A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

moongazer

  • 5 years ago

The distance of point (-1, 4) and the line 4x – 2y = 4 is ___.

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    8/sqart 5

  2. moongazer
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    how did you do it?

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[8/\sqrt{5}\]

  4. moongazer
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    how?

  5. moongazer
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Please explain.

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait, wait

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    4x-2y-4=0 put the values of x and y in the LHS then take the modulus of that value and divide it by sqrt of sum of the squares of coeffients of x and y. i think, u know the proof why we do that

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry, co-efficients ^^

  9. moongazer
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Sorry but I didn't get it.

  10. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    since the distance from a point to a line is along a perpendiular line; if we can construct a perpendicular line equation that goes thru the given point we can use a system of equations to determine the point where they meet. then use the distance formula to determine the distance from the meeting point and the given point.

  11. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1326977726363:dw|

  12. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    one trick to a perp line is to swap coeffs and negate one of them: 4x – 2y = 4 perp line: 2x + 4y = n calibrate with the given point (-1,4): 2(-1) + 4(4) = n -2 + 16 = n = 14 perp line: 2x + 4y = 14

  13. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    system of equations gives us: 4x - 2y = 4 2x + 4y = 14 ; *-2 4x - 2y = 4 -4x -8y = -28 ------------ -10y = -24 y = 24/10 = 12/5 .............................................. 4x - 2y = 4 ;*2 2x + 4y = 14 8x - 4y = 8 2x + 4y = 14 ------------ 10x = 22 x = 22/10 = 11/5 ................................ we want the distance from (-1,4) to (12/5, 11/5) if i did that right

  14. moongazer
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I think this is the explanation that I am looking for.

  15. moongazer
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    could I use the formula d=(/Ax1+By1+C/)/sqrtof(a^2+B^2) I think this is the formula for this. :)

  16. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[dist.=\sqrt{a^2+b^2}\] is a compact form of the distance formula yes; personally I just subtract the points, square them, add them and sqrt them

  17. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ( 12/5, 11/5) - ( -5/5, 20/5) --------------- (17/5 , -9/5) ^2 (289+81)/25 sqrt(370)/5

  18. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    that should simplify to 8/sqrt(5)

  19. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    but then some texts hate a sqrt in the denom; so maybe we rewrite it: \(cfrac{8}{5}\sqrt{5}\)

  20. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    well that dint format good on my screen lol

  21. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\frac{8\sqrt{5}}{5}\] or \[\frac{8}{5}\sqrt{5}\]

  22. moongazer
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[ d= (\left| Ax _{1}+Bx _{2}+C \right|)/\sqrt{A ^{2}+B ^{2}}\] This is what my teacher taught us :)

  23. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    that might work :) ive never quite seen it like that tho.

  24. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.