Posted by **Anonymous** on Saturday, February 2, 2013 at 8:11am.

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

- maths -
**bobpursley**, Saturday, February 2, 2013 at 9:31am
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

- maths -
**Reiny**, Saturday, February 2, 2013 at 9:36am
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

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