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
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
To find the length of BC, we can use the distance formula. The distance formula is given by:
Distance = √[(x2 - x1)^2 + (y2 - y1)^2]
Let's calculate the length of BC:
BC = √[(-1 - 4)^2 + (2 - (-1))^2]
= √[(-5)^2 + (3)^2]
= √[25 + 9]
= √34
Next, let's find the equation of the line passing through points B and C, which is the line BC. We can use the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept.
The slope (m) can be determined using the formula:
m = (y2 - y1) / (x2 - x1)
Using the coordinates of B(4,-1) and C(-1,2):
m = (2 - (-1)) / (-1 - 4)
= 3 / (-5)
= -3/5
Now, we have the slope of line BC, which is -3/5. To find the equation of the line, we need a point on the line. Let's use point B(4, -1).
Using the point-slope form of a line, y - y1 = m(x - x1):
y - (-1) = (-3/5)(x - 4)
y + 1 = (-3/5)(x - 4)
y + 1 = (-3/5)x + (12/5)
y = (-3/5)x + (12/5) - 1
y = (-3/5)x + (12/5) - (5/5)
y = (-3/5)x + 7/5
Now, we have the equation of line BC, which is y = (-3/5)x + 7/5.
To find the perpendicular distance from point A(2,3) to line BC, we need to find the equation of the line perpendicular to BC that passes through point A.
The slope of a line perpendicular to line BC is the negative reciprocal of the slope of line BC. The slope of line BC is -3/5, so the slope of the perpendicular line is 5/3.
Using the point-slope form of a line, y - y1 = m(x - x1), and substituting the values x1 = 2, y1 = 3, and m = 5/3:
y - 3 = (5/3)(x - 2)
y - 3 = (5/3)x - (10/3)
y = (5/3)x - (10/3) + 3
y = (5/3)x - (10/3) + (9/3)
y = (5/3)x - (1/3)
Now, we have the equation of the perpendicular line from A, which is y = (5/3)x - (1/3).
To find the point where the perpendicular line intersects line BC, we need to solve the system of equations formed by the equations of line BC and the perpendicular line.
Solving the system of equations:
(-3/5)x + (7/5) = (5/3)x - (1/3)
To simplify the equation, let's clear the fractions by multiplying through by 15.
-9x + 21 = 25x - 5
Adding 9x and subtracting 21 from both sides:
0 = 34x - 26
Adding 26 to both sides:
34x = 26
Dividing by 34:
x = 26/34
Simplifying:
x = 13/17
Now, substitute this value back into the equation of line BC to find y:
y = (-3/5)(13/17) + 7/5
y = (-39/85) + (119/85)
y = 80/85
y = 16/17
So, we have found the point of intersection between line BC and the perpendicular line from A, which is (13/17, 16/17).
Finally, we can calculate the area of triangle ABC using the formula for the area of a triangle given the lengths of two sides and the included angle.
The length of BC is √34, and the perpendicular distance from A to BC is the distance from point A to the point of intersection between line BC and the perpendicular line passing through A. We can use the distance formula again:
Distance = √[(x2 - x1)^2 + (y2 - y1)^2]
Distance = √[(13/17 - 2)^2 + (16/17 - 3)^2]
= √[(13/17 - 34/17)^2 + (16/17 - 51/17)^2]
= √[(-21/17)^2 + (-35/17)^2]
= √[(21/17)^2 + (35/17)^2]
= √[(441/289) + (1225/289)]
= √(1666/289)
= √(1666)/√(289)
= √(1666)/17√(2)
Then, we can calculate the area of triangle ABC using the formula:
Area = 1/2 * Base * Height
Area = 1/2 * √34 * √(1666)/17√(2)
Area = √(34 * 1666)/34√(2)
Area = √(56156)/34√(2)
Area = 236.76/17.88
Area = 13.23
Therefore, the area of triangle ABC is approximately 13.23 square units.
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