# Thread: [SOLVED] Simple Co-Ordinate Geometry

1. ## [SOLVED] Simple Co-Ordinate Geometry

Q#1

Are the following equations of line same in sense of meaning and values? How are you supposed to arrange your final line's equation?

$\displaystyle 2x-3y+9=0$ (Answer in the book)

$\displaystyle 3y-2x-9=0$ (My answer)

OR

$\displaystyle 3y=2x+3$

Q#2

If the line of equation is

$\displaystyle 2x-3y+9=0$

Shouldn't we be able to find out the value of y for it's corresponding value of x? I mean if I am 100% sure that at a certain point, in my question where the above line is intersecting with the y-axis, thus x = 0, right?

So if I put x=0 in the above equation, my value for the corresponding y comes as 1. But in the answer of the question the examiner has taken the value of y as the y-intercept.

Please explain what am I doing wrong.

2. Originally Posted by unstopabl3
Q#1

Are the following equations of line same in sense of meaning and values? How are you supposed to arrange your final line's equation?

$\displaystyle 2x-3y+9=0$ (Answer in the book)

$\displaystyle 3y-2x-9=0$ (My answer)

OR

$\displaystyle 3y=2x+3$

Q#2

If the line of equation is

$\displaystyle 2x-3y+9=0$

Shouldn't we be able to find out the value of y for it's corresponding value of x? I mean if I am 100% sure that at a certain point, in my question where the above line is intersecting with the y-axis, thus x = 0, right?

So if I put x=0 in the above equation, my value for the corresponding y comes as 1. But in the answer of the question the examiner has taken the value of y as the y-intercept.

Please explain what am I doing wrong.

for the first question it is the same in that way , but if we write them as a functions it is different

$\displaystyle f(x,y) = 2x -3y +9$ and $\displaystyle g(x,y)=3y-2x-9$ are different functions

second question the line intersect with y-axis when x=0 and intersect with x-axis when y=0

so

$\displaystyle 2x-3y+9 =0$ intersect y-axis at

$\displaystyle 2(0) -3y +9 =0 \Rightarrow y =3$ so the line intersect with y-axis at (0,3)

in the same way you can find where the line intersect with x-axis set y=0 and then find x value

3. Thanks for the quick reply. I feel stupid about Q2 as I made a mistake when I solved it myself

For the first question, I want to ask is that how are we supposed to arrange our final equation. For example in my question you can see that I have three variations of the equation and all of them are correct but what I mean is what is the most appropriate way of writing our final equation?

For example if I get the following equation after solving:

5y=3x+5

Am I supposed to do this

5y-3x-5=0

OR

-y+3x+5=0

OR

5y-3x=0

OR

y=(3x+5)/5

OR

Leave the equation as it is

5y=3x+5

You get my point? How do you know your final equation should look like?

Thanks!

4. Originally Posted by unstopabl3
Thanks for the quick reply. I feel stupid about Q2 as I made a mistake when I solved it myself

For the first question, I want to ask is that how are we supposed to arrange our final equation. For example in my question you can see that I have three variations of the equation and all of them are correct but what I mean is what is the most appropriate way of writing our final equation?

For example if I get the following equation after solving:

5y=3x+5

Am I supposed to do this

5y-3x-5=0

OR

-y+3x+5=0

OR

5y-3x=0

OR

y=(3x+5)/5

OR

Leave the equation as it is

5y=3x+5

You get my point? How do you know your final equation should look like?

Thanks!
Hi

it depends on the question .. If it asks to be an intercept form , then by all means put it in an intercept forms , and same goes for general form or standard form .

But if the question did not specify , then you can just put it in any of those forms .

5. Thanks, what would be the standard/general form and the intercept form?

Thanks!

6. Originally Posted by unstopabl3
Thanks, what would be the standard/general form and the intercept form?

Thanks!
3y+2x-4=0 . this would be standard form

y=-2/3 x + 4/3 , general form .

In general ,

Ax+By+C=0 .. standard

y=mx+c .. general

7. Thanks