Question

How do I find the equation of an straight line given two points of another line that is at a perpendicular distance of 4. Points of known line are: (2,3) and (4,7)

Answer 1

Ok. From the diagram, we can see that the 2 lines are parallel, right?

Answer 2

What is the equation of the line that passes through the 2 points?

Answer 3

Thats what I understand from my question

Answer 4

I'll work it out now

Answer 5

I believe the equation is: y=2x-1

Answer 6

Good.

Answer 7

Because the slopes of 2 parallel line are same, the other line can be written as y=2x+b, right?

Answer 8

Yep

Answer 9

Now, because the distance between 2 lines are always 4, if you pick a random point from either line and use point-line distance formula, you'll get 4.

Answer 10

I'm not fully aware what the point-line distance formula is.. sorry :/

Answer 11

It's fine. Then we'll go traditional, first, then I'll teach you the shorcut.

Answer 12

First, let's choose a very very easy point on the 1st line since we have to find b.
How does (0,b) sound to you?

Answer 13

Yep sounds good

Answer 14

Ok. Now, let's say that a point on the second line is (a,2a-1).

Answer 15

And also say that a line (0,b) and (a,2a-1) is perpendicular to y=2x-1

Answer 16

Ok

Answer 17

So the slope of (0,b) and (a,2a-1) is -1/2, right?

Answer 18

Also, the distance between the 2 points is 4, right?

Answer 19

Sorry I'm a little confused as to where you got that second set of points from

Answer 20

especially (a,2a-1)

Answer 21

Ok

Answer 22

Ok so (0,b) is y-intercept of perpendicular line and (a,2a-1) is a random intersection point of perpendicular line to known line?

Answer 23

Exactly!

Answer 24

Ok I get where you got those values now

Answer 25

Now, if I put the 2 sentences I said above into equations....\[\frac{ (2a-1)-b }{ a-0 }=\frac{ -1 }{ 2 }\]

Answer 26

and

Answer 27

\[\sqrt{(a-0)^2+((2a-1)-b)^2}=4\]

Answer 28

or

Answer 29

\[(a-0)^2+((2a-1)-b)^2=16\]

Answer 30

@thesecret20111
Do you know how to solve the system of these 2 equations?

Answer 31

Do you mean solve them like simultaneous equations?

Answer 32

yeah.

Answer 33

I'd need to use elimination right?

Answer 34

Yeah.
Make sure to eliminate a since we're looking for b

Answer 35

it can be short if you know either `cross product` or `distance between point/line formula`,
but working it this way is a good practice for ur coordinate geometry :)

and also notice that there will be two lines that are equidistant from a given line in 2d

Answer 36

I'm planning to teach thesecret20111 the distance between point/line formula after he gets the basic idea.

Answer 37

Sorry its been a while since I did simultaneous equations.. Do I multiply both equations by (a-0)?

Answer 38

For the "slope equation" solve for a in terms of b and substitute it in the "distance equation:

Answer 39

correction: "distance equation"

Answer 40

I'm having trouble solving for a in the slope equation I can't see a way to get 'a' by itself

Answer 41

What did you do so far?

Answer 42

Can you show me?

Answer 43

Its pretty bad its a=((-1/2)+(b/a)/2) +1

Answer 44

Ok. Let's start over.
I'll show all step for isolating a

Answer 45

cross multiply, first.

Answer 46

\[-a=2(2a-1-b)\]
\[-a=4a-2b-2\]
\[5a=2b+2\]
\[a=\frac{ 2 }{ 5 }b+\frac{ 2 }{ 5 }\]

Answer 47

Ok I get it so far

Answer 48

Now substitute that in the "distance equation" to get b
Remember that b has to be positive.

Answer 49

Ok I'll give it a go. BTW I had forgotten about cross multiplying before :/

Answer 50

Alright.

Answer 51

(2/5b+2/5)^2 +((2(2/5b+2/5) -1)-b)^2 = 16

Answer 52

Right.
Keep going.

Answer 53

Tip: Don't foil it right away.

Answer 54

from the first term, factor out (2/5)^2 and from the second term, factor out (-1/5)^2

Answer 55

Do you mean make the first term 2/5^2(b+1)?

Answer 56

\[(\frac{ 2 }{ 5 }b+\frac{ 2 }{ 5 })^2+(-\frac{ 1 }{ 5 }b-\frac{ 1 }{ 5 })^2=(\frac{ 2 }{ 5 }(b+1))^2+(-\frac{ 1 }{ 5 }(b+1)^2=16\]

Answer 57

Correction: \[(-\frac{ 1 }{ 5 }(b+1))^2\]