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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:XXXXXX="XXXXXXXXXXXX"> </p> <p> </p> <p><b>m=X</b> </p> <p> </p> <p><b>XXXXXXXX in general is y=XX+b where b XX XXXXXXXXX</b> </p> <p> </p> <p><b>For XXXXXXXXXXX it XXX XX use point B(6,11)</b> </p> <p> </p> <p><b>So we XXXX XXXX y=11 and x=6 XXXX put in XXX equation</b> </p> <p> </p> <p><b>Y=XX+b</b> </p> <p> </p> <p><b>11=2×X+b</b> </p> <p> </p> <p><b>b=-X</b> </p> <p> </p> <p> </p> <p><b>so XXXXXXXX XX</b> <b>= 2 − 1</b> </p> <p> </p> <ul><li><b>XXXX the XXXXXXXX of the XXXXXXXXXXXXX XXXXXXXX XX line AB (XXXX)</b> </li></XX> <p> </p> <p> </p> <p><b>For doing so first XXXXXXXXX slope of the perpendicular XXXXXXXX</b> </p> <p> </p> <table> <tbody><tr> <td> <p><b>Midpoint XX AB are (</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>Slope of XXX XX XXX XXXXXXXXX from above points as</b> </p></td><td> <p> </p></td><td> <p><b>= = X</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><img width="9" height="X" src="XXXX:///X:/XXXXX/XX%XXXXXXXXXX/XXXXXXX/XXXXX/XXXXXXXX/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/AC/XXXX/XXXXXXXXXXXX/01/XXXXXXXXXXXXX.XXX" v:XXXXXX="XXXXXXXXXXXX"> </p> <p> </p> <p><b>and that XX the Perpendicular XXXXXXXX XX−<img width="XX" height="4" src="XXXX:///X:/Users/XX%20computer/XXXXXXX/Local/Packages/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/XX/XXXX/XXXXXXXXXXXX/XX/clip_image007.XXX" v:XXXXXX="XXXXXXXXXXXX"> of AB so will XX</b> <b>-<img width="X" XXXXXX="4" src="file:///C:/Users/XX%20computer/AppData/XXXXX/XXXXXXXX/XXXXXXXXXXXXXXXXXXXXXXXXXXXXX/AC/Temp/msohtmlclip1/01/XXXXXXXXXXXXX.XXX" v:shapes="_x0000_i1027"></b> </p> <p> </p> <p> </p> <p> </p> <p><b>XXX we XXX XXXXXXXXX XXX intercept using (X, 9) and m XX -1/X</b> </p> <p> </p> <p> </p> <p> </p> <p><b>we XXX</b> </p> <p> </p> <p><b>9=− <img width="9" height="X" XXX="file:///C:/Users/XX%XXXXXXXXXX/AppData/XXXXX/Packages/oice_16_974fa576_32c1d314_f69/XX/XXXX/msohtmlclip1/01/XXXXXXXXXXXXX.XXX" v:shapes="_x0000_i1028">×5+b</b> <b><XX> </b> </p> <table> <tbody><tr> <td> <p><b>and y =-12hence these XXX XXX coordinate XX point C</b> </p></td></tr> </tbody></table> <p><b>b=11.X</b> </p> <p> </p> <p><b>now equation XX</b> </p> <p> </p> <p> </p> <p> </p> <p><b>X= −<img width="X" XXXXXX="X" src="file:///X:/Users/XX%XXXXXXXXXX/AppData/XXXXX/XXXXXXXX/oice_16_974fa576_32c1d314_f69/XX/XXXX/msohtmlclip1/01/XXXXXXXXXXXXX.jpg" v:XXXXXX="_x0000_i1029"> + XX.5 is</b> </p> <p> </p> <p> </p> <p> </p> <ul><li><b>XXXXX XXXX AC is XXXXXXXXXXXXX to XX and the equation of XXXX BC is y=-3x+XX, find XXX co-ordinates of</b> </li></ul> <p><b>(XXX)</b> </p> <p> </p> <p><b>Let’s XXXX XXX equation of XXXX XX XXXXX is XXXXXXXXXXXXX XX XXXX AB having equation XXXXX is y=2x-X XXXXXXXXXXXXX line XX one XXXX XXX a negative, reciprocal XXXXX to another.</b> </p> <p> </p> <p><b>So AC will have XXXXXXXX y=-1/2x-1</b> </p> <p> </p> <p> </p> <p> </p> <p><b>XX y will XX XXXX for the point c either we XXX XXXXXXXX of BC or XXXXXXXX XX so XX can XXX XXXX XXXXX</b> </p> <p> </p> <p><b>-X/2x-1=-XX+45</b> </p> <p> </p> <p><b>X.5x=46</b> </p> <p> </p> <p><b>X=92/5</b> </p>">
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