maths
posted by Anonymous on .
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

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 
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
y3 = (5/3)(x2)
3y  9 = 5x  10
5x  3y = 1  #1
equation of BC , using C
y2 = (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
= √(213/17)^ + (316/17)^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