# Write an Equation? I dont know how to write an equation..

• January 2nd 2007, 04:59 PM
JeeFong
Write an Equation? I dont know how to write an equation..
I'm in the 9th grade and i haven't been going to school.. I just started going again and my teacher gave me a ishload of homework to make up.. so i need help since i dont know how to do any of it..

Write an equation of a line through the given point with the given slope.
1. (3,-5);M = 2
2. (1,2); M = -3
3. (3,0); M = -1

All help is appreciated.. Thanks in advance..
• January 2nd 2007, 05:07 PM
ThePerfectHacker
Quote:

Originally Posted by JeeFong

Write an equation of a line through the given point with the given slope.
1. (3,-5);M = 2

Let me just give you a general formula.
The equation of a line (non-vertical) is:
$y-y_0=m(x-x_0)$
Where,
$(x_0,y_0)$ a point on the line.
Und $m$ is the slope.

Thus,
$y-(-5)=2(x-3)$
$y+5=2x-6$
$y=2x-5-5$
$y=2x-10$
• January 2nd 2007, 05:16 PM
JeeFong
Thanks .. but i still dont really get it.. could you break it down for me..
• January 2nd 2007, 05:35 PM
ThePerfectHacker
Quote:

Originally Posted by JeeFong
Thanks .. but i still dont really get it.. could you break it down for me..

What do you not understand?
The formula?
• January 2nd 2007, 05:46 PM
JeeFong
yea dont really understand it..
• January 2nd 2007, 06:00 PM
topsquark
Quote:

Originally Posted by JeeFong
I'm in the 9th grade and i haven't been going to school.. I just started going again and my teacher gave me a ishload of homework to make up.. so i need help since i dont know how to do any of it..

Write an equation of a line through the given point with the given slope.
1. (3,-5);M = 2
2. (1,2); M = -3
3. (3,0); M = -1

All help is appreciated.. Thanks in advance..

M is the symbol we are using for the slope, which tells how "steep" the line is on the graph. (3, -5) is an "ordered pair" where the first number is the x value of the point and -5 is the y value of the point.

Any line can be expressed in the form:
$y - y_0 = m(x - x_0)$
where $(x_0, y_0)$ is a point on the line, m is the slope of the line, x is the independent variable, and y is the dependent variable.

So for problem 1 we have that the slope m = 2, and $(x_0, y_0) = (3, -5)$. Thus:
$y - (-5) = 2(x - 3)$

Now solve this for y:
$y + 5 = 2x - 6$

$y = 2x - 11$ (TPH had a slight typo in his example here.)

The other two are done exactly the same way.

-Dan
• January 2nd 2007, 10:43 PM
earboth
Quote:

Originally Posted by JeeFong
yea dont really understand it..

Hello,

to set up the equation of a straight line you need 2 points: One fixed point $P_0$ with the coordinates $x_0$ and $y_0$ and a point P which represents every possible point of the straight line. It has the coordinates x and y.

The slope of the line is calculated by the ratio of the legs of a right triangle $m=\frac{red}{blue}$. You can calculate the length of those legs:
${red}=y-y_0$
${blue}=x-x_0$
That means:
$m=\frac{red}{blue}=\frac{y-y_0}{x-x_0}$. Now multiply by the denominator:

$m \cdot (x-x_0)=y-y_0$

And that's exactly the formula TPH has used.

EB
• January 3rd 2007, 03:55 AM
Soroban
Hello, JeeFong!

Let me give you my baby-step approach . . . .

Quote:

Write an equation of a line through the given point with the given slope.

$1.\;\;(3,-5),\;m = 2$

If you are given a point on a line and the slope of the line,
. . there is a formula we can use.
It's called appropriately the "Point-Slope Formula".

Given a point $(x_1,y_1)$ and the slope $m$
. . the equation of the line is: . $y - y_1 \:=\:m(x - x_1)$

It's simply a matter of dropping in the facts we are given into the right slots.

Look at the formula again: . $y - y_1 \:=\:m(x - x_1)$
. . . . . . . . . . . . . . . . . . . . . . $\uparrow\qquad\;\uparrow\qquad\;\;\;\uparrow$
. . . . . . . . . . . . . . . .
y-coordinate . . slope . . . x-coordinate

We are given: . $(3,-5),\:m = 2$
. . . . . . . . . . . . $\uparrow\quad\uparrow\qquad\quad\;\;\nwarrow$
. . . . . . .x-coordinate .y-coordinate . slope

Drop these into the formula: . $y - (-5) \:=\:2(x - 3)$

And simplify: . $y + 5 \:=\:2x - 6\quad\Rightarrow\quad \boxed{y \:=\:2x - 11}$

Got it?