# Thread: midpoint y axis ius this correct

1. ## midpoint y axis ius this correct

Which value of p makes the midpoint between E(3p, 8) and F(9, 4p) locate on y-axis?
-5
-3
2
3

2. Originally Posted by sanee66
Which value of p makes the midpoint between E(3p, 8) and F(9, 4p) locate on y-axis?
-5
-3
2
3
On y-axis means x=0. We want midpoint of x's to be zero.

$\displaystyle \frac{3p+9}{2} = 0$

3. ## reply

Originally Posted by ThePerfectHacker
On y-axis means x=0. We want midpoint of x's to be zero.

$\displaystyle \frac{3p+9}{2} = 0$
um, yeah. i don't get what you just did ThePerfectHacker. sorry if that offends you, but really, i don't get what you wrote.

yeah. um. use this wonderfully self made drawing I'm giving to you in this reply (for free! How generous of me! I mean, just look at it! Think of the skill and talent i needed to draw it in MS Paint! the bonus is that I've color coded it too!)

if you still don't get it reply! but this drawing should be pretty easy to understand. You'll figure it out... i hope.

4. Originally Posted by blurgh
um, yeah. i don't get what you just did ThePerfectHacker. sorry if that offends you, but really, i don't get what you wrote.

yeah. um. use this wonderfully self made drawing I'm giving to you in this reply (for free! How generous of me! I mean, just look at it! Think of the skill and talent i needed to draw it in MS Paint! the bonus is that I've color coded it too!)

if you still don't get it reply! but this drawing should be pretty easy to understand. You'll figure it out... i hope.
Nice diagram! But TPH's method is better (and easier) I think. Let me try to explain it in more detail.

Given two points, $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2, y_2)$, we can find the midpoint of the line connecting such points using the formula,

$\displaystyle \mbox { Midpoint } = \left( \frac {x_1 + x_2}{2}, \frac {y_1 + y_2}{2} \right)$

Now, our two points are $\displaystyle (3p,8)$ and $\displaystyle (9,4p)$

Thus, the midpoint is given by: $\displaystyle M = \left( \frac {3p + 9}{2}, \frac {4p + 8}{2} \right)$

Now we want the midpoint to be on the y-axis. the x-coordinate for all points on the y-axis is zero. Therefore, to find the p that satisfies this condition, we must set the x-coordinate of the midpoint to zero. Hence, we end up with TPH's equation,

$\displaystyle \frac {3p + 9}{2} = 0$

$\displaystyle \Rightarrow p = -3$

In light of the actual value of p, your diagram, while it exemplifies great artistic skill, is somewhat misleading

5. yeah i understood... right after i posted my diagram! it's just i read his signature as being part of the explanation, so i didn't understand how that had any relevance to the question.
i like drawing diagrams! they help me solve things, i just thought it would help. But thanks for the explanation anyway!

6. Originally Posted by blurgh
yeah i understood... right after i posted my diagram! it's just i read his signature as being part of the explanation, so i didn't understand how that had any relevance to the question.
i like drawing diagrams! they help me solve things, i just thought it would help. But thanks for the explanation anyway!
oh, yeah. i think you mentioned that you read his signature, and i laughed at you (we're friends so i can do that, right?) but one of the moderators must have deleted those two posts