My Math Forum Find points on a line of equal distance from certain point

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 December 17th, 2009, 12:56 AM #1 Newbie   Joined: Dec 2009 Posts: 3 Thanks: 0 Find points on a line of equal distance from certain point Hi Gents, We have line L1: 3 X + 4 Y + 12 = 0 I got the X-intercept (M) (-4,0) Y-intercept (A) (0,-3) Distance (d) between A and M (d=4) Then I got line, L2, which is perpendiclar to L1 through M L2: 3 Y = 4 X +16 Anyone can confirm answers? THEN I STUCK ON: - Finding 2 points B & C on L2 whose distance from A = d* sq. root of 2 (d ?2) - Finding a point D that makes ABCD a square Thanks alot
 December 17th, 2009, 08:23 AM #2 Senior Member   Joined: Dec 2009 From: Las Vegas Posts: 209 Thanks: 0 Re: Find points on a line of equal distance from certain point Hi horees; A) is correct M) is correct d is not correct, the distance between A and M is 5 L2 is correct. The 2 points you want that are 5 sqrt(2) away from (0,-3) and on L2 are (-7,-4) and (-1,4)
 December 18th, 2009, 10:39 AM #3 Newbie   Joined: Dec 2009 Posts: 3 Thanks: 0 Re: Find points on a line of equal distance from certain point how u got (-7,-4) & (-1,4)???
 December 18th, 2009, 01:22 PM #4 Senior Member   Joined: Dec 2009 From: Las Vegas Posts: 209 Thanks: 0 Re: Find points on a line of equal distance from certain point Hi horees; I would be glad to. Here is how I did it. It is a little long and tedious. I will provide the overview and you can ask me for a further explanation if you want it. Here is how I got the distance from A to M. $\sqrt{(4-(-3))^2+(-1-0)^2}= 5 \sqrt{2}$ This is how I found the 2 points that are on L2 and that are 5 sqrt(2) away from (0,-3) $\sqrt{\left(-3-\left(\frac{4}{3}x+\frac{16}{3}\right)\right)^2+(0-x)^2}=5\sqrt{2}$ By squaring both sides the above equation becomes a quadratic with x = - 1 and x = - 7. Substituting for both x's into L2: $y=\frac{4}{3}x+\frac{16}{3} \text{ with }x = -7 \ \ \ y= -4$ $y=\frac{4}{3}x+\frac{16}{3} \text{ with }x = -1 \ \ \ y= 4$ So the points are (-7,-4) and (-1, 4)

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