## anonymous one year ago I don't know where I went wrong, please help! The question: 7. The diagram (which I will post a picture of) shows a triangle whose vertices are A(-2,1), B(1,7) and C(3,1). The point L is the foot of the perpendicular line from A to BC, and M is the foot of the perpendicular from B to AC. (vi) Find the area of the triangle BLH. So what I did was to find L by finding the midpoint of BC. I got L=(2,4). Then I found the length of the base, BL, which equaled the square root of 10. I also found the length of the height of the triangle, HL, which equaled the square root of 5.

1. anonymous

I then multiplied the height and the base together and divided the answer by two to get Area=3.536 however the answer says it is actually 3.75. Please tell me what I did wrong!

2. anonymous

3. anonymous

That is the triangle given to me. I did not draw my own.

4. anonymous

5. anonymous

This is my working out.

6. welshfella

i agree that L = (2,4) and BL = sqrt(10)

7. welshfella

- but I don't see how you get H to be (1,2). I see why the x coordinate = 1, but how did you get y = 2?

8. anonymous

I put the x coordinate into the equation for line AL and it said when x=1 y would equal 2

9. anonymous

The equation I put it into was x-3y+5=0

10. welshfella

slope of AL = 4-1 / (2--2) = 3/4 y - y1 = 0.75(x - x1) y - 1 = 0.75(x - (-2)) y - 1 = 0.75x + 1.5 y = 0.75x + 2.5 when x = 1 y = 3.25

11. welshfella

looks like that is your problem

12. anonymous

This is actually really confusing because I am working through a textbook I have and this is a question from the book that I am stuck on and in the answers section it says that (1,2) is right, so they are wrong? because then that is slightly annoying (although I thank you for telling me otherwise I would have spent a lot of time on that)

13. welshfella

textbooks are sometimes wrong, i'm afraid

14. welshfella

(but so am i sometimes!)

15. welshfella

have you checked my calculations ?

16. anonymous

Will do that now. :)

17. welshfella

i used decimals when i can - I hate fractions!

18. anonymous

I can't quite see what you did with the equation y-y1=m(x-x1) , why plug in H ?

19. welshfella

I plugged in A (-2,1)

20. anonymous

Aha! Oh I see.

21. anonymous

Ok, I think I understand now and see where I went wrong. Thank you very much for your help!

22. welshfella

yw

23. welshfella

the only thing is that gives a different answer HL = sqrt(( 4 - 3.25)^2 + 1^2) = 1.25 and area of triangle BLH = (sqrt10 * 1.25) / 2 = 1.98 sq units

24. anonymous

Very interesting. Oh well I guess your right, the book won't always be right!

25. welshfella

yea - a bit frustrating though - I don't like mysteries! - It makes me fell like - 'Was I wrong, after all?' lol

26. welshfella

anyway gotta go good luck with your studies

27. phi

Using points (1,7) to (3,1),l the slope= -3 and the equation of the line is $y = -3x+10$ check: x=1 gives y= -3+10= 7 the line perpendicular to line BC will have slope 1/3 (negative reciprocal) going through point A (-2,1) we get the equation $y= \frac{x+5}{3}$ using this 2nd equation, at x= 1 we get y= 2. i.e. H (1,2) point L is the intersection of the two lines $-3x+10= \frac{x+5}{3}$ from which we find x= 5/2, and then y= 5/2 L (5/2, 5/2)

28. phi

we now have B (1,7) L (5/2 , 5/2) H (1,2) we could use vectors or we could use the distance formula to find the area of this triangle. distance formula: find base HL and height LB

29. phi

*** find L by finding the midpoint of BC.*** We are not justified in assuming the altitude bisects its base. In might, but not in general, and not in this problem.

30. phi

$|HL|= \sqrt{ \left(\frac{5}{2}-1 \right)^2 + \left(\frac{5}{2}-2 \right)^2 } = \frac{\sqrt{10}}{2}$ $|BL|= \sqrt{ \left(\frac{5}{2}-1 \right)^2 + \left(\frac{5}{2}-7 \right)^2 } = \frac{3\sqrt{10}}{2}$ $Area= \frac{1}{2} \cdot \frac{\sqrt{10}}{2} \cdot \frac{3\sqrt{10}}{2} = \frac{30}{8}=3.75$

31. phi

If you know vectors, we can use this approach: let u = <1,7> - < 1,2> = <0,5> and v= <2.5, 2.5> - <1,2>= <1.5, 0.5> extend these to 3 dimensions (set z coord to 0), u= <0,5,0> v= <1.5,0.5,0> so we can use $Area = \frac{1}{2} | u \times v|$ $= \frac{1}{2} | u \times v| = \frac{1}{2}\left|\begin{matrix}i & j & k \\ 0 & 5 & 0 \\ 1.5 & 0.5 & 0\end{matrix}\right|\\ \frac{1}{2}\left| 0i +0j +7.5k\right|= \frac{7.5}{2}= 3.75$

32. anonymous

Oh! No, I have not done vectors yet.

33. anonymous

Ok so I got L incorrect then?

34. anonymous

Thanks!

35. phi

yes, you got L wrong. This problem is solved using these ideas: 1) two points define a line (to find BC) 2) perpendicular lines have negative reciprocal slopes 3) point-slope formula to find the equation for line AL 4) line BM is x=1 now find the intersection points H and L once we have the points of the vertices of triangle BHL we can find its area as legs HL and LB form the base and height , we can use the distance formula if they were not perpendicular, we could find the lengths of all 3 legs and then use heron's formula. (that would be messy)

36. anonymous

Ok. One thing though, I assumed that L was the midpoint of BC and thats how I found it. I don't really see how thats wrong but since you found it the other way I assume it is wrong? Could you explain why? Sorry for taking so much of your time!

37. phi

I posted the reasons up above. In general, an altitude does not bisect its base. (in fact in may not even intersect the base) |dw:1441805596641:dw|

38. phi

in other words, it is wrong to assume L is the midpoint of segment BC if you knew you had an isosceles (or equilateral ) triangle, the altitude would bisect the base. But we don't know that here (and in fact it turns out that the altitude does not bisect the base in this case)

39. welshfella

ah I assumed that it bisected the base as well. I don't know why!!???

40. welshfella

my only excuse is that i'd just got up!

41. anonymous

Oh Ok! Thanks so much for all the help given to me on this question. (Oh and if you ask me, just getting up is a perfectly good excuse :) )