completed equation now what

Kiran · 2006-11-24 06:20:54

find the distance between the point (5,2) and the line 3x - 4y = 2

y = 3x + b
2 = 3(5) + b
2 = 15 + b
b = -13
y = 3x - 13 < completed equation

now what?

John E. Franklin · 2006-11-24 06:42:32

Well, the first mistake I see is that 3x is not right, it's 3/4 of x and the negative reciprocal is -4/3 of x for the perpendicular line.
First I would put the equation of the line given in y=mx+b form. Though there are other ways I bet.
So
-4y = -3x + 2
4y = 3x - 2
y = (3/4)x - (1/4)2
y = (3/4)x - 1/2
Now we need the equation for a line that is perpendicular to this line.
See next post...

John E. Franklin · 2006-11-24 06:46:46

If first line is y = (3/4)x - 1/2, then the
perpendicular line to that line is slope of minus 4/3.
So the equation of that line is y = (-4/3)x + b.
To find b, we sub in the points given above.
(5,2) is the point of interest.
So 2 = (-4/3)5 + b
2 = (-20/3) + b
2 + 20/3 = b
26/3 = b
I'm not checking my answer yet, so it could be wrong.
So y = (-4/3)x + 26/3
Now you can find the intersection of the two lines so you get the other point of interest, which is how far away from the original point, we will find out soon!
...

John E. Franklin · 2006-11-24 06:54:21

So to find this point on the original line, solve the system of two lines.
Write down both lines and solve them together.
This will yield the other point you want.
Eq#1: y = (-4/3)x + 26/3
Eq#2: y = (3/4)x - 1/2
Subtract these equations: Eq#1 - Eq#2.
y - y = (-4/3)x - (3/4)x + 26/3 + 1/2
0 = Do it with 12 in the denominator.
I gotta run and do some errands. Sorry.
When you get the other point.
Then use pythagorean theorm to get the hypotenuse, which is the distance between the points.

Kiran · 2006-11-24 07:22:01

John E. Franklin wrote:

If first line is y = (3/4)x - 1/2, then the
perpendicular line to that line is slope of minus 4/3.
So the equation of that line is y = (-4/3)x + b.
To find b, we sub in the points given above.
(5,2) is the point of interest.
So 2 = (-4/3)5 + b
2 = (-20/3) + b
2 + 20/3 = b
26/3 = b
I'm not checking my answer yet, so it could be wrong.
So y = (-4/3)x + 26/3
Now you can find the intersection of the two lines so you get the other point of interest, which is how far away from the original point, we will find out soon!
...

(-4/3 )5 = -6 2/3 plus 2 which makes it -4 2/3
so b = -4 2/3 or we can say -10/3

Kiran · 2006-11-24 07:36:02

ok now i got....
y - y = (-4/3)x - (3/4)x + (-10/3) + 1/2
y - y = (-2 1/12)x + (-2 5/6)
nowwwwwwwwwwww

John E. Franklin · 2006-11-27 03:35:09

I'll check your work Soha...
Eq#1: y = (-4/3)x + 26/3
Eq#2: y = (3/4)x - 1/2
Subtract these equations: Eq#1 - Eq#2.
y - y = (-4/3)x - (3/4)x + 26/3 + 1/2
( 0 = Do it with 12 in the denominator. )

0 = (-16/12)x - (9/12)x + 26/3 + 1/2

0 = (-25/12)x + 52/6 + 3/6

(25/12)x = 55/6

(25/12)x = 110/12

25x = 110

x = 440/100

x = 4.4

Now we have the x coordinate where both lines cross.

...

John E. Franklin · 2006-11-27 03:48:21

Now plug the 4.4 into x in both equations for a double check.
Should be the same for each one.
Each equation is a line; they are perpendicular and intersect.
Eq#1: y = (-4/3)x + 26/3
Eq#2: y = (3/4)x - 1/2
Eq# 1 Eq#2
y = (-4/3)4.4 + 26/3 y=(3/4)4.4 - 1/2

y = 14/5 y = 2.8

Both get same y.

SO THE POINT that is the closest on the line given is (4.4, 2.8), if I did it right.

NOW FIND THE distance between (5,2) originally given and (4.4, 2.8).

Use pythagoreans theorm. Draw a triangle. a squared plus b squared is c squared.

We want to know c squared, the slanted length.

a = x distance apart = 5 - 4.4, which is 0.6

b = y distance apart = 2.8 - 2, order of numbers doesn't matter here, it's a distance, not a vector direction.
So b = 2.8 - 2, which is 0.8

This is a classic 3,4, 5 triangle because 0.6 is double 3, and 0.8 is double 4, so the hypotenuse is
1.0, which is double 5.

So the answer is 1.0

3 squared plus 4 squared is 5 squared.
9 + 16 is 25. Take the square root and you get 5.

John E. Franklin · 2006-11-27 03:49:45

Or if you didn't see the 3-4-5 stuff, then just do this:

0.6 squared plus 0.8 squared is .36 + .64, which is one.

Take the square root of one, and you still have one, which is the answer to the problem how far is the point from the line!!!!

John E. Franklin · 2006-11-27 03:51:18

And the is really = (

mathsyperson · 2006-11-27 05:08:07

Options --> 'Never show smilies as icons for this post' --> *Tick* = =(

numen · 2006-11-27 13:41:37

I got confused reading John's posts (sorry ;P) so I did it by using vectors to check his result and got the exact answer

Which is approximately 1.016, but 1 might be good enough for some people This is what I did, maybe someone could check it (I usually don't do this in 2D, and it's really late ).

-----------------

We may easily find two points on the line 3x-4y=2 (meaning: find two solutions to the equation). Just pick a value for x and y so it "fits". Two solutions are Q = (2,1) and P = (6,4), check these. These points are on the line, which we will call L. The direction of this vector can be found through QP = (6-2,4-1) = (4,3). Thus, we can write L in the parametric form

x=2+4t
y=1+3t

We have the point S = (5,2), the distance to L is given by |SR|, where R is a point on L. SR is given by SR=(2+4t,1+3t)-(5,2). Do you see why? This gives SR=(-3+4t,-1+3t). Also note that SR is perpendicular to QP.
This means that 4(2+4t)+3(1+3t) has to be 0. This gives us t=7/11. Let us insert this value for t into SR. We get SR=1/11(-5,10)
The distance is the absolute value for SR. Let us find out what it is!

|SR|=1/11*√(5²+10²) = 1/11*√125

You can apply this technique in 3D systems as well (with x,y,z). It's pretty good to know

Math Is Fun Forum

#1 2006-11-24 06:20:54

completed equation now what

#2 2006-11-24 06:42:32

Re: completed equation now what

#3 2006-11-24 06:46:46

Re: completed equation now what

#4 2006-11-24 06:54:21

Re: completed equation now what

#5 2006-11-24 07:22:01

Re: completed equation now what

#6 2006-11-24 07:36:02

Re: completed equation now what

#7 2006-11-27 03:35:09

Re: completed equation now what

#8 2006-11-27 03:48:21

Re: completed equation now what

#9 2006-11-27 03:49:45

Re: completed equation now what

#10 2006-11-27 03:51:18

Re: completed equation now what

#11 2006-11-27 05:08:07

Re: completed equation now what

#12 2006-11-27 13:41:37

Re: completed equation now what

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