# coordinates

• Feb 13th 2010, 04:29 AM
coolubi
coordinates
http://www.emaths.co.uk/sats/2009/2009_68_Paper1.pdf

• Feb 13th 2010, 04:41 AM
mathemagister
Quote:

Originally Posted by coolubi
http://www.emaths.co.uk/sats/2009/2009_68_Paper1.pdf

This looks a lot like a test, so here is just a hint:
Notice they have told you the two equations you need to deal with. You have 2 variables and 2 equations. Do you know how to solve simultaneous equations? When you solve for x and y, make sure you pair up the right numbers with each other for the 2 points.

Feel free to ask if you need any more help.
• Feb 13th 2010, 04:52 AM
coolubi
Quote:

Originally Posted by mathemagister
This looks a lot like a test, so here is just a hint:
Notice they have told you the two equations you need to deal with. You have 2 variables and 2 equations. Do you know how to solve simultaneous equations? When you solve for x and y, make sure you pair up the right numbers with each other for the 2 points.

Feel free to ask if you need any more help.

pls could you help me a little more as i am new to it(help with some working out) and only have one more week to my exam thanx
• Feb 13th 2010, 05:44 AM
mathemagister
Quote:

Originally Posted by coolubi
pls could you help me a little more as i am new to it(help with some working out) and only have one more week to my exam thanx

The information tells you that at the points A and B, [b]both[/both] the equations \$\displaystyle y=x-2\$ and \$\displaystyle y=x^2-2\$ are correct for the same x and y values.

So, to solve for the intercepts, you need to simultaneously solve the equations
\$\displaystyle y=x-2\$ and \$\displaystyle y=x^2-2\$.

You know that \$\displaystyle y=x-2\$, so anywhere you see a \$\displaystyle y\$, you are allowed to substitute it for \$\displaystyle x-2\$ because they are both the same thing. (Think of it like if you know \$\displaystyle 3=5-2\$, you can put \$\displaystyle 5-2\$ wherever you see a \$\displaystyle 3\$.

Well, you see a \$\displaystyle y\$ in the second equation; you can substitute that \$\displaystyle y\$ for something else that equals it: \$\displaystyle x-2\$.

This gives you:
\$\displaystyle x-2=x^2-2\$

\$\displaystyle x=x^2\$

\$\displaystyle x-x^2=0\$

\$\displaystyle x(1-x)=0\$

\$\displaystyle x=0\$ and \$\displaystyle 1-x=0\$
\$\displaystyle x=0,1\$

to find the y value for these points simply plug them into one of the original equations:
\$\displaystyle y=x-2\$
\$\displaystyle y(0) = 0 - 2 = -2\$
\$\displaystyle y(1) = 1 - 2 = -1\$

These give you the points (0,-2) and (1,-1).

Since you can see on the graph that point A lies on the y-axis (where x is always 0), you know that point A is (0,-2) and point B has to therefore be (1,-1).

Hope that helps :)