Math Help Forum

Math Help Forum Feed Site Feed

Go Back   Math Help Forum > Pre-University Math Help > Pre-Calculus
Reply
 
Thread Tools Display Modes
  #1  
Old December 2nd, 2008, 02:22 PM
Member
 
Join Date: Oct 2008
Posts: 94
Country:
Thanks: 40
Thanked 1 Time in 1 Post
db5vry is on a distinguished road
Default Finding something in co-ordinate geometery

The question said that the points of A, B and C were (-2,-3), (6,1) and (k,3) respectively, and that AB is perpendicular to BC.

It asked to work out the gradient of AB:

y2-y1 / x2-x1 = 4/8 --> 1/2.

I think that might be right but then it says to prove that the value of k equals 5. How could I do that?
Reply With Quote
Advertisement
 
  #2  
Old December 2nd, 2008, 02:37 PM
masters's Avatar
He's dead, Jim

 
Join Date: Jan 2008
Location: Big Stone Gap, Virginia
Posts: 2,045
Country:
Thanks: 511
Thanked 1,769 Times in 1,211 Posts
masters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant future
Default

Quote:
Originally Posted by db5vry View Post
The question said that the points of A, B and C were (-2,-3), (6,1) and (k,3) respectively, and that AB is perpendicular to BC.

It asked to work out the gradient of AB:

y2-y1 / x2-x1 = 4/8 --> 1/2.

I think that might be right but then it says to prove that the value of k equals 5. How could I do that?
A(-2, -3)
B(6, 1)
C(k, 3)

Your slope of AB is correct. Since AB is perpendicular to BC, this means that the slope of BC must be the negative reciprocal of the slope of AB.

Therefore, the slope of BC must be -2.

Solve this: \frac{3-1}{k-6}=-2 for k.
__________________
He who knows not and knows not that he knows not is a fool, shun him. He who knows not and knows that he knows not is a child, teach him. He who knows and knows not that he knows is asleep, wake him. And he who knows and knows that he knows is wise, follow him.
-- Persian Proverb



To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Reply With Quote
  #3  
Old December 2nd, 2008, 02:39 PM
vincisonfire's Avatar
Senior Member
 
Join Date: Oct 2008
Location: Sainte-Flavie
Posts: 412
Country:
Thanks: 55
Thanked 156 Times in 151 Posts
vincisonfire has a spectacular aura aboutvincisonfire has a spectacular aura about
Send a message via Skype™ to vincisonfire
Default

If AB is perpendicular to BC then the dot product is equal to 0.
AB = (6,1)-(-2,-3)=(8,4) and BC = (k,3) - (6,1) = (k-6,2)
*AB \cdot BC =  (8,4)\cdot (k-6,2) =8k - 48 +8 =0 \implies k =5
Reply With Quote
  #4  
Old December 2nd, 2008, 03:02 PM
Member
 
Join Date: Oct 2008
Posts: 94
Country:
Thanks: 40
Thanked 1 Time in 1 Post
db5vry is on a distinguished road
Default

Quote:
Originally Posted by masters View Post
A(-2, -3)
B(6, 1)
C(k, 3)

Your slope of AB is correct. Since AB is perpendicular to BC, this means that the slope of BC must be the negative reciprocal of the slope of AB.

Therefore, the slope of BC must be -2.

Solve this: \frac{3-1}{k-6}=-2 for k.
Thanks for that. It was really helpful.

The next question says that the line L is parallel to BC and passes through A, asking you to figure out an equation for L:

I used the formula "y-y1=m(x-x1)", A(-2,-3) and the -2 gradient from earlier used to put into values as this:

y--3 = -2(x--2)
y+3 = -2(x+2)
y+3 = -2x-4
y+2x-7=0 [L]

Is this right? Thanks again for your help.
Reply With Quote
  #5  
Old December 2nd, 2008, 03:16 PM
masters's Avatar
He's dead, Jim

 
Join Date: Jan 2008
Location: Big Stone Gap, Virginia
Posts: 2,045
Country:
Thanks: 511
Thanked 1,769 Times in 1,211 Posts
masters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant future
Default

Quote:
Originally Posted by db5vry View Post
Thanks for that. It was really helpful.

The next question says that the line L is parallel to BC and passes through A, asking you to figure out an equation for L:

I used the formula "y-y1=m(x-x1)", A(-2,-3) and the -2 gradient from earlier used to put into values as this:

y--3 = -2(x--2)
y+3 = -2(x+2)
y+3 = -2x-4
y+2x+7=0 [L]

Is this right? Thanks again for your help.
One change. See sign change in red above. And, you might want to put the equation in one of the more recognizable forms:

Standard Form: 2x+y=-7
Slope-Intercept Form: y=-2x-7
__________________
He who knows not and knows not that he knows not is a fool, shun him. He who knows not and knows that he knows not is a child, teach him. He who knows and knows not that he knows is asleep, wake him. And he who knows and knows that he knows is wise, follow him.
-- Persian Proverb



To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Reply With Quote
The following users thank masters for this useful post:
Donate to MHF
  #6  
Old December 2nd, 2008, 04:16 PM
Member
 
Join Date: Oct 2008
Posts: 94
Country:
Thanks: 40
Thanked 1 Time in 1 Post
db5vry is on a distinguished road
Default

Quote:
Originally Posted by masters View Post
One change. See sign change in red above. And, you might want to put the equation in one of the more recognizable forms:

Standard Form: 2x+y=-7
Slope-Intercept Form: y=-2x-7
I'll use the slope-intercept form there then. It makes more sense! And I see how you get +7 because you need to take the 4 over to the other side, I didn't see that.

There's only one more part to the question, it says that the line intersects the y-axis at D and asks to calculate the length of CD.

Now here, I'm very stuck and not sure what formula to use. Can you help at all? If you can thanks again!
Reply With Quote
  #7  
Old December 3rd, 2008, 06:07 AM
masters's Avatar
He's dead, Jim

 
Join Date: Jan 2008
Location: Big Stone Gap, Virginia
Posts: 2,045
Country:
Thanks: 511
Thanked 1,769 Times in 1,211 Posts
masters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant futuremasters has a brilliant future
Default

Quote:
Originally Posted by db5vry View Post
I'll use the slope-intercept form there then. It makes more sense! And I see how you get +7 because you need to take the 4 over to the other side, I didn't see that.

There's only one more part to the question, it says that the line intersects the y-axis at D and asks to calculate the length of CD.

Now here, I'm very stuck and not sure what formula to use. Can you help at all? If you can thanks again!
The y-intercept of this line is (0, -7).

Use the distance formula to find the distance from C (5, 3) to D (0, -7)

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
__________________
He who knows not and knows not that he knows not is a fool, shun him. He who knows not and knows that he knows not is a child, teach him. He who knows and knows not that he knows is asleep, wake him. And he who knows and knows that he knows is wise, follow him.
-- Persian Proverb



To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
Reply With Quote
The following users thank masters for this useful post:
Donate to MHF
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off
Forum Jump


All times are GMT -7. The time now is 07:05 PM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2009, Jelsoft Enterprises Ltd.
SEO by vBSEO 3.2.0 ©2008, Crawlability, Inc.
©2005 - 2009 Math Help Forum


Math Help Forum is a community of maths forums with an emphasis on maths help in all levels of mathematics.
Register to post your math questions or just hang out and try some of our math games or visit the arcade.