To find the x and y-intercepts of the line $ax+by+c=0$

Question

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 $(-\frac ca,0)$

Answer 1

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:

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

Answer 2

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)

Answer 3

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 $y=-c/b$.