We are given y=3-x passes through (1,2) and that line L, y = (2-a) + ax, also does, for any value "a".
We are then told to:
i) find the equation of the line perpendicular to line L which passes through (1,2)
ii) Find equations of lines which pass through the point (1,2) and have perpendicular distance 1 from origin
So i) is not a problem.. y = -1/a(x-1) + 2b
but for ii) i thought of using the equation L to be a tangent to the circle x^2 + y^2 = 1
so substitute y = (2-a) + ax into the equation of the circle, this gives a = 3/4
so the equations of the lines are x=1 and 4y = 3x + 5
Is this correct? Are there any other methods?
Also, do we have to consider the case for the equation we got in i) to be a tangent to the circle? Substituting y = -1/a(x-1) + 2b into the circle gives me a quartic equation, with complicated roots.
1. The given line has the equation that means the slope of the line is . If
Then the line L has the equation
2. If the equation of a line is then the distance of a point P(p, q) to this line is calculated by
With your example: and P(0, 0) you'll get:
. Solve for a.
Spoiler:
Thanks, I got the same answer by substituting y = (2-a) + ax into the equation of the circle with centre origin and unit radius.
This gives a = 3/4, which then by inspection shows that x=1 is also a solution.
However, when using this circle method, don't we also have to consider the other line, which is perendicular to L, is a tangent to the circle instead?
i.e. substituting y = -1/a (x-1) + 2 into the circle?
Hi, I don't think you got my question. So basically, we have 2 lines which are perpendicular to each other and both pass through (1,2) for any value of a:
L : y = (2 - a) + ax
G: y = -1/a (x - 1) + 2
Now the question is the find the equation of the lines which pass through the point (1,2) and have perpendicular distance 1 from the origin.
Now in your solution, you took the perpendicular distance of L from the origin to be 1.
However, do we have to consider the case where we take perpendicular distance to be 1 from G, rather than line L? The part before this question got us to work out G, by saying to work out the equation of the line perpendicular to L passing through (1,2).
Then in this part, we don't use G at all. Shouldn't G have a connection to it? Or don't we at least consider the case when G has perpendicular distance 1 from the origin as well?
Thank you
Friend, sorry for misreading you question the first time.
Now, let’s go back to square one.
When you are given a line equation , to which the perpendicular line equation is .
These two lines intersect at a point at the coordinate (a, b). Knowing and coordinate (a, b), you can plug them into and find .
Then <---This is the answer to you part 1 of your assignment.
Next we will find the equation a distance = 1 unit from the origin (0,0):
Distance equation is given by
Since we know and the origin (0,0), we can plug them into the distance equation so that it becomes:
and obtain
Now we have the line equations perpendicular to L:1 and passing through the point at coordinate (a, b) and 1 unit away from the origin:
<---this is the answer to part 2 of your assignment. One line is below the origin, and the other above the origin.
Note: In your assignment, is given as . The (a,b) is (1,2), and the equation is only a decoy that was used to confuse you. It’s only another line that passes through (1,2) that bears no relation with
No, Aquafina,
I did not use line G in my last post. I used line L. As I have noted that line G is only a decoy. It's only a line that passed (1,2), and it bears no useful relation to line L.
The slope of line L can vary until the line perpendicular to L formed a distance of one unit from the origin. If you draw a circle of 1 unit radius about the (0,0), you can draw two lines tangent to the circle and cross (1,2). Both tangent lines have different slopes, and both lines are perpendicular to line L. In other words, there are two line L's--one provides a perpendicular line tangent and intersects the circle on the upper side of (0,0); the other line L provides a perpendicular line tangent and intersects the circle on the lower side of (0,0).
You might wonder how could there be two line L's. The answere is "As stated in you assignment, 'a' is any number." The constant "a" is a variable to provide a changable slope to make the intersection of the circle possible.
Remark: Your teach is excellent. I have never seen such tough algebra teaching while growing up.
Please try as I suggested first, and I will catch up with you later with a complete solution.
Aquafina,
I am giving the solution in each step and explanations, but you must do your own algebra to save me time.
Solution:
We were given , which we have no use, so we will ignore it.
Let denotes line , and denotes the perpendicular line to it.
--------------------------------eq (1)
-------------------eq (2)
We can rearrange eq (2) to one side:
and using it with the known point (0,0), we plug all into the distance equation .
We get , which gives us that we can plug back into eq (2) and turn it into as follows:
---------eq (3)
Now to solve for the variable in the line equations, we can plug in the coordinate (1,2), where both lines intersect, into eq (3).
So eq (3) at (1,2) becomes which gives us ---------------------eq (4).
We solve for a in eq (4), and the results are and , which tells us that there 2 sets of and
Now plug to one set, and to the other.
There results should be answers to all parts 1 and 2 of your assignment:
For : , and .
For : , and
Will one of you explain to me why you all seem to think that part ii) has anything to do with line .
ii) Find equations of lines which pass through the point (1,2) and have perpendicular distance 1 from origin.
Where does it say or even imply that line is part of the question?
Well, Plato, my guru,
Your point is well taken. I cannot disagree with you. Now, for our frined's, Aqufina's, sake, would you please show him the short cuts, which I would much appreciate to learn from you since I still have many years ahead of me in catching up with you.
Aquafina, I am presenting you our guru, Professor Plato. Well come professor.