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Co-ordinate Geometry-Please Help

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by bblast » Thu Aug 11, 2011 9:52 am
Touseef wrote:A triangle has vertices A(4,1), B(-2,3) and C(2,-3).
What is the equation of the altitude through A i.e. the line which passes through A and is perpendicular to BC?
Draw the triangle in space :
1>find out the slope of the line BC. (y2-y1/x2-x1)
2>take negative reciprocal of this slope.
3>make an equation -> y = mx + c. where m is the -ve reciprocal obtained from 2.
4>Put the x and y co-ordinates of A in above equation to obtain value of C.
5>Then rewrite the equation in the form y = mx + c, with the m and C we got from 2 and 4.
Cheers !!

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by lazymonster » Sat Aug 13, 2011 10:17 pm
as the line is perpendicular to BC product of slopes = -1.slope of BC=(Y2-Y1)/(X2-X1)=-3/2.so slope of AD=2/3.substitute in y=mx+c.as AD passes through A(4,1) 1=2/3x+c ,c=-5/3 substitute in line eqn which yields 2x-3y-5=0
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