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 19-12-2010, 12:06 AM #7 ctbram Moderator     Join Date: Jan 2004 Location: Michigan, USA Posts: 2,994 Thanks: 42 Thanked 582 Times in 532 Posts the next part is also easy. given p3(-500,400) then the shortest distance to L1 is a line that from p3 to L1 is perpendicular to L1 and we will call it L2. If the slope of L1 is m then the slope of L2 (m2) will be -(1/m) or m2 = -(140/285). Now just use point-slope with p3 and m2 to find the equation of L2: L2: y-400 = m2(x + 500) y = -(140/285)(x+500)-400 I'll let you reduce this. Now with the equation of L1 and L2 you can find the x intersection by setting L1 = L2 and solving for x. L1 = L2 (285/140)x-(750/140) = -(140/285)(x+500)-400 solve for x now plug x into either L1 or L2 and get y. P4(x,y) is the perp intersection of L1 and L2. Now use P3 and P4 to compute the shortest distance from P3 to the line L1 where dist = sqrt(dy^2 + dx^2). dist_p3.p4 = sqrt((y4 -y3)^2 + (x4-x3)^2) <===== this is your second answer you just need to solve for p4(x4,y4) using the stuff above. __________________ "If I have seen further it is by standing on the shoulders of giants." Sir Isaac Newton, 1675 Last edited by ctbram : 20-12-2010 at 12:37 PM.