# Co-ordinate Geometry - Applications of a straight line

• Feb 22nd 2011, 09:04 PM
ebonez
Co-ordinate Geometry - Applications of a straight line
Hi, guys.

Can someone please explain to me how to work this out? It's one question out of my homework. I have a bunch of problems similar to this. If someone can explain how to do this I should be able to manage the rest myself.

"Find the equation of the line joining each of these pairs of given points."
(4,1) and (7,3)

Thank you.
• Feb 22nd 2011, 09:06 PM
Prove It
The equation of a line is $\displaystyle y = mx + c$, where $\displaystyle m$ is the gradient, and $\displaystyle c$ is the $\displaystyle y$ intercept.

First, find the gradient $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$.

Then substitute this value, and one of the points, into the equation $\displaystyle y = mx + c$ to find $\displaystyle c$.
• Feb 22nd 2011, 09:21 PM
ebonez
Sorry, I don't fully understand. Could you help me from here?

(4,1) and (7,3)

m = (3-1)/(7-4)
m = 2/3

y = 2/3(4) + c
y = 2.66666667 + c <<<?
• Feb 22nd 2011, 09:23 PM
Prove It
You also need to let $\displaystyle y=1$
• Feb 22nd 2011, 09:35 PM
ebonez
Ahh, still not getting it. The answer is 2x-3y-5=0 but I really don't have a clue how to get there.

...

1 = 2.67 + c
So c = -1.67?
• Feb 23rd 2011, 04:04 AM
mr fantastic
Quote:

Originally Posted by ebonez
Sorry, I don't fully understand. Could you help me from here?

(4,1) and (7,3)

m = (3-1)/(7-4)
m = 2/3

y = 2/3(4) + c
y = 2.66666667 + c <<<?

From the given point (4, 1) substitute x = 4 and y = 1 into $\displaystyle y = \frac{2}{3} x + c$. Solve for c as an exact fraction (do not use decimal approximations). Then substitute this value back into $\displaystyle y = \frac{2}{3} x + c$. Now get rid of the fractins by multiplying both sides by 3.

Show all your working to the above if you're still stuck getting the books final answer.