# Thread: Co-ordinate Geometry - Applications of a straight line

1. ## 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.

2. The equation of a line is $\displaystyle \displaystyle y = mx + c$, where $\displaystyle \displaystyle m$ is the gradient, and $\displaystyle \displaystyle c$ is the $\displaystyle \displaystyle y$ intercept.

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

Then substitute this value, and one of the points, into the equation $\displaystyle \displaystyle y = mx + c$ to find $\displaystyle \displaystyle c$.

3. 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 <<<?

4. You also need to let $\displaystyle \displaystyle y=1$

5. 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?

6. 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 $\displaystyle \displaystyle y = \frac{2}{3} x + c$. Solve for c as an exact fraction (do not use decimal approximations). Then substitute this value back into $\displaystyle \displaystyle y = \frac{2}{3} x + c$. 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.