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.

- anonymous

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- 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

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- anonymous

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

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- anonymous

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- anonymous

This is my working out.

- welshfella

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

- 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

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

- anonymous

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

- 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

looks like that is your problem

- 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

textbooks are sometimes wrong, i'm afraid

- welshfella

(but so am i sometimes!)

- welshfella

have you checked my calculations ?

- anonymous

Will do that now. :)

- welshfella

i used decimals when i can - I hate fractions!

- anonymous

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

- welshfella

I plugged in A (-2,1)

- anonymous

Aha! Oh I see.

- anonymous

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

- welshfella

yw

- 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

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

- welshfella

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

- welshfella

anyway gotta go
good luck with your studies

- 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

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

*** 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

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

- phi

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

Oh! No, I have not done vectors yet.

- anonymous

Ok so I got L incorrect then?

- anonymous

Thanks!

- 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

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

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

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

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

- welshfella

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

- 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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