# Math Help - Equation of line

1. ## Equation of line

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.

2. Originally Posted by Aquafina
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 I got from i) to be a tangent to the circle x^2 + y^2 = 1

so substitute y = y = -1/a(x-1) + 2b into the equation of the circle.

Is this correct? Are there any other methods?
It's algebra. Line equation y=mx+c, where m is slope, c is the intersection of line with y-axis.

The perpendicular line to $y_1=mx+c_1$ is the line $y_2=\frac{-1}{m}x+c_2$

The perpendicular line to $y_1=-1x+3$ is

$y=ax+(2-a) = x+c_2$

Since $a=1, c_2=1$, which gives $y=x+1$

3. Originally Posted by Aquafina
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 I got from i) to be a tangent to the circle x^2 + y^2 = 1

so substitute y = y = -1/a(x-1) + 2b into the equation of the circle.

Is this correct? Are there any other methods?
1. The given line has the equation $G: y = -x +3$ that means the slope of the line is $m_G = -1$. If $G \perp L~\implies~(-1) \cdot a = -1 ~\implies~ a = 1$

Then the line L has the equation $L: y = x+1$

2. If the equation of a line is $Ax + By + C = 0$ then the distance of a point P(p, q) to this line is calculated by
$d = \dfrac{Ap + Bq +C}{\sqrt{A^2+B^2}}$

With your example: $L: ax - y + (2-a)=0$ and P(0, 0) you'll get:

$d = 1 = \left| \dfrac{2-a}{\sqrt{a^2+1}} \right|$. Solve for a.

Spoiler:
I've got 2 lines: $L_1: x = 1$ and $L_2: y = \dfrac34 x+\dfrac54$

4. Originally Posted by earboth
1. The given line has the equation $G: y = -x +3$ that means the slope of the line is $m_G = -1$. If $G \perp L~\implies~(-1) \cdot a = -1 ~\implies~ a = 1$

Then the line L has the equation $L: y = x+1$

2. If the equation of a line is $Ax + By + C = 0$ then the distance of a point P(p, q) to this line is calculated by
$d = \dfrac{Ap + Bq +C}{\sqrt{A^2+B^2}}$

With your example: $L: ax - y + (2-a)=0$ and P(0, 0) you'll get:

$d = 1 = \left| \dfrac{2-a}{\sqrt{a^2+1}} \right|$. Solve for a.

Spoiler:
I've got 2 lines: $L_1: x = 1$ and $L_2: y = \dfrac34 x+\dfrac54$
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?

5. Originally Posted by Aquafina
...

However, when using this circle method, don't we also have to consider the other line, <<<<<<< since the perpendicular line corresponds with the given line you can determine the slope of the perpendicular line which novice had already done. (to a given line there is only one perpendicular direction in R²)
which is perendicular to L, is a tangent to the circle instead?

i.e. substituting y = -1/a (x-1) + 2 into the circle?
...

6. Originally Posted by earboth
1. The given line has the equation $G: y = -x +3$ that means the slope of the line is $m_G = -1$. If $G \perp L~\implies~(-1) \cdot a = -1 ~\implies~ a = 1$

Then the line L has the equation $L: y = x+1$

2. If the equation of a line is $Ax + By + C = 0$ then the distance of a point P(p, q) to this line is calculated by
$d = \dfrac{Ap + Bq +C}{\sqrt{A^2+B^2}}$

With your example: $L: ax - y + (2-a)=0$ and P(0, 0) you'll get:

$d = 1 = \left| \dfrac{2-a}{\sqrt{a^2+1}} \right|$. Solve for a.

Spoiler:
I've got 2 lines: $L_1: x = 1$ and $L_2: y = \dfrac34 x+\dfrac54$
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

7. Originally Posted by novice
It's algebra. Line equation y=mx+c, where m is slope, c is the intersection of line with y-axis.

The perpendicular line to $y_1=mx+c_1$ is the line $y_2=\frac{-1}{m}x+c_2$

The perpendicular line to $y_1=-1x+3$ is

$y=ax+(2-a) = x+c_2$

Since $a=1, c_2=1$, which gives $y=x+1$
Friend, sorry for misreading you question the first time.

Now, let’s go back to square one.

When you are given a line equation $L_1: y = m_1 x + c_1$, to which the perpendicular line equation is $L_2: y = \frac{-1}{m_1} x + c_2$.
These two lines intersect at a point at the coordinate (a, b). Knowing $m_1$ and coordinate (a, b), you can plug them into $L_2$ and find $c_2$.

$L_2: b = \frac{-1}{m_1}(a)+c_2 \rightarrow c_2 = b + \frac{a}{m_1}$

Then $L_2: y = \frac{-1}{m_1} x + ( b + \frac{a}{m_1})$ <---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 $d= \frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$

Since we know $d =1, A = \frac{-1}{m_1}, B = 1$ and the origin (0,0), we can plug them into the distance equation so that it becomes:

$\pm 1= \frac{|A(0)+B(o)+C|}{\sqrt{(\frac{-1}{m_1})^2+1}}$ and obtain

$C= \pm \sqrt{(\frac{-1}{m_1})^2+1}$

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:

$y = \frac{-1}{m_2} x \pm \sqrt{(\frac{-1}{m_1})^2+1}$<---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, $L:1$ is given as $y=ax + (2-1)$. The (a,b) is (1,2), and the equation $y = -x+3$ 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 $L_1$

8. Originally Posted by novice
Friend, sorry for misreading you question the first time.

Now, let’s go back to square one.

When you are given a line equation $L_1: y = m_1 x + c_1$, to which the perpendicular line equation is $L_2: y = \frac{-1}{m_1} x + c_2$.
These two lines intersect at a point at the coordinate (a, b). Knowing $m_1$ and coordinate (a, b), you can plug them into $L_2$ and find $c_2$.

$L_2: b = \frac{-1}{m_1}(a)+c_2 \rightarrow c_2 = b + \frac{a}{m_1}$

Then $L_2: y = \frac{-1}{m_1} x + ( b + \frac{a}{m_1})$ <---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 $d= \frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$

Since we know $d =1, A = \frac{-1}{m_1}, B = 1$ and the origin (0,0), we can plug them into the distance equation so that it becomes:

$\pm 1= \frac{|A(0)+B(o)+C|}{\sqrt{(\frac{-1}{m_1})^2+1}}$ and obtain

$C= \pm \sqrt{(\frac{-1}{m_1})^2+1}$

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:

$y = \frac{-1}{m_2} x \pm \sqrt{(\frac{-1}{m_1})^2+1}$<---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, $L:1$ is given as $y=ax + (2-1)$. The (a,b) is (1,2), and the equation $y = -x+3$ 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 $L_1$
Thanks. Now, you used the line G to have a perpenicular distance of 1 from the origin, and earboth used the line L.

Whats the thing behind that?

9. Originally Posted by Aquafina
Thanks. Now, you used the line G to have a perpenicular distance of 1 from the origin, and earboth used the line L.

Whats the thing behind that?
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.

10. Originally Posted by Aquafina
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.
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 $y=3-x$, which we have no use, so we will ignore it.

Let $L$ denotes line $y=(2-a)+ax$, and $\perp L$ denotes the perpendicular line to it.

$L: y=ax+(2-a)$--------------------------------eq (1)
$\perp L: y=\frac{-1}{a}x+C$-------------------eq (2)

We can rearrange eq (2) to one side:

$\frac{1}{a}x+ y-C=0$ and using it with the known point (0,0), we plug all into the distance equation $D=\frac{Ax+By+C}{\sqrt{A^2+B^2}}$.

We get $1=\frac{0+0+C}{\sqrt{(\frac{1}{a})^2+1}}$, which gives us $C=\frac{\sqrt{1+a^2}}{a}$ that we can plug back into eq (2) and turn it into as follows:

$\perp L: y=\frac{\sqrt{1+a^2}-x}{a}$---------eq (3)

Now to solve for the variable $a$ 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 $2=\frac{\sqrt{1+a^2}-1}{a}$ which gives us $a(3+4)=0$ ---------------------eq (4).

We solve for a in eq (4), and the results are $a=0$ and $a=-\frac{4}{3}$, which tells us that there 2 sets of $L's$ and $\perp L's$

Now plug $a=0$ to one set, and $a=-\frac{4}{3}$ to the other.

There results should be answers to all parts 1 and 2 of your assignment:

For $a=0$: $L: y=2$, and $\perp L: x=1$.

For $a=-\frac{4}{3}$: $L: y=\frac{2}{3}(1-2x)$, and $\perp L: y=-\frac{3}{4}(\frac{5}{3}-x)$

11. Originally Posted by Aquafina
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
Will one of you explain to me why you all seem to think that part ii) has anything to do with line $L$.

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 $L$ is part of the question?

12. Originally Posted by Plato
Will one of you explain to me why you all seem to think that part ii) has anything to do with line $L$.

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 $L$ is part of the question?

Plato,

It's implied in this quote which says "also does, for any value "a".'; otherwise, it would be a no brainer.

13. Originally Posted by Aquafina
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
Originally Posted by novice
It's implied in this quote which says "also does, for any value "a".'; otherwise, it would be a no brainer.
I think that you are completely mistaken about that.
The phrase ‘also does, for any value "a".’ is purely parenthetical referring to the equation of the line $L$

Exactly as stated, that is nice problem without bringing $L$ into it.

14. Originally Posted by Plato
I think that you are completely mistaken about that.
The phrase ‘also does, for any value "a".’ is purely parenthetical referring to the equation of the line $L$

Exactly as stated, that is nice problem without bringing $L$ into it.
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.

15. Originally Posted by Plato
Will one of you explain to me why you all seem to think that part ii) has anything to do with line $L$.

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 $L$ is part of the question?
The question was originally about L. So it can't just be something completely different...

I will be most interested to see your solution. I have had many various answers for this question from different people.

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