The sides of the triangle are on the of x+y+6=0, x+2y-4=0, 3x+y+2=0. Find the vertices of the triangle
Given three non-parallel lines.
We are to find the vertices of a triangle formed by the intersection of the lines.
The intersections are found by solving for the three systems of two equations at a time.
For example:
L1: x+y+6=0
L2: x+2y-4=0
Subtract L1 from L2 gives:
y-10=0, this means that
y=10
Now substitute y=10 in equation L1 gives x+10+6=0
x=-16.
So (-16,10) is one of the vertices.
Repeat the same for lines L2 & L3, and then L1 & L3 will give the two other vertices.
To find the vertices of the triangle, we need to solve the system of equations formed by the given equations:
1. x + y + 6 = 0
2. x + 2y - 4 = 0
3. 3x + y + 2 = 0
Let's solve the system of equations step-by-step:
Step 1: Solve equations 1 and 2 for x:
From equation 1, we have x = -y - 6.
Substituting this value of x into equation 2:
(-y - 6) + 2y - 4 = 0
Simplifying, we get:
y - 10 = 0
y = 10
Step 2: Substitute the value of y into equation 1:
x + 10 + 6 = 0
x + 16 = 0
x = -16
So, one vertex of the triangle is (-16, 10).
Step 3: Solve equations 1 and 3 for x:
Substituting -y - 6 for x in equation 3:
3(-y - 6) + y + 2 = 0
Simplifying, we get:
-3y - 18 + y + 2 = 0
-2y - 16 = 0
y = -8
Step 4: Substitute the value of y into equation 1:
x - 8 + 6 = 0
x - 2 = 0
x = 2
So, another vertex of the triangle is (2, -8).
Step 5: Solve equations 2 and 3 for x:
Substituting x + 2y - 4 for x in equation 3:
3(x + 2y - 4) + y + 2 = 0
Simplifying, we get:
3x + 6y - 12 + y + 2 = 0
3x + 7y - 10 = 0
Step 6: Substitute the value of x from equation 1 into equation 2:
(-y - 6) + 2y - 4 = 0
y - 10 = 0
y = 10
Substituting y = 10 into equation 3:
3x + 10 + 2 = 0
3x + 12 = 0
3x = -12
x = -4
So, the third vertex of the triangle is (-4, 10).
Therefore, the vertices of the triangle are (-16, 10), (2, -8), and (-4, 10).
To find the vertices of the triangle, we need to solve the system of equations formed by the given equations.
Step 1: Write the given equations in the form of ax + by + c = 0.
The given equations are already in this form:
1) x + y + 6 = 0
2) x + 2y - 4 = 0
3) 3x + y + 2 = 0
Step 2: Solve the system of equations simultaneously.
To solve the system of equations, there are different methods like substitution, elimination, or matrix methods. Here, we'll use the elimination method.
Multiply equation 1 by 3:
3(x + y + 6) = 0
3x + 3y + 18 = 0
3x + y + 18 = 0 -- Equation 4
Now, subtract Equation 2 from Equation 4:
(3x + y + 18) - (x + 2y -4) = 0
2x - y + 22 = 0 -- Equation 5
Now, we have two equations:
Equation 5: 2x - y + 22 = 0
Equation 3: 3x + y + 2 = 0
Add Equation 5 and Equation 3:
(2x - y + 22) + (3x + y + 2) = 0
5x + 24 = 0
5x = -24
x = -24/5
Now substitute x = -24/5 into Equation 5 to find y:
2(-24/5) - y + 22 = 0
-48/5 - y + 22 = 0
-48/5 + 22 = y
(-48 + 110)/5 = y
62/5 = y
So, we have found x = -24/5 and y = 62/5.
Step 3: Substitute the values of x and y into any of the given equations to find the third variable.
Let's substitute the values into Equation 1:
x + y + 6 = 0
(-24/5) + (62/5) + 6 = 0
(-24 + 62 + 30)/5 = 0
68/5 = 0
Since 68/5 is not equal to 0, these values do not satisfy Equation 1.
Step 4: Repeat Steps 2 and 3 with different combinations of equations.
We can repeat Steps 2 and 3 using Equation 2 and Equation 3, as well as Equation 1 and Equation 3, to find the other two vertices of the triangle.
By solving these combinations, you will find the three pairs of x and y values that represent the vertices of the triangle, denoting the x-coordinate and y-coordinate of each vertex.