Question

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.

Answer 1

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!

Answer 2

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

Answer 3

This is my working out.

Answer 4

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

Answer 5

- 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?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

looks like that is your problem

Answer 10

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)

Answer 11

textbooks are sometimes wrong, i'm afraid

Answer 12

(but so am i sometimes!)

Answer 13

have you checked my calculations ?

Answer 14

Will do that now. :)

Answer 15

i used decimals when i can - I hate fractions!

Answer 16

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

Answer 17

I plugged in A (-2,1)

Answer 18

Aha! Oh I see.

Answer 19

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

Answer 20

yw

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

anyway gotta go
good luck with your studies

Answer 25

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)

Answer 26

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

Answer 27

*** 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.

Answer 28

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

\[ Area= \frac{1}{2} \cdot \frac{\sqrt{10}}{2} \cdot \frac{3\sqrt{10}}{2} = \frac{30}{8}=3.75\]

Answer 29

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
\[ Area = \frac{1}{2} | u \times v| \]

\[ = \frac{1}{2} | u \times v| = \frac{1}{2}\left|\begin{matrix}i & j & k \\ 0 & 5 & 0 \\ 1.5 & 0.5 & 0\end{matrix}\right|\\
\frac{1}{2}\left| 0i +0j +7.5k\right|= \frac{7.5}{2}= 3.75
\]

Answer 30

Oh! No, I have not done vectors yet.

Answer 31

Ok so I got L incorrect then?

Answer 32

Thanks!

Answer 33

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)

Answer 34

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!

Answer 35

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

Answer 36

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)

Answer 37

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

Answer 38

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

Answer 39

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 :) )