Thread: iterated integral and point slope form

1. iterated integral and point slope form

so i am given a picture of a triangle, with its vertices labeled as ordered pairs.
(0,3) (2,0) (0,-6)

now i need to find the equations of these lines in order to know the bounds of my integrals.
my solution was to find to first find the slope. so for hte line from (0,3) to (2,0),
that would be (0-3)/(2-0) = -3/2. then plug in one of those points to find C, and i get 3.

so the line equation is y=-3/2x +3. i was marked off and told to use point slope form, and my teacher came up with the equation y=-x+2. what is point slope and how did i do this wrong? i don't understand why my equation is not correct.

2. Hello, isuckatcalc!

i am given a picture of a triangle with vertices: .$\displaystyle A(0,3),\;B(2,0),\;C(0,\text{-}6)$

i need to find the equations of these lines in order to know the bounds of my integrals.
My solution was to first find the slope.
So for the line from A(0,3) to B(2,0), that would be (0-3)/(2-0) = -3/2.
Then plug in one of those points to find C, and i get 3.

So the line equation is y = -(3/2)x +3.
i was marked off and told to use point-slope form.
And my teacher came up with the equation y = -x + 2 .??

What is point-slope and how did i do this wrong?
i don't understand why my equation is not correct.
Code:
      |A
(0,3)o
| *
|   *
|     * B
- - + - - - o - - -
|      *(2,0)
|     *
|    *
|   *
|  *
| *
|*
(0,-6)o
|C

Your teacher's line has a slope of $\displaystyle -1$ and a $\displaystyle y$-intercept of $\displaystyle 2$.
. . Neither feature is on the graph. .The error seems to be his/hers.

The Point-Slope Formula is one I recommend to all my students.
. . However, I do NOT insist that they use it.

Given a point $\displaystyle P(x_1,y_1)$ and a slope $\displaystyle m$
. . the line through $\displaystyle P$ with slope $\displaystyle m$ has the equation:
. . . . . . . . $\displaystyle y - y_1 \;=\;m(x-x_1)$

If you're using the Slope-Intercept form, $\displaystyle y \:=\:mx + b$
. . and you are given the slope and the y-intercept, you're golden!

If you are not given the y-intecept, you must perform
. . some extra acrobatics to get the equation.

The Point-Slope formula eliminates those extra steps.
. . It works for any point, not just the y-intercept.

3. but i'm not given the slope, i have to find it (and the y intercept) so how could i use point slope? i don't understand. don't i have to solve for the y=mx+b equation? all i have are points, no slope given, thus i must find it.

4. ** Alarms **

Stop. Take a deep breath. Change your screen name.

Really, if you can't make a natural connection between two points and the slope of a linear equation, you need to run, not walk, out of the calculus class and do some serious refreshing on basic algebra and geometry. By the time you see an interated integral, you should have forgotten the word "slope", except to teach it to 7th graders. (not really, but kind of)

You need an equation. Use ALL your prior experience to get one. Why do you care what someone else produced? Prove it right or wrong. Don't have a panic attack because someone else managed an incorrect result.

These are prime candidates for the Intercept Form!

The first two $\displaystyle \;\;\frac{x}{2} + \frac{y}{3} = 1$

The last two $\displaystyle \;\;\frac{x}{2} + \frac{y}{-6} = 1$

Solve for y. Find the intersection.

No kidding. Change your screen name.