## moongazer 5 years ago The distance of point (-1, 4) and the line 4x – 2y = 4 is ___.

1. anonymous

8/sqart 5

2. moongazer

how did you do it?

3. anonymous

$8/\sqrt{5}$

4. moongazer

how?

5. moongazer

6. anonymous

wait, wait

7. anonymous

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

sorry, co-efficients ^^

9. moongazer

Sorry but I didn't get it.

10. amistre64

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

|dw:1326977726363:dw|

12. amistre64

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

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

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

15. moongazer

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

16. amistre64

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

17. amistre64

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

18. amistre64

that should simplify to 8/sqrt(5)

19. amistre64

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

20. amistre64

well that dint format good on my screen lol

21. amistre64

$\frac{8\sqrt{5}}{5}$ or $\frac{8}{5}\sqrt{5}$

22. moongazer

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

23. amistre64

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