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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_image003.XXX" v:shapes="XXXXXXXXXXXX"> </p> <p> </p> <p><b>m=X</b> </p> <p> </p> <p><b>XXXXXXXX in XXXXXXX is y=XX+b XXXXX b is XXXXXXXXX</b> </p> <p> </p> <p><b>For calculating it let XX use point B(6,11)</b> </p> <p> </p> <p><b>XX we have XXXX y=XX and x=X lets XXX in XXX XXXXXXXX</b> </p> <p> </p> <p><b>X=XX+b</b> </p> <p> </p> <p><b>11=2×6+b</b> </p> <p> </p> <p><b>b=-X</b> </p> <p> </p> <p> </p> <p><b>so equation XX</b> <b>= X − X</b> </p> <p> </p> <ul><li><b>XXXX the XXXXXXXX of XXX perpendicular bisector of XXXX AB (XXXX)</b> </li></XX> <p> </p> <p> </p> <p><b>XXX XXXXX so XXXXX calculate slope of the perpendicular bisector</b> </p> <p> </p> <table> <tbody><tr> <td> <p><b>XXXXXXXX of AB XXX (</b> </p></td><td> <p> </p></td><td> <p><b>,</b> </p></td><td> <p> </p></td><td> <p><b>)=(X,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>XXXXX XX XXX AB XXX calculate XXXX XXXXX XXXXXX 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=X</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="X" XXX="XXXX:///C:/XXXXX/jp%XXXXXXXXXX/XXXXXXX/XXXXX/Packages/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/XX/Temp/msohtmlclip1/01/XXXXXXXXXXXXX.XXX" v:shapes="_x0000_s1027"> </p> <p> </p> <p><b>and that XX XXX XXXXXXXXXXXXX bisector is−<img width="XX" height="4" XXX="XXXX:///C:/Users/jp%20computer/XXXXXXX/XXXXX/Packages/oice_16_974fa576_32c1d314_f69/XX/XXXX/msohtmlclip1/01/clip_image007.jpg" v:XXXXXX="_x0000_i1026"> of XX so XXXX XX</b> <b>-<XXX width="X" XXXXXX="X" XXX="XXXX:///X:/Users/XX%XXXXXXXXXX/XXXXXXX/XXXXX/XXXXXXXX/oice_16_974fa576_32c1d314_f69/XX/XXXX/msohtmlclip1/01/clip_image009.XXX" v:XXXXXX="_x0000_i1027"></b> </p> <p> </p> <p> </p> <p> </p> <p><b>XXX XX can XXXXXXXXX XXX intercept using (5, X) and m XX -1/2</b> </p> <p> </p> <p> </p> <p> </p> <p><b>XX get</b> </p> <p> </p> <p><b>X=− <XXX width="X" height="X" src="file:///C:/XXXXX/jp%20computer/XXXXXXX/Local/XXXXXXXX/oice_16_974fa576_32c1d314_f69/XX/Temp/msohtmlclip1/XX/XXXXXXXXXXXXX.jpg" v:XXXXXX="XXXXXXXXXXXX">×X+b</b> <b><br> </b> </p> <table> <tbody><tr> <td> <p><b>XXX y =-12hence these are the XXXXXXXXXX of XXXXX X</b> </p></td></tr> </tbody></table> <p><b>b=XX.X</b> </p> <p> </p> <p><b>now XXXXXXXX is</b> </p> <p> </p> <p> </p> <p> </p> <p><b>Y= −<img width="9" XXXXXX="X" src="file:///C:/Users/jp%XXXXXXXXXX/XXXXXXX/XXXXX/XXXXXXXX/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/XX/Temp/XXXXXXXXXXXX/01/clip_image005.jpg" v:XXXXXX="_x0000_i1029"> + 11.5 XX</b> </p> <p> </p> <p> </p> <p> </p> <ul><li><b>XXXXX that XX is perpendicular XX AB XXX XXX equation of line BC is y=-3x+XX, XXXX the co-XXXXXXXXX XX</b> </li></XX> <p><b>(XXX)</b> </p> <p> </p> <p><b>Let’s find XXX equation XX line AC XXXXX is XXXXXXXXXXXXX to line AB having equation XXXXX is y=2x-X perpendicular XXXX is XXX XXXX XXX a negative, reciprocal XXXXX to XXXXXXX.</b> </p> <p> </p> <p><b>So XX will XXXX equation y=-1/2x-X</b> </p> <p> </p> <p> </p> <p> </p> <p><b>As y XXXX be same XXX XXX XXXXX c XXXXXX we use XXXXXXXX of XX or XXXXXXXX XX so XX XXX put XXXX equal</b> </p> <p> </p> <p><b>-1/2x-1=-3x+45</b> </p> <p> </p> <p><b>X.XX=XX</b> </p> <p> </p> <p><b>X=92/X</b> </p>">
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