0

In coordinate geometry, whenever we solve a problem we see that if the resulting solution is a line, then all the lines which are parallel to y - axis are left out since their slope will be $\infty$ and thus can't be calculated. This is a big problem when solving questions since very often we tend to miss out on specific solutions or sometimes get no solution at all when the only solution is a line parallel to y - axis.

My way: I usually check out specifically for this case(taking the line to be $x = \lambda$) when I feel that there can be a potential solution.

Is there any better method of dealing with this problem ?

user2369284
  • 2,231

1 Answers1

1

Any line can be represented as $$ax+by+c=0$$ where $a,b,c\in\mathbb R.$

Set $b=0$ in the equation.

Edit : In your 'new' question, it is clear that $x=1$ is one of the tangents. So, we may suppose that $y=mx+n$.

mathlove
  • 139,939
  • Usually we take the line to be $y = mx + x$ and then solve for variables "m" and "c". And when we solve for them ,we will never get $m = \infty$ and thus all the lines parallel to y -axis will be left out. If we take line as $ax + by + c = 0$, then we have 3 variables which will be more difficult to solve. – user2369284 Dec 28 '13 at 15:19
  • You have misinterpreted my problem. – user2369284 Dec 28 '13 at 15:20
  • Well, I don't think so. If we take line as $y=mx+n$, the lines parallel to $x$-axis cannot be include as you mentioned. If we use the equation I wrote, such things never happen. I don't think this makes it more difficult to solve. – mathlove Dec 28 '13 at 15:25
  • Questions are not so lenient so as to provide you with 3 equations when it can be solved with just 2 equations. – user2369284 Dec 28 '13 at 15:27
  • Also it is not parallel to x-axis but y-axis – user2369284 Dec 28 '13 at 15:28
  • When you write $y=mx+k$ you usually are talking about a function $y\colon \Bbb R \to \Bbb R$, so it's no surprise you can't get a vertical line. In analytic geometry, we study more general structures, and you can write this line by a number of methods, e.g. changing coordinates $x\to y$ or setting $b=0$ in the equation above. If you are uncomfortable with three letters, you can just write $m=-a/b$ and $k=-c/b$, but you'll lose some generality. – Ian Mateus Dec 28 '13 at 15:33
  • @IanMateus Suppose you have to find out the equation of tangent from (1,5) to $x^2 + y^2 = 1$.Please(if possible) post your solution or just check it out yourself. – user2369284 Dec 28 '13 at 15:40
  • @user2369284 using a good deal of geometry, the lines are $x=1$ and $5x-2y+5=0$ or, if you prefer, $y=\frac52 x+ \frac 52$. Why? – Ian Mateus Dec 28 '13 at 15:45
  • @IanMateus The point I'm emphasizing is how you you calculated x = 1. Did you use $ax + by +c = 0$ or $y = mx + c$ as your tangent line. – user2369284 Dec 28 '13 at 15:48
  • @user2369284 to be honest, none. I just looked at the picture and saw this line, no algebra. – Ian Mateus Dec 28 '13 at 15:51
  • @IanMateus It was a quite easy problem and even if it had been a little difficult you must have been forced to use coordinate geometry. – user2369284 Dec 28 '13 at 16:10
  • @IanMateus Please try it once using coordinate geometry. – user2369284 Dec 28 '13 at 16:11
  • @mathlove Please try to solve a simple problem I posted in the comments using coordinate geometry and see if you can understand what I am trying to ask. – user2369284 Dec 28 '13 at 16:22
  • @user2369284 this problem is a pain to do using coordinates, but here it is. I did it wrong last time by the way, sorry. Choose a point $(a, b)$ in the circumference at the origin with radius one. The tangent line passing through it is $ax+by=1$. We want it to pass through $(1, 5)=K$, so $a+5b=1$. Further, we necessarily have $a^2+b^2=1$, and this system gives $(a, b) = (-12/13, 5/13) = P$ or $(a,b)=(1,0)=Q$, so we have the lines that pass through $PK$ or $QK$. They are $x=1$ and $-\frac{12}{13}x+\frac{5}{13}y=1$. – Ian Mateus Dec 28 '13 at 17:05
  • @IanMateus Thanks for the idea that a line can also be represented as $ax + by = 1$.This has 2 variables and can represent lines parallel to y-axis. Now you must have understood what would have happened if you had taken the line to be $ax + by + c = 0$.Also, in my opinion this was a VERY EASY question. The brains are blown out if we get question of finding the locus with a thousand other things. Please post this(not the question's answer but the "idea" thing) as answer. – user2369284 Dec 29 '13 at 09:24
  • @mathlove This was a very easy question.When it comes to questions of locus dealing with parabola and ellipse you just cannot observe and tell.You HAVE TO calculate. So please answer after considering this into account. – user2369284 Dec 29 '13 at 09:26
  • @user2369284 be careful, not every line is of the form $ax+by=1$. What I meant is that the lines that are tangent to the circumference you gave at a point $(a,b)$ are of this form. You have to prove it. – Ian Mateus Dec 29 '13 at 15:38
  • @user2369284 looking better, they actually can, if $c\neq 0$. But we always lose some generality, you see. – Ian Mateus Dec 29 '13 at 20:37