# Find the equation of a line perpendicular to a line with a given equation

• June 2nd 2009, 04:50 PM
fcabanski
Find the equation of a line perpendicular to a line with a given equation

Find the equation perpendicular to 10x – 9y + 2 = 0 and passing through point (3,3) is given by...

Eqn. of reqd line is of form 9x+10y+k = 0; k is to be determined.
(3,3)belongs to line=> 27+30+k = 0 or k = -57
9x+10y-57 = 0
Ans: C

I set up the point-slope :

y=10/9x + 2/9

m for the perp line is -9/10, but then how do I solve for the intercept?

y = -9/10x +c

but then I set

10y=-9x+10c

10y+9x=10c

c= y + 9/10x - that isn't right!
• June 2nd 2009, 05:07 PM
TheAbstractionist
Quote:

Originally Posted by fcabanski

Find the equation perpendicular to 10x – 9y + 2 = 0 and passing through point (3,3) is given by...

Eqn. of reqd line is of form 9x+10y+k = 0; k is to be determined.
(3,3)belongs to line=> 27+30+k = 0 or k = -57
9x+10y-57 = 0
Ans: C

I set up the point-slope :

y=10/9x + 2/9

m for the perp line is -9/10, but then how do I solve for the intercept?

y = -9/10x +c

but then I set

10y=-9x+10c

10y+9x=10c

c= y + 9/10x - that isn't right!

Hi fcabanski.

What you are doing is correct. The $k$ to be determined is equal to $-10c.$
• June 2nd 2009, 05:26 PM
fcabanski
So how do I arrive at the answer as was given in the solution?
• June 2nd 2009, 06:12 PM
TheAbstractionist
I just told you. Your $c$ is equal to $\frac{57}{10}$ (set $(x,y)=(3,3)$ in $c=y+\frac9{10}x);$ hence $k=-10c.$
• June 2nd 2009, 06:26 PM
fcabanski
That makes no sense. Normally I don't have to set k=nc.

For example in this problem:

Write an equation of the line
perpendicular to 4y - x = 20 and
containing the point (2, -3).

y = 1/4x +20

slope perpendicular line = -4

y = -4x +b

Plug in x and y

-3 = -4(2) + b

-3 = -8 +b

5=b

Perp line is therefore

y = -4x +5

Or

4x + y - 5 = 0

there's no setting an arbitrary k = nc

• June 2nd 2009, 06:31 PM
TheAbstractionist
Quote:

Originally Posted by fcabanski
That makes no sense.

What isn’t making sense? You just have not understood it yet (though I thought you were so close to doing so).
• June 2nd 2009, 06:33 PM
fcabanski
Please explain it. In the second example I showed, there was no setting k=nc.

Why is K=-10c set? How do you know to do it before you have the solution?
• June 2nd 2009, 06:37 PM
TheAbstractionist
Quote:

Originally Posted by fcabanski
Please explain it. In the second example I showed, there was no setting k=nc.

Why is K=-10c set? How do you know to do it before you have the solution?

Forget your second example. It was just a red herring.

Let me put it this way. The equation of the line can be written as either

$10y+9x-57\ =\ 0$

or

$y\ =\ -\frac9{10}x+\frac{57}{10}$

They are the same. Yes? If you write it the first way, the constant term appears as $-57.$ If the second way, it appears as $\frac{57}{10}.$ Does this now make sense?
• June 2nd 2009, 06:51 PM
fcabanski
Yes, but how do I get there from the slope point form?

Find the equation perpendicular to 10x – 9y + 2 = 0 and passing through point (3,3) is given by...

-9y =-10x -2

y = 10/9x +2/9

Now we have the slope of the perp line = -9/10

Perp Line y = -9/10x + b

So we can find the y int given the point (3,3)

3 = -9/30 +b

What, without having the solution, would tell us to set k = -10 b? (I don't even know where I got c )

We'd just solve:

b = 3 + 9/30

So that's what I'm asking, without that solution, how do we know to set some arbitrary k?
• June 2nd 2009, 06:53 PM
wytiaz
Quote:

m for the perp line is -9/10, but then how do I solve for the intercept?
y - y1 = m (x - x1)

y - 3 = -9/10 (x - 3)

y - 3 = -9x/10 + 27/10

y = -9x/10 + 57/10

When you know a point on a line and the slope of the line, just grind them into point-slope form for lines and get your line. No reason to get all wordy.
• June 2nd 2009, 06:57 PM
TheAbstractionist
Quote:

Originally Posted by fcabanski
So we can find the y int given the point (3,3)

3 = -9/30 +b

You made a typo this time. It should be $3\ =\ -\frac{27}{10}+b.$
• June 2nd 2009, 07:08 PM
fcabanski
OK, so

3 = -27/10+ b

b = 3 +27/10 = 57/10

so

y = -9/10x +57/10

10y = -9x +57

9x + 10y -57 = 0

I was forgetting to multiply the 57/10 by 10 to get the final equation.

Now I see the mistake I was making since yesterday. Thanks.