1/01/clip_image005.XXX" v:XXXXXX="_x0000_s1027"> </p> <p> </p> <p><b>and that of XXX Perpendicular XXXXXXXX XX−<img width="12" height="X" src="file:///X:/XXXXX/jp%20computer/XXXXXXX/Local/Packages/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/XX/Temp/XXXXXXXXXXXX/XX/clip_image007.jpg" v:shapes="_x0000_i1026"> XX AB so will XX</b> <b>-<XXX width="X" XXXXXX="4" src="XXXX:///X:/Users/jp%XXXXXXXXXX/AppData/Local/Packages/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/AC/Temp/msohtmlclip1/XX/XXXXXXXXXXXXX.jpg" v:XXXXXX="_x0000_i1027"></b> </p> <p> </p> <p> </p> <p> </p> <p><b>XXX we can XXXXXXXXX the XXXXXXXXX using (5, X) and m as -1/2</b> </p> <p> </p> <p> </p> <p> </p> <p><b>XX XXX</b> </p> <p> </p> <p><b>9=− <XXX width="X" height="X" XXX="file:///C:/XXXXX/XX%20computer/XXXXXXX/Local/Packages/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/XX/XXXX/XXXXXXXXXXXX/XX/clip_image005.XXX" v:XXXXXX="_x0000_i1028">×X+b</b> <b><XX> </b> </p> <table> <tbody><tr> <td> <p><b>and y =-12hence XXXXX are the coordinate XX XXXXX C</b> </p></td></tr> </tbody></table> <p><b>b=11.5</b> </p> <p> </p> <p><b>now XXXXXXXX XX</b> </p> <p> </p> <p> </p> <p> </p> <p><b>Y= −<img width="9" height="3" src="XXXX:///C:/Users/XX%XXXXXXXXXX/XXXXXXX/Local/Packages/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/AC/XXXX/XXXXXXXXXXXX/XX/clip_image005.jpg" v:shapes="XXXXXXXXXXXX"> + 11.5 is</b> </p> <p> </p> <p> </p> <p> </p> <XX><li><b>XXXXX XXXX AC XX perpendicular XX XX and the equation of XXXX BC is y=-3x+45, find the co-XXXXXXXXX of</b> </li></XX> <p><b>(3ks)</b> </p> <p> </p> <p><b>XXX’s find XXX equation XX XXXX AC which is XXXXXXXXXXXXX to line XX having XXXXXXXX XXXXX is y=XX-X XXXXXXXXXXXXX XXXX XX XXX that has a XXXXXXXX, XXXXXXXXXX slope to another.</b> </p> <p> </p> <p><b>So AC XXXX have equation y=-1/XX-X</b> </p> <p> </p> <p> </p> <p> </p> <p><b>As y XXXX be same XXX the point c either we use XXXXXXXX of XX or equation XX so XX can XXX XXXX XXXXX</b> </p> <p> </p> <p><b>-X/2x-X=-XX+XX</b> </p> <p> </p> <p><b>X.5x=XX</b> </p> <p> </p> <p><b>X=XX/5</b> </p>">
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