Help with task

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swag312

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Hey, not sure how to translate this from my native language, I hope you understand what I mean.
Write down for the line
y = 2x + 3 perpendicular and parallel lines passing through the point
(1; 1) equations.
 
I've moved this thread to a more appropriate forum.

1.) What will be the slope of a line parallel to the given line? What will be the slope of a line perpendicular to the given line?

2.) Knowing the slope of a line and a point on that line, what formula for a ine would be appropriate to get the equation of that line?
 
I'm not sure but the answer section requires 2 things:
perpendicular line: y =
parallel line: y =
I'm not sure how the problem translates to english since I can't really understand it in my own language, sadly.
 
Yes, as I understand it, you are to find a line parallel to the given line and a line perpendicular to the given line, where both lines pass though the given point.

As you have been asked to answer this question, you should've been taught that two lines having the same slope are parallel and two lines whose slopes have a product of -1 are perpendicular. Does this sound familiar?
 
It does sound familiar, but I'm still kind of lost. I'm pretty sure that I just have to somehow put everything into correct formula but I am unable to find the right path in my studying material..
 
No, it is not correct.

Let's work with the parallel line 1st: The slope of the given line y=2x+3 is 2.
OK, so now we know the slope of the line that is parallel to y=2x+3 is 3 (So m=3) and it passes through the point (1,1).
The equation of this line (in fact any line) can be written as y=mx+b.
You need to find m and b. m shouldn't be hard to find as we already found it. So try to figure out what b must be. Please post back.

We will tackle the 2nd part (perpendicular) after this problem.
 
swag312 said:
I guess b is = 3 then
Click to expand...
How did you arrive at b=3? To get good help you must show us your work so we see where you are making any mistakes.
The original equation was y=2x+b----so b=3 in this equation.
Sure a line is parallel to itself but lets see if it contains the point (1,1). 1=2*1 +3, ie 1=5. This is not true, so the point (1,1) is NOT on the line y=2x+3
 
For any line in the form:

[MATH]y = mx + b[/MATH]​

... the value of [MATH]m[/MATH] describes how far the line travels "up" for each 1 that the line travels "right". This is known as the slope. If any two lines have the same slope [MATH]m[/MATH], they will be parallel, meaning they differ only by [MATH]b[/MATH].

As @MarkFL pointed out, two lines have perpendicular slopes if the product of their slopes is -1. In other words, you can calculate a perpendicular slope as [MATH]-\frac{1}{m}[/MATH] (as long as [MATH]m[/MATH] is not zero).

That [MATH]b[/MATH] value describes where the line crosses the Y axis, the point where X is zero. They call this the y-intercept. A feature of this kind of linear equation is that shifting a line up and down can produce all of the lines that you could get by shifting a line left and right. After all, if you shift a line to the right, its y-intercept changes accordingly.

The original problem asks two things:
  1. Find a line that changes the [MATH]b[/MATH] of the given line so that it passes through the point (1, 1)
  2. Find a line that changes both the [MATH]m[/MATH] and the [MATH]b[/MATH] to produce a perpendicular line that passes through the point (1, 1)
This is simpler than it might seem. The line in your example is [MATH]y = 2x + 3[/MATH], meaning its position at point (1, [MATH]y[/MATH]) is given by [MATH]y = 2(1) + 3 = 5[/MATH]. What needs to happen in order for that to cross (1, 1) instead?
 
In informal terms, you need to figure out how much you need to change the position of your line to fit the point.
 
Mr. Bland said:
For any line in the form:

[MATH]y = mx + b[/MATH]​

... the value of [MATH]m[/MATH] describes how far the line travels "up" for each 1 that the line travels "right". This is known as the slope. If any two lines have the same slope [MATH]m[/MATH], they will be parallel, meaning they differ only by [MATH]b[/MATH].

As @MarkFL pointed out, two lines have perpendicular slopes if the product of their slopes is -1. In other words, you can calculate a perpendicular slope as [MATH]-\frac{1}{m}[/MATH] (as long as [MATH]m[/MATH] is not zero).

That [MATH]b[/MATH] value describes where the line crosses the Y axis, the point where X is zero. They call this the y-intercept. A feature of this kind of linear equation is that shifting a line up and down can produce all of the lines that you could get by shifting a line left and right. After all, if you shift a line to the right, its y-intercept changes accordingly.

The original problem asks two things:
  1. Find a line that changes the [MATH]b[/MATH] of the given line so that it passes through the point (1, 1)
  2. Find a line that changes both the [MATH]m[/MATH] and the [MATH]b[/MATH] to produce a perpendicular line that passes through the point (1, 1)
This is simpler than it might seem. The line in your example is [MATH]y = 2x + 3[/MATH], meaning its position at point (1, [MATH]y[/MATH]) is given by [MATH]y = 2(1) + 3 = 5[/MATH]. What needs to happen in order for that to cross (1, 1) instead?
Click to expand...
Mr Bland, I disagree with one part of your post. I do not think that two parallel lines need to differ in b. I am a believer that the values of b can be the same and that the TWO lines are parallel. I guess that I am saying that given two lines that are the same I believe that you have two lines, not one. Obviously I then believe that two lines can intersect and when they do they intersect everywhere on the line.
 
swag312 said:
I understand this, but where do I go from there ?
Click to expand...

Are you familiar with the point-slope formula?

[MATH]y=m(x-x_1)+y_1[/MATH] ?

This is useful for when you know the slope of a line and a point on the line...
 
swag312 said:
I understand this, but where do I go from there ?
Click to expand...
I will give you this one but you have to try the perpendicular problem yourself and show us your work so we can see if it is correct.

The slope of the original line is m=2. Since you want the equation of a parallel line that passes through (1,1) the slope will be 2 as well.
So we can update y=mx+b with y=2x+b. Since (x=1, y=1) is a point on this new line we must have 1=2*1=b. That is 1=2+b. So b=-1. The equation of the line is y=2x-1.
 
swag312 said:
I understand this, but where do I go from there ?
Click to expand...
I may have buried it in my post a bit, but it's this part:
  1. Find a line that changes the [MATH]b[/MATH] of the given line so that it passes through the point (1, 1)
  2. Find a line that changes both the [MATH]m[/MATH] and the [MATH]b[/MATH] to produce a perpendicular line that passes through the point (1, 1)
After you find the slope of the perpendicular line, then as @firemath stated, you need only adjust the positions of the lines to ensure they pass through the point (1, 1).

The line in your example is [MATH]y = 2x + 3[/MATH]. For any value of X, you can calculate the corresponding value of Y. A point is written in the form [MATH](x, y)[/MATH], and you know that X should be 1. Therefore, you can calculate what Y currently is when X=1 and adjust the y-intercept accordingly.

Jomo said:
I do not think that two parallel lines need to differ in b.
Click to expand...
I've always conceptualized coinciding objects to be the same, regardless of how many times they're expressed. To use a simpler example, if [MATH]a = 5[/MATH] and [MATH]b = 5[/MATH], I don't think of that as two distinct yet equivalent values--I see them as the same value multiple times. I apply the same philosophy to more complicated functions.
 
Mr. Bland said:
For any line in the form:

[MATH]y = mx + b[/MATH]​

... the value of [MATH]m[/MATH] describes how far the line travels "up" for each 1 that the line travels "right". This is known as the slope. If any two lines have the same slope [MATH]m[/MATH], they will be parallel, meaning they differ only by [MATH]b[/MATH].

As @MarkFL pointed out, two lines have perpendicular slopes if the product of their slopes is -1. In other words, you can calculate a perpendicular slope as [MATH]-\frac{1}{m}[/MATH] (as long as [MATH]m[/MATH] is not zero).

That [MATH]b[/MATH] value describes where the line crosses the Y axis, the point where X is zero. They call this the y-intercept. A feature of this kind of linear equation is that shifting a line up and down can produce all of the lines that you could get by shifting a line left and right. After all, if you shift a line to the right, its y-intercept changes accordingly.

The original problem asks two things:
  1. Find a line that changes the [MATH]b[/MATH] of the given line so that it passes through the point (1, 1)
  2. Find a line that changes both the [MATH]m[/MATH] and the [MATH]b[/MATH] to produce a perpendicular line that passes through the point (1, 1)
This is simpler than it might seem. The line in your example is [MATH]y = 2x + 3[/MATH], meaning its position at point (1, [MATH]y[/MATH]) is given by [MATH]y = 2(1) + 3 = 5[/MATH]. What needs to happen in order for that to cross (1, 1) instead?
Click to expand...
I don't know, maybe put 1 instead of y as well. I'm kinda lost right not, to be honest..
 
Considering [MATH]y = 2x + 3[/MATH]...

If X is 1, then the function becomes [MATH]y = 2(1) + 3[/MATH]. Therefore, when X is 1, Y is 5. You want to change the function so that when X is 1, Y is also 1.

Let us modify the function slightly, to say [MATH]y = 2x + 4[/MATH] instead. Now, when X is 1, Y becomes 6. Notice that when we increased the y-intercept by 1, the result was also increased by 1. Since you want the result to change from 5 to 1, what must be done to the y-intercept?
 
Yeah I understand it now, y and x has to be 1 and 1, so you have to modify b in order for it to happen which leads to y=2x-1
 
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