Find the area of a triangle whose vertices have coordinates (2,-5), (6,1), and (-3,-4)
let the triangle be ABC with A,B, and C matching the order of coordinate points the way they are written above.
Step 1: find equation for BC
slope = 5/9
equation is 5x-9y = 21
length of BC = √(9^2 + 5^2) = √106, I will use that as the base of the triangle.
distance for A to line BC (my height of triangle)
=│5(2) - 9(-5) - 21│/√106
= 34/√106
area of triangle = 1/2 base*height
= 1/2(√106)(34/√106) = 17
In addition to several methods for calculating the area of a triangle given the coordinates of its vertices, the area can be determined using only the coodinates themselves.
Given the rectangular coordinates of the three vertices, the area of a triangle whose vertices are P1(x1,y1), P2(x2,y2), P3(x3,y3) is given by Area triangle P1P2P3 = 1/2[(x1y2 - x2y1) + (x2y3 - x3y2) + (x3y1 - x1y3)].
Method:
First step-Write down the vertices in two columns, abscissas in one, ordinates in the other, repeating the coordinates of the first vertex. They must be in order around the triangle, clockwise preferably.
Second step-Multiply each abscissa by the ordinate of the next row, and add the results. This gives x1y2+x2y3+x3y1.
Third step-Multiply each ordinate by the abscissa of the next row, and add the results. This gives y1x2+y2x3+y3x1.
Fourth step-Subtract the result of the third step from that of the second step, and divide by 2. This gives the required area. (If you list the vertices in clockwise order the result will be positive. If you list them is counterclockwise order the result will be negative.)
For example, consider the coordinates (0, 4), (4, 2), (-4, -6) from which our table looks like:
x1 y1 0 +4
x2 y2 +4 +2
x3 y3 -4 -6
x1 y2 0 +4
For the second step--0x3 + 4x(-6) + (-4)x4 = 0-24-16 = -40
For the third step------4x4 + 2x(-4) + (-6)x0 = 16-8+0 = 8
For the fourth step--[8-(-40)]/2 = 24 sq units. That was easy.
This method applies to any multiple sided figure that a set of vertices produces.
To find the area of a triangle with given vertices, you can use the formula for the area of a triangle using coordinates.
Step 1: Use the given coordinates to find the length of the base of the triangle.
- The given coordinates are (2, -5), (6, 1), and (-3, -4).
- Find the distance between (2, -5) and (6, 1) using the distance formula:
- Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
- Distance = sqrt((6 - 2)^2 + (1 - (-5))^2)
- Distance = sqrt(4^2 + 6^2)
- Distance = sqrt(52)
- Distance ≈ 7.21 (rounded to two decimal places)
Step 2: Use the formula for the area of a triangle using coordinates:
- Area = 0.5 * base * height
- The height of the triangle is the perpendicular distance from the third vertex to the base.
- Find the equation of the line passing through (2, -5) and (6, 1) using the slope-intercept form:
- Slope (m) = (y2 - y1) / (x2 - x1)
- Slope = (1 - (-5)) / (6 - 2)
- Slope = 6 / 4 = 3/2
- Use the point-slope form to find the equation of the line:
- y - y1 = m(x - x1)
- y - (-5) = (3/2)(x - 2)
- y + 5 = (3/2)(x - 2)
- y + 5 = (3/2)x - 3
- y = (3/2)x - 8
- Now we need to find the coordinates of the intersection point of the line y = (3/2)x - 8 and the line passing through (-3, -4).
- Substitute the value of y from the equation of the line into the equation of the line passing through (-3, -4):
- (3/2)x - 8 = -4
- (3/2)x = 4
- x = 4 * (2/3) = 8/3
- Substitute the value of x back into the equation of the line passing through (-3, -4):
- y = (3/2)(8/3) - 8
- y = 12/2 - 8
- y = 6 - 8
- y = -2
- Therefore, the coordinates of the intersection point are (8/3, -2).
Step 3: Now, find the perpendicular distance from the intersection point to the line passing through (2, -5) and (6, 1).
- Use the distance formula to find the distance between the intersection point and the line:
- Distance = |(Ax + By + C) / sqrt(A^2 + B^2)|
- The equation of the line passing through (2, -5) and (6, 1) is (3/2)x - y - 8 = 0
- Substitute the values of A, B, C, and the coordinates of the intersection point into the distance formula:
- (|((3/2)(8/3) - (-2) - 8) / sqrt((3/2)^2 + (-1)^2)|)
- (|(12/2 + 2 - 8) / sqrt((9/4) + 1)|)
- (|6 - 6/4| / (sqrt(9/4 + 1) = 2.39 (rounded to two decimal places)
Step 4: Now that we have the base and the height of the triangle, we can find the area using the formula:
- Area = 0.5 * base * height
- Area = 0.5 * 7.21 * 2.39
- Area ≈ 8.62 (rounded to two decimal places)
Therefore, the area of the triangle whose vertices have coordinates (2, -5), (6, 1), and (-3, -4) is approximately 8.62 square units.
To find the area of a triangle, you can use the formula for the area of a triangle given its vertices' coordinates. Let's call the coordinates A, B, and C, with A being (2, -5), B being (6, 1), and C being (-3, -4).
To calculate the area of a triangle, you can use the formula:
Area = 1/2 * |(x1 × y2 + x2 × y3 + x3 × y1) - (x2 × y1 + x3 × y2 + x1 × y3)|
To calculate the area, you need to substitute the vertex coordinates into this formula. Let's break it down step by step:
Step 1: Assign the coordinates to variables
Let's assign the coordinates of each vertex to variables:
A = (2, -5)
B = (6, 1)
C = (-3, -4)
Step 2: Calculate the area
Now, substitute the coordinate values into the area formula:
Area = 1/2 * |(x1 × y2 + x2 × y3 + x3 × y1) - (x2 × y1 + x3 × y2 + x1 × y3)|
= 1/2 * |(2 × 1 + 6 × -4 + -3 × -5) - (6 × -5 + -3 × 1 + 2 × -4)|
= 1/2 * |(2 + -24 + 15) - (-30 + -3 - 8)|
= 1/2 * |-7 - (-41)|
= 1/2 * |-7 + 41|
= 1/2 * |34|
= 17
Therefore, the area of the triangle with vertices (2, -5), (6, 1), and (-3, -4) is 17 square units.