I need help on 2 questions. Don't have much to offer so I'll offer free hosting to someone who can help me out. OK, here goes... 1. From the mid point of BC a perpendicular to AC is drawn, meeting AC at Q. Find the equation of this perpendicular and hence the co-ordinates of Q. 2. Find the equation that goes through the origin (0,0) and in perpendicular to the line 8x + 6y - 50 = 0 Find where the 2 lines intersect and hence find the perpendicular distance of the line 8x + 6y - 50 = 0 from the origin.

thats easy.. you are better off learning it than having someone do your homework.. itll be in your exam and then you are fucked.. i found you help.. http://www.analyzemath.com/line/Tutorials.html read the examples.. its pretty easy.. find the slope and everything else is easy.. one of the examples answers your question.. #2 is even easier.. read the examples

if you can't figure out how to do it mathematically, graph it do it manually. This is pretty basic stuff and without it you won't make it far in math.

Interesting... I remember my old school days... I give u clues: 1. mid point of BC is ((1+7)/2 ,(11+13)/2). Now equation of line passing thru that point will be , in general , (y-12)=m(x-4). Now slope of AC is (13-1)/(7-1)=2 so slope of that line that u want to know is -1/2 as they are perendicular. so m=-1/2. Done part one 2. now line that goes thru (0,0) is always of the form y=mx; as the slope of given line in part 2 is -6/8 so slope of line perpendicular to it is 8/6 so m=4/3. Part 2 done now part 3: solve the equation y=4/3*x and given line, that point is point of intersection (25/8,25/6). and so the distance of 8x+6y-50=0 is distance of that point from origin = sqrt(sqr(25/8)+sqr(25/6)). Now u can solve this, can't u?. DONE No need to thanks. You sud have learned this by urself.

I have taken many math classes and physics and good stuff like that and every now and then I will get stuck(or lazy) and need some help. When that happened, I posted it on yahoo answers and waited a few minutes and just like that the answer was there. Some of the problems were fairly complex and people would still answer them and show the work too. Some would even draw diagrams for me and post them. Actually, a lot of times, this helped me more than you might think. But anyway, if you want to get answers, that is a great spot.

2. Find the equation that goes through the origin (0,0) and in perpendicular to the line 8x + 6y - 50 = 0 Find where the 2 lines intersect and hence find the perpendicular distance of the line 8x + 6y - 50 = 0 from the origin. Solution: Slope of the line (8x + 6y - 50 = 0) = -4/3 Therefore, slope of perpendicular = 3/4 Now, perpendicular passes through (0,0) and have slope 3/4 Equation of perpendicular-> y = (3/4)(x) To find point of intersection, Solve the two equations. Distance of line 8x + 6y - 50 = 0 from the origin = |8x0 + 6x0 -50|/10 = 5 |