Hi, need a little C1 coordinate geometry help please.

Hi, i am doing some homework and i have come upone this question, i have no idea how to do it, however it is probably really easy, any tips would be much appreciated. Rep for most help :smile:

Question:The points A and B have coordinates (k,1) and (8,2k-1) respectively, where K is a constant. Given that the gradient of AB is 1/3, Show that k=2.
I have to go have dinner know, but any help would be much appreciated.
Thanks in advance.

Reply 1

Gaz031

15

Nutchopper

Hi, i am doing some homework and i have come upone this question, i have no idea how to do it, however it is probably really easy, any tips would be much appreciated. Rep for most help :smile:

Question:The points A and B have coordinates (k,1) and (8,2k-1) respectively, where K is a constant. Given that the gradient of AB is 1/3, Show that k=2.
I have to go have dinner know, but any help would be much appreciated.
Thanks in advance.

What is the gradient of AB in terms of k?

Reply 2

Malik

12

Nutchopper

Hi, i am doing some homework and i have come upone this question, i have no idea how to do it, however it is probably really easy, any tips would be much appreciated. Rep for most help :smile:

Question:The points A and B have coordinates (k,1) and (8,2k-1) respectively, where K is a constant. Given that the gradient of AB is 1/3, Show that k=2.
I have to go have dinner know, but any help would be much appreciated.
Thanks in advance.

the general formula to get gradient between two points is:
gradient=(y1-y2)/(x1-x2)
Where (x1,y1) and (x2,y2) are the points.

So applying that to the question we have:
A(k,1) B(8,2k-1)
Gradient_AB=(1-(2k-1))/(k-8)
=(1-2k+1)/(k-8)
=(2-2k)/(k-8)

We know that is equal to 1/3.

.:. 1/3=2-2k/(k-8)
Cross multiply
k-8=3(2-2k)
k-8=6-6k
-8-6=-6k-k
-14=-7k
k=2

Reply 3

Heat_Seeker

Since gradient is y2-y1/x2-x1 therefore,
[(2k-1)]-1/(8-k) = 1/3

Substitute and then u can find k

--------------

Since gradient is y2-y1/x2-x1 therefore,
[(2k-1)-1]/(8-k) = 1/3

make k the subject and then calculate the value

Reply 4

mullet

(k-2)-1/8-k=1/3
2k-2=1/3(8-k0
2k-2=8/3-1/3k
7/3k=14/3
k=2

Reply 5

mhines87

I did it like:

General formula: y - y₁ = m(x - x₁)

For A(k,1)...
y = 1 + (1/3)x - (1/3)k

For B(8,2k-1)...
y = (1/3)x - (8/3) + 2k - 1

Then set equal to eliminate x and solve for k...
1 + (1/3)x - (1/3)k = (1/3)x - (8/3) + 2k - 1
7k = 14
k = 2

:biggrin:

Reply 6

~NC~

OP

11

Wow thanks for the influx of help, and gaz that was all the info i was given, ill spread some rep around to those who helped :smile:

Thanks very much for the help :P

Reply 7

Zen 07

So applying that to the question we have:
A(k,1) B(8,2k-1)
GradientAB=(1-(2k-1))/(k-8)
=(1-2k+1)/(k-8)
=(2-2k)/(k-8)

We know that is equal to 1/3.

.:. 1/3=2-2k/(k-8)
Cross multiply
k-8=3(2-2k)
k-8=6-6k
-8-6=-6k-k
-14=-7k
k=2

Reply 8

Malik

12

Zen 07

So applying that to the question we have:
A(k,1) B(8,2k-1)
GradientAB=(1-(2k-1))/(k-8)
=(1-2k+1)/(k-8)
=(2-2k)/(k-8)

We know that is equal to 1/3.

.:. 1/3=2-2k/(k-8)
Cross multiply
k-8=3(2-2k)
k-8=6-6k
-8-6=-6k-k
-14=-7k
k=2

You just copied and pasted my solution :p:

Reply 9

mhines87

Zen 07

So applying that to the question we have:
A(k,1) B(8,2k-1)
GradientAB=(1-(2k-1))/(k-8)
=(1-2k+1)/(k-8)
=(2-2k)/(k-8)

We know that is equal to 1/3.

.:. 1/3=2-2k/(k-8)
Cross multiply
k-8=3(2-2k)
k-8=6-6k
-8-6=-6k-k
-14=-7k
k=2

Well done, you can copy and paste somebody elses work. :thumpdown