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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=X</b> </p> <p> </p> <p><b>Equation in XXXXXXX XX y=mx+b XXXXX b is XXXXXXXXX</b> </p> <p> </p> <p><b>For calculating it let XX use XXXXX X(X,XX)</b> </p> <p> </p> <p><b>So XX have XXXX y=XX and x=6 XXXX put in XXX XXXXXXXX</b> </p> <p> </p> <p><b>Y=mx+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 XXXXXXXX XX</b> <b>= X − X</b> </p> <p> </p> <ul><li><b>Find the XXXXXXXX XX XXX perpendicular bisector of line XX (XXXX)</b> </li></XX> <p> </p> <p> </p> <p><b>For doing so first calculate slope of the XXXXXXXXXXXXX bisector</b> </p> <p> </p> <table> <tbody><tr> <td> <p><b>XXXXXXXX of XX 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 of the XX can calculate from above 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=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="9" height="3" XXX="file:///X:/Users/XX%20computer/AppData/Local/XXXXXXXX/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/XX/Temp/msohtmlclip1/XX/XXXXXXXXXXXXX.XXX" v:XXXXXX="_x0000_s1027"> </p> <p> </p> <p><b>and that XX XXX XXXXXXXXXXXXX XXXXXXXX is−<XXX width="XX" height="4" src="file:///X:/Users/XX%20computer/XXXXXXX/XXXXX/Packages/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/XX/XXXX/msohtmlclip1/01/clip_image007.jpg" v:shapes="_x0000_i1026"> XX XX so will XX</b> <b>-<img width="8" XXXXXX="X" XXX="file:///C:/XXXXX/jp%20computer/AppData/Local/XXXXXXXX/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/AC/XXXX/msohtmlclip1/XX/clip_image009.XXX" v:XXXXXX="XXXXXXXXXXXX"></b> </p> <p> </p> <p> </p> <p> </p> <p><b>now XX XXX XXXXXXXXX XXX intercept using (X, X) and m XX -1/2</b> </p> <p> </p> <p> </p> <p> </p> <p><b>XX get</b> </p> <p> </p> <p><b>9=− <img width="X" XXXXXX="X" XXX="XXXX:///X:/Users/jp%XXXXXXXXXX/XXXXXXX/XXXXX/Packages/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/AC/Temp/msohtmlclip1/01/clip_image005.jpg" v:shapes="_x0000_i1028">×X+b</b> <b><XX> </b> </p> <table> <tbody><tr> <td> <p><b>and y =-XXXXXXX these XXX XXX XXXXXXXXXX XX point X</b> </p></td></tr> </tbody></table> <p><b>b=XX.X</b> </p> <p> </p> <p><b>now XXXXXXXX XX</b> </p> <p> </p> <p> </p> <p> </p> <p><b>Y= −<img width="9" height="X" XXX="XXXX:///X:/XXXXX/jp%XXXXXXXXXX/AppData/Local/XXXXXXXX/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/AC/XXXX/msohtmlclip1/XX/clip_image005.jpg" v:shapes="_x0000_i1029"> + XX.5 XX</b> </p> <p> </p> <p> </p> <p> </p> <ul><li><b>XXXXX XXXX XX is XXXXXXXXXXXXX XX AB and the equation XX XXXX BC XX y=-3x+XX, XXXX the co-ordinates of</b> </li></XX> <p><b>(3ks)</b> </p> <p> </p> <p><b>XXX’s XXXX XXX XXXXXXXX XX XXXX XX which is XXXXXXXXXXXXX XX XXXX AB having XXXXXXXX which is y=XX-1 perpendicular line is one that XXX a XXXXXXXX, reciprocal slope XX XXXXXXX.</b> </p> <p> </p> <p><b>So AC XXXX XXXX equation y=-X/2x-X</b> </p> <p> </p> <p> </p> <p> </p> <p><b>XX y XXXX be same for XXX XXXXX c XXXXXX XX XXX equation of BC or equation XX so XX can put XXXX XXXXX</b> </p> <p> </p> <p><b>-1/2x-1=-XX+45</b> </p> <p> </p> <p><b>2.5x=XX</b> </p> <p> </p> <p><b>X=92/5</b> </p>">
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