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

1. ## 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..

2. 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,
$y-(-5)=2(x-3)$
$y+5=2x-6$
$y=2x-5-5$
$y=2x-10$

3. Thanks .. but i still dont really get it.. could you break it down for me..

4. 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?

5. yea dont really understand it..

6. 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

7. 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

8. Hello, JeeFong!

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

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?