anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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!
anonymous
  • anonymous
anonymous
  • anonymous
That is the triangle given to me. I did not draw my own.

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anonymous
  • anonymous
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anonymous
  • anonymous
This is my working out.
welshfella
  • welshfella
i agree that L = (2,4) and BL = sqrt(10)
welshfella
  • 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?
anonymous
  • anonymous
I put the x coordinate into the equation for line AL and it said when x=1 y would equal 2
anonymous
  • anonymous
The equation I put it into was x-3y+5=0
welshfella
  • 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
welshfella
  • welshfella
looks like that is your problem
anonymous
  • 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)
welshfella
  • welshfella
textbooks are sometimes wrong, i'm afraid
welshfella
  • welshfella
(but so am i sometimes!)
welshfella
  • welshfella
have you checked my calculations ?
anonymous
  • anonymous
Will do that now. :)
welshfella
  • welshfella
i used decimals when i can - I hate fractions!
anonymous
  • anonymous
I can't quite see what you did with the equation y-y1=m(x-x1) , why plug in H ?
welshfella
  • welshfella
I plugged in A (-2,1)
anonymous
  • anonymous
Aha! Oh I see.
anonymous
  • anonymous
Ok, I think I understand now and see where I went wrong. Thank you very much for your help!
welshfella
  • welshfella
yw
welshfella
  • 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
anonymous
  • anonymous
Very interesting. Oh well I guess your right, the book won't always be right!
welshfella
  • welshfella
yea - a bit frustrating though - I don't like mysteries! - It makes me fell like - 'Was I wrong, after all?' lol
welshfella
  • welshfella
anyway gotta go good luck with your studies
phi
  • 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)
phi
  • 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
phi
  • 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.
phi
  • phi
\[ |HL|= \sqrt{ \left(\frac{5}{2}-1 \right)^2 + \left(\frac{5}{2}-2 \right)^2 } = \frac{\sqrt{10}}{2}\] \[ |BL|= \sqrt{ \left(\frac{5}{2}-1 \right)^2 + \left(\frac{5}{2}-7 \right)^2 } = \frac{3\sqrt{10}}{2}\] \[ Area= \frac{1}{2} \cdot \frac{\sqrt{10}}{2} \cdot \frac{3\sqrt{10}}{2} = \frac{30}{8}=3.75\]
phi
  • 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 \[ Area = \frac{1}{2} | u \times v| \] \[ = \frac{1}{2} | u \times v| = \frac{1}{2}\left|\begin{matrix}i & j & k \\ 0 & 5 & 0 \\ 1.5 & 0.5 & 0\end{matrix}\right|\\ \frac{1}{2}\left| 0i +0j +7.5k\right|= \frac{7.5}{2}= 3.75 \]
anonymous
  • anonymous
Oh! No, I have not done vectors yet.
anonymous
  • anonymous
Ok so I got L incorrect then?
anonymous
  • anonymous
Thanks!
phi
  • 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)
anonymous
  • 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!
phi
  • 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|
phi
  • 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)
welshfella
  • welshfella
ah I assumed that it bisected the base as well. I don't know why!!???
welshfella
  • welshfella
my only excuse is that i'd just got up!
anonymous
  • 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 :) )

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