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## anonymous 5 years ago What is the distance from the point (6,0) to the line y=3x+2? Delete Cancel Submit

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1. amistre64
• 5 years ago
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which distance you want? shortest most likly

2. amistre64
• 5 years ago
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the shortest distance is a perpendicular line....

3. amistre64
• 5 years ago
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so to find it, we need the line that is perp to this and the point they intersect

4. anonymous
• 5 years ago
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Possible answers are 4, $2\sqrt{10} , \sqrt{26} , and\ 3\sqrt{6}$ so I guess which ever equals one of those

5. amistre64
• 5 years ago
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the perp has a product of slopes that equals -1 3x = -1 x = -1/3 0 = -1/3(6) +b b = 2 y = (-1/3)x +2

6. amistre64
• 5 years ago
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these lines meet at: 3x+2 = (-1/3)x +2 3x + 1/3x = 0 10/3x = 0 x = 0

7. amistre64
• 5 years ago
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they appear to meet at the origin...

8. amistre64
• 5 years ago
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they meet at (0,2) if anything lol

9. amistre64
• 5 years ago
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id go with 4

10. anonymous
• 5 years ago
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Per distance formula is $d= \frac{ \left| ax +by +c \right| }{ \sqrt{a^2 +b^2}}$

11. amistre64
• 5 years ago
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im prolly wrong since i guessed that lol

12. amistre64
• 5 years ago
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y = 3x+2 ..... 0 = -6/3 +b 0 = -2 +b b = 2 y = (-1/3)x +2.... that seems to work out

13. anonymous
• 5 years ago
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y= 3x+2 get into general form 3x - y + 2 =0 so this means that a=3 , b=-1 and c=2 in our formulas above, and (x,y) are the coordinates of the point ( 6,0) so x=6 , y=0

14. anonymous
• 5 years ago
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d= (9 +2 ) / 2 = 4.5 units

15. anonymous
• 5 years ago
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wait thats wrong

16. amistre64
• 5 years ago
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3x+2 = (-1/3)x +2 3x = -1/3x when x=0 right?

17. amistre64
• 5 years ago
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(0,2) (6,0) 36 + 4 = 40 = sqrt(40)

18. amistre64
• 5 years ago
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2sqrt(10)

19. amistre64
• 5 years ago
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got it ;)

20. anonymous
• 5 years ago
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d = $\frac { \left| 3(6) + 0 + 2 \right|}{\sqrt{10} }$

21. amistre64
• 5 years ago
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22. anonymous
• 5 years ago
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20/ sqrt (10)

23. anonymous
• 5 years ago
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wtf

24. amistre64
• 5 years ago
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20sqrt(10) --------- 10

25. anonymous
• 5 years ago
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yes I know

26. amistre64
• 5 years ago
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2 sqrt(10)

27. amistre64
• 5 years ago
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need a calculator? :)

28. anonymous
• 5 years ago
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Very puzzling question lol

29. amistre64
• 5 years ago
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just had to get the sleep of my ears to see what i was doin

30. anonymous
• 5 years ago
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Well thank you

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