Equation of the line that has $x$ and $y$ intercepts at $a$ and $b$.

Question

Please can anyone help me with proving the following problem:

Show that the line that crosses the $X$-axis at $a \neq 0$ and the $Y$-axis at $b \neq 0$ has the equation $$\dfrac{x}a + \dfrac{y}b -1=0$$

I don't have an idea of where I should start from for proving the problem.

amWhy · Accepted Answer · 2013-05-11T21:06:17.960

5

The $x$ intercept is given by $(a, 0)$, and the $y$ intercept is given by $(0, b)$. And we know $a \neq 0, b\neq 0$, so the points are distinct.

All you need to do is substitute these values into your equation to see that equality holds: i.e., that they lie on the line defined by $$\frac xa + \frac yb - 1 = 0$$

$$(a, 0): x = a, y = 0 \implies \frac aa + \frac 0b - 1 = 0\implies \frac aa = 1\implies 1 = 1\quad \checkmark$$

$$(0, b): x = 0, y = b \implies \frac 0a + \frac bb - 1 = 0 \implies \frac bb = 1 \implies 1 = 1 \quad \checkmark$$

Indeed: These are unique points lying on the line given by $$\frac xa + \frac yb - 1 = 0$$

edited May 11 '13 at 21:06

answered May 11 '13 at 20:48

amWhy

209,954

Let me know, Samama, if you have any questions about this! :-) – amWhy May 11 '13 at 21:01
Yeah, sure. Thanks :) – Samama Fahim May 11 '13 at 21:22
I've no questions indeed. The explanation is quite comprehensible, and I just needed an idea. But you guys have really proved the whole problem. However, thanks for that. – Samama Fahim May 11 '13 at 21:35
1

@amWhy: very nice write up +1 – Amzoti May 12 '13 at 00:50
in question where is it state that the line will only cut X and Y axis on their positive side?so why every one thinking only 1 case that interception points are only (a,0) and (0,b).I don't agree with this. – iostream007 May 12 '13 at 14:29

score 2 · Answer 2 · answered May 11 '13 at 20:45

Clearly, the equation $\dfrac{x}a + \dfrac{y}b - 1 =0 $ defines a straight line.
Also, two distinct points uniquely define a straight line.
Now check that $(a,0)$ and $(0,b)$ satisfy the equation.
This would imply $(a,0)$ and $(0,b)$ lie on the straight line.
Since $a \neq 0 \neq b$, these are two distinct distinct points.

Alex Wertheim · Answer 3 · 2013-05-11T20:54:43.287

1

Hint: you have two points from the problem description: $(a, 0)$, and $(0, b)$. Can you write the equation of a line connecting them?

If you are still stuck:

Using slope-intercept form, we have:

$$y = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}x + b$$ where $b$ is the $y$-intercept and $(x_{1}, y_{1}), (x_{2}, y_{2})$ are two points on the line. Evaluating, then, we see:

$$y = \frac{b - 0}{0-a}x + b = \frac{-b}{a}x + b$$ $$\implies y + \frac{b}{a}x - b = 0$$ $$\implies \frac{1}{b}y + \frac{1}{a}x - 1 = 0$$ $$\implies \frac{x}{a} + \frac{y}{b} - 1 = 0$$

edited May 11 '13 at 20:54

answered May 11 '13 at 20:45

Alex Wertheim

20,278

how did you get $\frac {1}{b}y + \frac {1}{a}x - 1 = 0$? – Samama Fahim May 11 '13 at 21:28
1

I divided both sides by $b$ (or if you'd like, multiplied both sides by $\frac{1}{b})$, which is fine as $b \neq 0$. Let me know if you have any more questions! – Alex Wertheim May 11 '13 at 21:30
Thanks. That is also a good answer and a comprehensible answer. No questions indeed. – Samama Fahim May 11 '13 at 21:32

score -1 · Answer 4 · answered May 11 '13 at 22:03

-1

Actually there are set of 4 lines. because intercepts are length and its coordinates have set of 4 coordinates.(a,0)(0,b),(a,0)(0,-b),(-a,0)(0,b),(-a,0)(0,-b). so lines equation will be.. $$ \frac xa + \frac yb -1 =0$$

$$ \frac xa - \frac yb -1 =0$$

$$ -\frac xa + \frac yb -1 =0$$ $$ -\frac xa - \frac yb -1 =0$$

answered May 11 '13 at 22:03

iostream007

4,529

1

The equation of the line is given. It is the one equation you list first. And the points given where the lines crosses the x-axis is when $x = a$ and the y-axis is when $y = b$. That means it intersects the x-axis at $(a, 0)$, and it intersects the y-axis at $(0, b)$. So I'm afraid your answer may just confuse the OP, not help. – amWhy May 11 '13 at 23:35
@amWhy please read carefully the question it is saying "that Show that the line that crosses the X-axis at a≠0 and the Y-axis at b≠0 has given equation" means OP has to prove that line equation.I'm just saying that here only stated that a≠0,b≠0 doesn't say that these point are on positive axis so they can be on any axis.from just interceptor length we can't decide which point has to taken. – iostream007 May 12 '13 at 12:12
peoples are too fast to downvote without consider any option – iostream007 May 12 '13 at 12:13
@SamamaFahim it is exactly same in your problem.Their is not mention that a and b must be positive their is only mention that a and b are not equal to 0. – iostream007 May 12 '13 at 18:35

Equation of the line that has $x$ and $y$ intercepts at $a$ and $b$.

4 Answers4