# I need help with these 2 problems i just can't figure out for the life of me

• Sep 18th 2009, 08:46 PM
tyonna
I need help with these 2 problems i just can't figure out for the life of me
are these lines perpendicular, parallel, the same, or neither

(1.) y =2/3 x + 3 and y = -3x + 2

(2.) 2x + 5y = 1 and y = 5/2x +4
• Sep 18th 2009, 09:30 PM
red_dog
Let $\displaystyle d_1:y=m_1x+n_1$ and $\displaystyle d_2:y=m_2x+n_2$.

$\displaystyle d_1\parallel d_2\Leftrightarrow m_1=m_2, \ n_1\neq n_2$

$\displaystyle d_1\perp d_2\Leftrightarrow m_1\cdot m_2=-1$

$\displaystyle d_1=d_2\Leftrightarrow m_1=m_2, \ n_1=n_2$
• Sep 18th 2009, 09:31 PM
Rapha
Hello tyonna!

Quote:

Originally Posted by tyonna
are these lines perpendicular, parallel, the same, or neither

(1.) y =2/3 x + 3 and y = -3x + 2

(2.) 2x + 5y = 1 and y = 5/2x +4

You have to check some criteria

Let two lines given by

y1 = ax+b
y2 = cx+d

they are perpendicular, if a*c = -1

they are parallel, if a=c (and b is not equal to d)

(1)y =2/3 x + 3 and y = -3x + 2

It is a = 2/3 and c = -3. It is obvious that these lines are not parallel

EDIT:
But it is a*c = 2/3 * (-3) = -2

This is not equal to -1, because $\displaystyle -2 \not= -1$

So they are not perp.

Number 2
2x + 5y = 1 and y = 5/2x +4

Solve the first "line" for y
2x + 5y = 1

5y = 1-2x

y =1/5 * (1-2x) = 1/5 - 2/5x = -2/5x+1/5

Okay, and the other line is defined by y = 5/2x +4

It is a = -2/5 and c = +5/2 => they are not parallel, but perpendicular.

Yours
Rapha
• Sep 18th 2009, 09:35 PM
tyonna
Quote:

Originally Posted by red_dog
Let $\displaystyle d_1:y=m_1x+n_1$ and $\displaystyle d_2:y=m_2x+n_2$.

$\displaystyle d_1\parallel d_2\Leftrightarrow m_1=m_2, \ n_1\neq n_2$

$\displaystyle d_1\perp d_2\Leftrightarrow m_1\cdot m_2=-1$

$\displaystyle d_1=d_2\Leftrightarrow m_1=m_2, \ n_1=n_2$

what?
• Sep 18th 2009, 09:38 PM
Rapha
Quote:

Originally Posted by tyonna
what?

These are the criteria I mentioned

Two lines are given by

Quote:

Originally Posted by red_dog
Let $\displaystyle d_1:y=m_1x+n_1$ and $\displaystyle d_2:y=m_2x+n_2$.

They are parallel, if

Quote:

Originally Posted by red_dog
$\displaystyle d_1\parallel d_2\Leftrightarrow m_1=m_2, \ n_1\neq n_2$

They are perpendicular/orthogonal, if
Quote:

Originally Posted by red_dog
$\displaystyle d_1\perp d_2\Leftrightarrow m_1\cdot m_2=-1$

and they are the same/identical, if
Quote:

Originally Posted by red_dog
$\displaystyle d_1=d_2\Leftrightarrow m_1=m_2, \ n_1=n_2$

• Sep 18th 2009, 09:38 PM
tyonna
Quote:

Originally Posted by Rapha
Hello tyonna!

You have to check some criteria

Let two lines given by

y1 = ax+b
y2 = cx+d

they are perpendicular, if a*c = -1

they are parallel, if a=c (and b is not equal to d)

(1)y =2/3 x + 3 and y = -3x + 2

It is a = 2/3 and c = -3. It is obvious that these lines are not parallel

But it is a*c = 2/3 * (-3) = -1

So they are perpendicular.

Number 2
2x + 5y = 1 and y = 5/2x +4

Solve the first "line" for y
2x + 5y = 1

5y = 1-2x

y =1/5 * (1-2x) = 1/5 - 2/5x = -2/5x+1/5

Okay, and the other line is defined by y = 5/2x +4

It is a = -2/5 and c = +5/2 => they are not parallel, but perpendicular.

Yours
Rapha

okay, so both are perp....I thought so but I was not sure and when I asked a friend they said I was wrong.
• Sep 18th 2009, 09:41 PM
Rapha
Quote:

Originally Posted by tyonna
okay, so both are perp....I thought so but I was not sure and when I asked a friend they said I was wrong.

Ooooops, Sorry, I'm wrong. I made a mistake.

In (1) I wrote:

But it is a*c = 2/3 * (-3) = -1

a*c = -2, so they are not perp. And not parallel.
Sorry for that :(
• Sep 18th 2009, 09:56 PM
tyonna
Quote:

Originally Posted by Rapha
Ooooops, Sorry, I'm wrong. I made a mistake.

In (1) I wrote:

But it is a*c = 2/3 * (-3) = -1

a*c = -2, so they are not perp. And not parallel.
Sorry for that :(

So are they neither or the same? Dang, I was happy there for a minute. I thought I could tell my friend, "told you so" glad I hadn't done that yet! Thanks
• Sep 18th 2009, 10:01 PM
Rapha
Quote:

Originally Posted by tyonna
So are they neither or the same? Dang, I was happy there for a minute.

They can't be the same, because $\displaystyle a \not= c$.

They are the same, if
a) the lines are parallel, that means a=c, e. g. $\displaystyle m_1 = m_2$
AND
b) b = d, e. g. in red_dog 's post the criteria is $\displaystyle n_1 = n_2$

'Neither' is the right answer for (1)

Quote:

Originally Posted by tyonna
I thought I could tell my friend, "told you so" glad I hadn't done that yet! Thanks

I'm really sorry for that (Crying)
I'm glad you didn't go offline and checked the forum again.
• Sep 18th 2009, 10:05 PM
tyonna
[QUOTE=Rapha;367561]
Quote:

Originally Posted by tyonna
So are they neither or the same? Dang, I was happy there for a minute.

They can't be the same, because $\displaystyle a \not= c$.

They are the same, if
a) the lines are parallel, that means a=c, e. g. $\displaystyle m_1 = m_2$
AND
b) b = d, e. g. in red_dog 's post the criteria is $\displaystyle n_1 = n_2$

'Neither' is the right answer for (1)

I'm really sorry for that (Crying)
I'm glad you didn't go offline and checked the forum again.

ok thanks