## anonymous 4 years ago Find the distance from the point (2,-1) to the line y = 2x + 3

1. amistre64

y = -1/2 x +(2)/2 - 1 y = -1/2 x seems to be the line with the point on it system of equation to find the intersection then distance it from point to point

2. amistre64

2x+3 = -1/2 x 5/2 x = -3 x = -3/(5/2) = -6/5 maybe :)

3. amistre64

-1/2 * -6/5 = 6/10 = 3/5 then 2(-6/5) + 3 = -12/5 + 15/5 = 3/5 so thats a match

4. anonymous

My correct answer says : 8/5 * sqrt(5), could you help me to get there?

5. amistre64

(10/5, -5/5) - (-6/5 , 10/5) --------------- (16/5)^2 + (-15/5)^2 sqrt(16^2+15^2)/5 should be it then

6. amistre64

seeing how i thought my first explanation was suffieicnt; youll have to address where it is your following goes amiss at

7. amistre64

other than typos :)

8. amistre64

|dw:1327864212303:dw|

9. amistre64

y = 2x + 3 needs a perp line to it; perp line is just flip and negate the slope 2 flips to -1/2 y = -1/2 x + b ; use (2,-1) to calibrate 2 = -1(-1)/2 + c 2 = 2 + c ; c = 0 y = -1/2 x is equation of line perp to and containing (2,-1)

10. amistre64

if we know where the lines meet, we have a point of reference to measure distance with ...|dw:1327864376411:dw|

11. amistre64

y = 2x + 3 -( y = -1/2 x) ------------- 0 = 5/2 x + 3 ; when x = -3/5

12. amistre64

y = -1/2 * -3/5 = 3/10 therefore, point of intersection is: (-3/2 , 3/10) looks better

13. amistre64

(-3/5 , 3/10) that is

14. amistre64

|dw:1327864617337:dw|

15. amistre64

use distance formula for distance

16. amistre64

ugh!!, lost my x ... 5/2 * -2/5 * 3 = -6/5 0 = 5*-6/2*5 + 3 0 = -30/10 + 3 = -3+3 x = -6/5 ....

17. amistre64

y = -1/2*-6/5 = 6/10 = 3/5 (-6/5, 3/5) is the point of intersection .....

18. amistre64

$d=\sqrt{(2-\frac{-6}{5})^2+(-1-\frac{3}{5})^2}$ $d=\sqrt{(\frac{10+6}{5})^2+(\frac{-5-3}{5})^2}$ $d=\sqrt{(\frac{16}{5})^2+(\frac{-8}{5})^2}$ $d=\frac{\sqrt{16^2+8^2}}{5}$

19. amistre64

that simplifies to 8sqrt(5)/5

20. anonymous

|dw:1327864993315:dw|

21. anonymous

that is what we found?

22. amistre64

yes

23. anonymous

Why doesn't using the pythagorean theorem work here?

24. amistre64

it does, but first you need to know 2 points. the given point (2,-1) and the intersection of the perped lines

25. amistre64

once you know the 2 points, the distance formula IS the pythag thrm

26. anonymous

" the intersection of the perped lines" confused me, I am really stumped because, I still don't understand why we can't just draw the right triangle from the y axis to the point 2,-1

27. amistre64

because you dont want the distance from the y axis, you want the distance to the line and the point; that distance is defined as the shortest distance between the line and the point. which just so happens to make a 90degree angle when it hits|dw:1327865470057:dw|

28. amistre64

teh y axis has nothing to do with it at all

29. anonymous

OH, THE POINT OF PERPENDICULAR(ness)? to the point 2,-1?

30. amistre64

yes

31. anonymous

which is the part where we found the perp. point? that is the next step correct?

32. anonymous

we do the negative inversse, OH, then we go from the point 2,-1 towards the other line, using the perpendicular slope, is that right?

33. anonymous

until there is an intersection

34. amistre64

yep

35. anonymous

is there an equation to find that point?

36. amistre64

its solving a system of equations; y = 2x + 3 y = -1/2x we can sub, eliminate, or any other method to solve the sytem

37. amistre64

it turns out that i used elimination and came up with those fractions

38. amistre64

y = -1/2 x -2y = x, sub in top y = 2(-2y) + 3 y = -4y + 3 5y = 3 y = 3/5

39. amistre64

knowing y, solve for x

40. amistre64

3/5 = -1/2 x -2*3/5 = x -6/5 = x

41. amistre64

those x and y parts are the point of intersection

42. anonymous

ahhh, i understand now... now i use the distance formula between the points (-6/5, 3/5) and (2,-1) Hey, i'd like to thank you for spending about an hour of your time to help me understand this, i appreciate it, much kudos to you.

43. amistre64

youre welcome :)