# Thread: how to find the ratio

1. ## how to find the ratio

Question: Given that ax+ 3y + 2 = 0 and 2x - by - 5 = 0 are perpendicular lines, find the ratio a:b.

My attempt:
$ax + 3y + 2 = 0$
$y = -\frac{2}{3} - \frac{a}{3}x$
$2x - by - 5= 0$
$y = \frac{2}{b}x - \frac{5}{b}$
$m X - \frac{1}{m} =-1$
therefore

$-\frac{a}{3} = -\frac{2}{b}.......(1)$

$m = -\frac{a}{3}$

thus $-\frac{1}{m} = -\frac{1}{-\frac{a}{3}} = \frac{3}{a}$
$\frac{3}{a}= \frac{2}{b}......(2)$

using simultaneous equation for 1 and 2
you get $a = \pm 3$
and

$b = \pm 2$

therefore the ratio i got is 3:2 or -3:-2

the problem is that the book only states 3:2

in the case of this question i check both positive and negative and my answrs are right but the book only states 3:2 is that just a mistake in the book?
or did my method allowed an extraneous result?

2. ## Re: how to find the ratio

Originally Posted by bigmansouf
Question: Given that ax+ 3y + 2 = 0 and 2x - by - 5 = 0 are perpendicular lines, find the ratio a:b
If neither $m_1~\&~m_2$ is zero and are the slopes of perpendicular lines $\ell_1~\&~\ell_2$ then $m_1\cdot m_2=-1$.

So in your case $\dfrac{-a}{3}\cdot\dfrac{-2}{-b}=-1$

3. ## Re: how to find the ratio

Originally Posted by bigmansouf
Question: Given that ax+ 3y + 2 = 0 and 2x - by - 5 = 0 are perpendicular lines, find the ratio a:b.

My attempt:
$ax + 3y + 2 = 0$
$y = -\frac{2}{3} - \frac{a}{3}x$
$2x - by - 5= 0$
$y = \frac{2}{b}x - \frac{5}{b}$
$m X - \frac{1}{m} =-1$
therefore

$-\frac{a}{3} = -\frac{2}{b}.......(1)$

$m = -\frac{a}{3}$

thus $-\frac{1}{m} = -\frac{1}{-\frac{a}{3}} = \frac{3}{a}$
$\frac{3}{a}= \frac{2}{b}......(2)$

using simultaneous equation for 1 and 2
you get $a = \pm 3$
and

$b = \pm 2$

therefore the ratio i got is 3:2 or -3:-2

the problem is that the book only states 3:2

in the case of this question i check both positive and negative and my answrs are right but the book only states 3:2 is that just a mistake in the book?
or did my method allowed an extraneous result?
-3:-2 =3:2

a=6 and b=4 also satisfies your equations.
6:4 = 3:2. You are asked for the ratio a:b, not a and b separately.

4. ## Re: how to find the ratio

Originally Posted by Plato
If neither $m_1~\&~m_2$ is zero and are the slopes of perpendicular lines $\ell_1~\&~\ell_2$ then $m_1\cdot m_2=-1$.

So in your case $\dfrac{-a}{3}\cdot\dfrac{-2}{-b}=-1$
i understand this but how do you get the ratio from this.

2a=3b
but how does it become 3:2 as a ratio thats my problem

5. ## Re: how to find the ratio

Originally Posted by bigmansouf
i understand this but how do you get the ratio from this.

2a=3b
but how does it become 3:2 as a ratio thats my problem
$2a=3b$

$a=\frac{3b}{2}$

$\frac{a}{b}=\frac{3}{2}$

$a:b = 3:2$

6. ## Re: how to find the ratio

Originally Posted by bigmansouf
i understand this but how do you get the ratio from this.
2a=3b but how does it become 3:2 as a ratio thats my problem
Oh come on! Are you really saying that if $2a=3b$ that you fail to understand that $\dfrac{a}{b}=\dfrac{3}{2}~\bf{??}$

7. ## Re: how to find the ratio

Perhaps bigmansouf is saying that he does not know what the word "ratio" means.

8. ## Re: how to find the ratio

Originally Posted by Plato
Oh come on! Are you really saying that if $2a=3b$ that you fail to understand that $\dfrac{a}{b}=\dfrac{3}{2}~\bf{??}$
i understood 2a = 3b but not how to get it to a ratio form. I understood the relationship which you get from m1 x m2 = -1 but from there i failed to rearrange it to give a/b = 3/2 instead i thought of using simultaneous equation to help me find the ratio hence when i did that as shown in my first post i thought i got an extra answer which i believed to be extraneous result.

The problem was i thought the question was far more complex than it was. i simply failed to spot it