Tell me, must I use m = (y2 - y1)/(x2 - x1), where m is 1/4?

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In summary: I am sorry if I offended you.In summary, RTCNTC is one of my coworkers who is trying to review precalculus through this website. He is planning to return to college next semester. I also help him tremendously through PM here and through personal email. I met RTCNTC at mathhelpforum.com. I joined this website to get help with calculus 3. Secondly, I love math as much as anyone here. I love the fact that math is black or white. After learning calculus 3, I plan to continue into linear algebra. Learning is fun. Learning from others who know more than I do is great.
  • #1
mathdad
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Find point P on the curve y = sqrt{x} such that the slope of the line passing through P and (1, 1) is 1/4.

Tell me, must I use m = (y2 - y1)/(x2 - x1), where m is 1/4?

Seeking point P means we need the coordinates of P or (x, y), right?
 
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  • #2
Yes, we have the points:

\(\displaystyle (x,\sqrt{x}),\,(1,1)\)

and we want to set the slope of the line segment joing those points to 1/4:

\(\displaystyle \frac{1-\sqrt{x}}{1-x}=\frac{1}{4}\)

Now solve for $x$...
 
  • #3
Ok. I knew I was on the right track.
 
  • #4
(1 - sqrt{x})/(1 - x) = 1/4

(1 - x) = 4(1 - √x)

(1 - x) = 4 - 4√x

1 - x - 4 = -4√x

(-x - 3) = -4√x

(-x - 3)/(-4) = √x

After further simplifying, I got x = 9.

Correct?

If so, then point P = (9, 3).
 
  • #5
$$\frac{1-\sqrt x}{1-x}=\frac14$$

$$\frac{1-\sqrt x}{(1-\sqrt x)(1+\sqrt x)}=\frac14$$

$$\frac{1}{1+\sqrt x}=\frac14$$

$$3=\sqrt x\implies y=3$$

$$9=x\implies x=9$$

$$P_{x,y}=(9,3)$$
 
  • #6
Taking a break from precalculus and math in general. Thank you everyone.
 
  • #7
greg1313 said:
$$\frac{1-\sqrt x}{1-x}=\frac14$$

$$\frac{1-\sqrt x}{(1-\sqrt x)(1+\sqrt x)}=\frac14$$

$$\frac{1}{1+\sqrt x}=\frac14$$

$$3=\sqrt x\implies y=3$$

$$9=x\implies x=9$$

$$P_{x,y}=(9,3)$$

Nicely done for him.
 
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  • #8
Harpazo said:
I tried to stay away from math but could not do it. I am stuck on math.

Erm... wasn't it RTCNTC that thought to stay away...?
 
  • #9
I like Serena said:
Erm... wasn't it RTCNTC that thought to stay away...?

What are you talking about?
 
  • #10
Harpazo said:
What are you talking about?

RTCNTC said he wanted to take a break from math.
And you responded that you didn't want to take a break after all.
That suggests that you are the same person.
That's okay really, although it annoys me that you pretend to be different people.
 
  • #11
Just for the record, math is about absolute truths.
We can disagree with each other, but in the context of math there really is only one truth - black and white.
This is what I like about math.
After only a couple of posts we can (usually) always get to the bottom of it, and agree on who is right and who is wrong.
It's not subjective, and it's not about who we like or dislike, the mathematical truth is absolute.
At times I will be wrong, and I will be glad to admit it when I'm confronted with proof that I am, and consider it a win because I learned something new.
What I dislike is people pretending that things are other than they are. It shows a disrespect to math in general, and it offends my sensibilities.
 
  • #12
I like Serena said:
RTCNTC said he wanted to take a break from math.
And you responded that you didn't want to take a break after all.
That suggests that you are the same person.
That's okay really, although it annoys me that you pretend to be different people.

Since you are so curious, I'll you tell who RTCNTC is. RTCNTC is one of my coworkers who is trying to review precalculus through this website. He is planning to return to college next semester. I also help him tremendously through PM here and through personal email. I met RTCNTC at mathhelpforum.com. I joined this website to get help with calculus 3.
 
  • #13
I like Serena said:
Just for the record, math is about absolute truths.
We can disagree with each other, but in the context of math there really is only one truth - black and white.
This is what I like about math.
After only a couple of posts we can (usually) always get to the bottom of it, and agree on who is right and who is wrong.
It's not subjective, and it's not about who we like or dislike, the mathematical truth is absolute.
At times I will be wrong, and I will be glad to admit it when I'm confronted with proof that I am, and consider it a win because I learned something new.
What I dislike is people pretending that things are other than they are. It shows a disrespect to math in general, and it offends my sensibilities.

Since you are so curious, I'll you tell who RTCNTC is. RTCNTC is one of my coworkers who is trying to review precalculus through this website. He is planning to return to college next semester. I also help him tremendously through PM here and through personal email. I met RTCNTC at mathhelpforum.com. I joined this website to get help with calculus 3.

Secondly, I love math as much as anyone here. I love the fact that math is black or white. After learning calculus 3, I plan to continue into linear algebra. Learning is fun. Learning from others who know more than I do is great.

I get hurt when people accuse me of being someone else. I am HARPAZO which is the Greek word for BEING CAUGHT UP VIOLENTLY. I have no clue what RTCNTC means. Why don't you ask him? Don't run to crazy conclusions without knowing the facts. I am learning calculus 3. Why on Earth would I post precalculus questions when I am beyond anything before calculus 3?
 

FAQ: Tell me, must I use m = (y2 - y1)/(x2 - x1), where m is 1/4?

What does the formula m = (y2 - y1)/(x2 - x1) represent?

The formula m = (y2 - y1)/(x2 - x1) represents the slope of a line passing through two points (x1,y1) and (x2,y2).

Why is the formula m = (y2 - y1)/(x2 - x1) used?

The formula m = (y2 - y1)/(x2 - x1) is used to calculate the slope of a line, which is a measure of how steep the line is. It is a commonly used formula in mathematics and physics.

When should I use the formula m = (y2 - y1)/(x2 - x1)?

The formula m = (y2 - y1)/(x2 - x1) should be used when you need to find the slope of a line passing through two points on a graph or in a coordinate system.

Can I use a different formula to calculate the slope of a line?

Yes, there are other formulas that can be used to calculate the slope of a line, such as the point-slope form or the slope-intercept form. However, the formula m = (y2 - y1)/(x2 - x1) is the most commonly used and easiest to apply.

Is the formula m = (y2 - y1)/(x2 - x1) applicable to all types of lines?

Yes, the formula m = (y2 - y1)/(x2 - x1) can be used to find the slope of any type of line, including straight lines, curved lines, and vertical or horizontal lines.

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