The graphs of 3x+2y=1, y=2, and 3x - 4y=-29 contain the sides of a triangle. Find the measure of the area of the triangle.
Eq1: 3X + 2Y = 1.
Eq2: 3x - 4y = -29.
Eq3: Y = 2. This is a hor. line.
IN Eq1, substitute 2 for y:
3X + 2*2 = 1,
X = -1.
Solution: (X,Y) = (-1,2). = An end point of line3.
In Eq2: substitute 2 for Y:
3X - 4*2 = -29,
X = -7.
Solution: (x,y) = (-7,2).
Solve Eq1 and Eq2 using Elimination method and get:
Solution: (X,Y) = (-3,5).
Therefore the 3 vertices are:
A(-7,2), B(-1,2), C(-3,5).
tanA = (5-2) / (-3-(-7)) = 0.75,
A = 36.9deg.
(AC)^2 = (-3-(-7))^2 + (5-2)^2 = 25,
AC = 5.
Base = AB = -1 -(-7) = 6.
h = AC*sinA = 5*sin36.9 = = 3.
Area = bh/2 = 6 * 3 / 2 = 9 sq. Units.
To find the measure of the area of the triangle, we need to determine the vertices of the triangle formed by the intersection of the given lines.
1. Let's start by solving the first two equations simultaneously.
a. 3x + 2y = 1
b. y = 2
Substituting equation b into equation a, we get:
3x + 2(2) = 1
3x + 4 = 1
3x = -3
x = -1
Substituting the value of x in equation b, we get:
y = 2
Therefore, the first intersection point is (-1, 2).
2. Now, let's solve the second and third equations simultaneously.
a. y = 2
b. 3x - 4y = -29
Substituting equation a into equation b, we get:
3x - 4(2) = -29
3x - 8 = -29
3x = -29 + 8
3x = -21
x = -7
Substituting the value of x in equation a, we get:
y = 2
Therefore, the second intersection point is (-7, 2).
3. Lastly, let's solve the first and third equations simultaneously.
a. 3x + 2y = 1
b. 3x - 4y = -29
Subtracting equation b from equation a, we get:
(3x + 2y) - (3x - 4y) = 1 - (-29)
3x + 2y - 3x + 4y = 30
6y = 30
y = 5
Substituting the value of y into equation a, we get:
3x + 2(5) = 1
3x + 10 = 1
3x = -9
x = -3
Therefore, the third intersection point is (-3, 5).
4. Now that we have the vertices of the triangle (-1, 2), (-7, 2), and (-3, 5), we can calculate the area of the triangle.
Using the Shoelace Formula, the area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by:
Area = 1/2 * |(x1 * (y2 - y3)) + (x2 * (y3 - y1)) + (x3 * (y1 - y2))|
Plugging in the values, we have:
Area = 1/2 * |((-1 * (2 - 5)) + (-7 * (5 - 2)) + (-3 * (2 - 2)))|
Area = 1/2 * |(-3 + (-21) + 0)|
Area = 1/2 * |-24|
Area = 12
Therefore, the measure of the area of the triangle is 12 square units.
To find the measure of the area of the triangle, we need to determine the coordinates of its vertices.
Let's start by finding the intersection points of the given equations:
1) 3x + 2y = 1
2) y = 2
3) 3x - 4y = -29
For equation (1) and (2), we can substitute y in equation (1) with 2:
3x + 2(2) = 1
3x + 4 = 1
3x = 1 - 4
3x = -3
x = -1
So, for (1) and (2), the intersection point is (-1, 2).
Now, let's find the intersection points for equation (1) and (3). We will perform a similar substitution:
3x + 2y = 1
3x - 4y = -29
Substitute 3x from equation (1) into equation (3):
(1) - (2) => (3):
(3x + 2y) - (3x - 4y) = 1 - (-29)
6y = 30
y = 30/6
y = 5
Substitute y = 5 in equation (1):
3x + 2(5) = 1
3x + 10 = 1
3x = 1 - 10
3x = -9
x = -3
So, for equations (1) and (3), the intersection point is (-3, 5).
Lastly, let's find the intersection points for equation (2) and (3).
Substitute y = 2 in equation (3):
3x - 4(2) = -29
3x - 8 = -29
3x = -29 + 8
3x = -21
x = -7
So, for equations (2) and (3), the intersection point is (-7, 2).
Now we have the coordinates of the three intersection points: (-1, 2), (-3, 5), and (-7, 2).
Using these points, we can calculate the distances between them and apply the formula for area calculation. Since the triangle is not specified as a right triangle, we will use Heron's formula to find the area.
Let's denote the three intersection points as A(-1, 2), B(-3, 5), and C(-7, 2).
Now, calculate the lengths of the triangle's sides:
Side AB:
Using the distance formula: AB = √((x2 - x1)^2 + (y2 - y1)^2)
AB = √((-3 - (-1))^2 + (5 - 2)^2)
AB = √((-2)^2 + 3^2)
AB = √(4 + 9)
AB = √13
Side BC:
BC = √((-7 - (-3))^2 + (2 - 5)^2)
BC = √((-4)^2 + (-3)^2)
BC = √(16 + 9)
BC = √25
BC = 5
Side AC:
AC = √((-7 - (-1))^2 + (2 - 2)^2)
AC = √((-6)^2)
AC = √36
AC = 6
Now we can calculate the semi-perimeter (s) of the triangle:
s = (AB + BC + AC) / 2
s = (√13 + 5 + 6) / 2
s = (√13 + 11) / 2
Using Heron's formula, the area (A) of the triangle is given by:
A = √(s(s-AB)(s-BC)(s-AC))
A = √((√13 + 11)/2 * ((√13 + 11)/2 - √13) * ((√13 + 11)/2 - 5) * ((√13 + 11)/2 - 6))
Simplifying the equation further will yield the measure of the area of the triangle.