Asked by Anonymous
the points A(2,3) B(4,-1) C(-1,2) are the vertices of a triangle. find the length and perpendicular from A to BC and hence the area of ABC
Answers
Answered by
bobpursley
a. Find the equation of the line BC, given two points (B and c)
Then, find the perpendicular (given the negative inverse slope from a), and the point A.
Area? Area=lengthBC*lengthperpendicular*1/2
Just finding the area, there are other ways easier.
Graph it, use Pick's Theorem http://en.wikipedia.org/wiki/Pick%27s_theorem
or
http://www.mathopenref.com/coordtrianglearea.html
or to check your answer ONLY: http://www.gottfriedville.net/mathtools/triarea.html
Then, find the perpendicular (given the negative inverse slope from a), and the point A.
Area? Area=lengthBC*lengthperpendicular*1/2
Just finding the area, there are other ways easier.
Graph it, use Pick's Theorem http://en.wikipedia.org/wiki/Pick%27s_theorem
or
http://www.mathopenref.com/coordtrianglearea.html
or to check your answer ONLY: http://www.gottfriedville.net/mathtools/triarea.html
Answered by
Reiny
consider BC as your base:
BC = √(5^2 + (-3)^2) = √34
slope of BC = -3/5
so the slope of the perpendicular = 5/3
equation of line from A to BC , using A as the point
y-3 = (5/3)(x-2)
3y - 9 = 5x - 10
5x - 3y = 1 ---- #1
equation of BC , using C
y-2 = -(3/5)(x+1)
5y - 10 = -3x -3
3x + 5y = 7 ----#2
5 times #1 --- 25x - 15y = 5
3 times #2 --- 9x + 15y = 21
add them
34x = 26
x = 26/34 = 13/17 .... ughh
in #2
3(13/17) + 5y = 7
..
y = 16/17 , yuk, the point on BC = (13/17 , 16/17)
so length of altitude
= √(2-13/17)^ + (3-16/17)^2)
= √(441/289 + 1125/289) = √(1666/289)
= 7√34/17
so the area = (1/2) base x heigh
= (1/2) * √34 * (7/17)√34 = 7 , yeahhhh
There are of course much easier ways to find the area of a triangle if you are given the three points.
The simplest way is to list the 3 points in a column repeating the first one you listed.
2 3
4 -1
-1 2
2 3
area = (1/2) | ( sum of downproducts - sum of upproducts)|
= (1/2)|(-2+ 8 - 3 -(12 + 1 + 4)|
= (1/2)| -14 |
= 7
BC = √(5^2 + (-3)^2) = √34
slope of BC = -3/5
so the slope of the perpendicular = 5/3
equation of line from A to BC , using A as the point
y-3 = (5/3)(x-2)
3y - 9 = 5x - 10
5x - 3y = 1 ---- #1
equation of BC , using C
y-2 = -(3/5)(x+1)
5y - 10 = -3x -3
3x + 5y = 7 ----#2
5 times #1 --- 25x - 15y = 5
3 times #2 --- 9x + 15y = 21
add them
34x = 26
x = 26/34 = 13/17 .... ughh
in #2
3(13/17) + 5y = 7
..
y = 16/17 , yuk, the point on BC = (13/17 , 16/17)
so length of altitude
= √(2-13/17)^ + (3-16/17)^2)
= √(441/289 + 1125/289) = √(1666/289)
= 7√34/17
so the area = (1/2) base x heigh
= (1/2) * √34 * (7/17)√34 = 7 , yeahhhh
There are of course much easier ways to find the area of a triangle if you are given the three points.
The simplest way is to list the 3 points in a column repeating the first one you listed.
2 3
4 -1
-1 2
2 3
area = (1/2) | ( sum of downproducts - sum of upproducts)|
= (1/2)|(-2+ 8 - 3 -(12 + 1 + 4)|
= (1/2)| -14 |
= 7
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