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Please check if I've solved the problem in the correct way:

The problem is as follows:

Find the points at which the line $ax+by+c=0$ crosses the x and y-axes. (Assume that $a \neq 0$ and $b \neq 0$.

My solution:

We have to find the x and y-intercepts of the line. At the 'x-intercept' the ordinate must be equal to $0$ and at the 'y-intercept' the abscissa must be equal to $0$.

Now we solve the equation $ax+by+c=0$ for $y$:

$ax + by + c=0$

$ax + by = -c$

$by = -ax -c$

$y = \frac {-ax-c}{b}$

$\because x = 0$ at y-intercept,

$\therefore y = \frac {-a(0)}{b} -\frac{c}{b}$

$y = -\frac cb$.

The point at which the line crosses the y-axis is $(0,-\frac cb)$

Now we solve the equation $ax+by+c=0$ for $x$:

$ax+by+c=0$

$ax+by=-c$

$ax = -by-c$

$x = \frac {-by-c}{a}$

$\because y = 0$ at x-intercept

$\therefore x = \frac {-b(0)}{a} -\frac{c}{a}$

$x = -\frac ca$

The point at which the line crosses the x-axis is $(-\frac ca,0)$

Samama Fahim
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    absolutely right – iostream007 May 11 '13 at 22:23
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    That's fine work, Samama! You are perfectly correct. – amWhy May 11 '13 at 23:12
  • @amWhy do $a$ and $b$ in this question represent the 'x' and 'y' intercepts respectively as they did in my previous question? I don't think so well. – Samama Fahim May 11 '13 at 23:17
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    In this case, we have we have a different equation, so we have different x, y intercepts. Those values (the intercepts) depend on the equation of the line. Since this equation is different than the equation in your earlier question, we are going to get different intercepts. You can use this same procedure to discover that in your last question, indeed, $x = a$ and $y = b$ are the x, y intercepts, respectively. – amWhy May 11 '13 at 23:28

3 Answers3

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Your work is exemplary, Samama. You know your definitions well, and your answers are entirely correct.

A nice "shortcut" is to take advantage of what you already know:

  • the $x$ intercept is the value of $x$ when $y = 0$, and
  • the $y$ intercept is the value of $y$ when $x = 0$.
  • $ax + by + c = 0 \iff ax + by = -c$

x intercept

We can use $\;ax + by = -c\;$ to solve for $x$ when $y = 0$, by plugging in $0$ for $y,\;$ right at the start:

$$ax + \underbrace{by}_{y = 0} = -c \iff ax = - c \iff x = \frac {-c}{a}$$

y intercept

And we can do the same to solve for $y$ when $x = 0$, by plugging in $0$ for $x,\,$ right at the start:

$$\underbrace{ax}_{x = 0} + by = -c \;\;\iff\;\; by = -c \;\;\iff \;\; y = \frac{-c}{b}$$

amWhy
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  • You all have led me to the best 'shortcut'. Actually, I'm inspired of the long equations in Physics in relation with some phenomena. That's the reason I expand my work so much. I hope that will not do me much harm. Well, +1 for the answer of yours and that of Andreas Blass. – Samama Fahim May 12 '13 at 01:59
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    No, not at all, Samama: it is good practice to do what you did...no harm at all. Better to be careful and practice than be careless and create errors! "Shortcuts" will come with time, when you're an "old hat" at solving problems like this. We just all like to show how you can use the full power of what you know, but there's nothing at all wrong with what you did, as I tried to make clear in my comment, and in my answer, at the start: you've been doing exemplary work, my friend. – amWhy May 12 '13 at 02:02
  • Thanks for such nice comments and admiration. I want to tell you that I've been dealing with pre-calculus for many days. Now, I'm heading toward the real calculus. I'm going to study HYPER-REAL NUMBERS, INFINITESIMALS, and I hope that you guys would help me in the same way as ever before because this is the first time I'm going to study it without a teacher and I'm afraid about that. – Samama Fahim May 12 '13 at 02:11
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    Yes, of course, we will be here to help! Our pleasure. You won't be alone! And given the care you take, I think you'll do fine. – amWhy May 12 '13 at 02:20
  • Thanks for the support. :) – Samama Fahim May 12 '13 at 02:23
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Being lazy, I'd simplify the equation first and do algebraic manipulations later. When looking for the $x$-intercept, I'd know that I'll need to plug in $0$ for $y$, so I'd do that before any complicated algebraic work. Plugging it into $ax+by+c0$, I'd get $ax+c=0$. Then I'd solve, getting $x=-c/a$. Similarly, to get the $y$-intercept, I'd plug in $x=0$ first, getting $by+c=0$, and then solve, getting $y=-c/b$.

Andreas Blass
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easier way is this equation of line intercept form $$ \frac xa+\frac yb=1$$ so write the given eqn in this form $$ax+by+c=0$$ $$ax+by=-c$$ $$\frac {ax}{-c}+\frac {by}{-c}=1$$ $$\frac {x}{\frac{-c}{a}}+\frac {y}{\frac {-c}{b}}=1$$

so intercept at X axis is (-c/a,0) and Y axis is (0,-c/b)

iostream007
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