http://sgforums.com/forums/2297/topics/316599 en-US 60 <p>length BC = sqrt (7^2 + 6^2) = sqrt(85)</p> <p>0.5 * length BC * perpendicular distance from A to BC = area of triangle ABC</p> <p>0.5 * sqrt(85) * perpendicular distance from A to BC = 9 (from part c)</p> <p>perpendicular distance from A to BC = 1.95 units</p> Mon, 05 May 2008 21:57:21 +0800 sgforums.com:2297:316599:8064168 eagle http://sgforums.com/forums/2297/topics/316599 <p>(a) AB : x = -4</p> <p>BC : y = 7/6 x + 8/3</p> <p>CA : y = 2/3x +11/3</p> <p>&nbsp;</p> <p>b. (0, 11/3)</p> <p>&nbsp;</p> <p>(c) 9 units 2</p> <p>(d) 1.95 units</p> Mon, 05 May 2008 21:54:22 +0800 sgforums.com:2297:316599:8064159 bonkysleuth http://sgforums.com/forums/2297/topics/316599 <p>pls give the answers for the first few parts so that there's no necessity to do extra work. If possible, provide the answer for part d too. Thanks.</p> Mon, 05 May 2008 21:40:50 +0800 sgforums.com:2297:316599:8064109 eagle http://sgforums.com/forums/2297/topics/316599 <p>----------</p> Mon, 05 May 2008 21:26:58 +0800 sgforums.com:2297:316599:8064044 FireIce http://sgforums.com/forums/2297/topics/316599 <p>the vertice of the triangleABC are A(-4,1), B(-4,-2) and C(2,5). find</p> <p>&nbsp;</p> <p>just answer question d. the earlier questions may well serve you as a reference .</p> <p>&nbsp;</p> <p>(a) the equations of the lines AB, BC, CA</p> <p>(b) coordinates of the point where AC cuts tge y0axis</p> <p>(c) area of triangle ABC</p> <p><strong>(d) perpendicular distance from A to BC</strong></p> Mon, 05 May 2008 21:22:26 +0800 sgforums.com:2297:316599:8064027 bonkysleuth http://sgforums.com/forums/2297/topics/316599