need help please:
the sides of the triangle on the lines 6x-7+11=0 2x+9y+11=0 and 9x-2y-11=0.
find the equation of the circle inscribed in the triangle.
To find the equation of the circle inscribed in a triangle, we need to determine the coordinates of the triangle's vertices. Since you have provided the equations of the lines on which the triangle's sides lie, we can find their intersection points to obtain the vertices.
Let's start with the equations of the lines:
1. 6x - 7y + 11 = 0
2. 2x + 9y + 11 = 0
3. 9x - 2y - 11 = 0
We can solve these equations simultaneously to find the intersection points. Here's how we can do that:
1. Find the intersection point of lines 1 and 2.
- Solve the system of equations formed by lines 1 and 2.
- One method to solve this is by substitution or elimination.
- Let's choose the elimination method. Multiply the first equation by 9 and the second equation by 7 to make the coefficients of y cancel out.
- After multiplying, we get: 54x - 63y + 99 = 0 and 14x + 63y + 77 = 0.
- Adding these equations eliminates the y term: 68x + 176 = 0.
- Solving for x, we find x = -2.5882.
- Substitute this value of x back into either the first or second equation to find y. Let's use the first equation:
6*(-2.5882) - 7y + 11 = 0
-15.529 - 7y + 11 = 0
-4.529 - 7y = 0
-7y = 4.529
y = -0.6470
- Therefore, the intersection point of lines 1 and 2 is (-2.5882, -0.6470).
2. Find the intersection point of lines 1 and 3.
- Solve the system of equations formed by lines 1 and 3.
- Again, we can use the elimination method. Multiply the first equation by 2 and the third equation by 7 to make the coefficients of y cancel out.
- After multiplying, we get: 12x - 14y + 22 = 0 and 63x - 14y - 77 = 0.
- Adding these equations eliminates the y term: 75x - 55 = 0.
- Solving for x, we find x = 0.7333.
- Substitute this value of x back into either the first or third equation to find y. Let's use the first equation:
6*(0.7333) - 7y + 11 = 0
4.3998 - 7y + 11 = 0
15.3998 - 7y = 0
-7y = -15.3998
y = 2.1999
- Therefore, the intersection point of lines 1 and 3 is (0.7333, 2.1999).
3. Find the intersection point of lines 2 and 3.
- Solve the system of equations formed by lines 2 and 3.
- We can again use the elimination method. Multiply the second equation by 9 and the third equation by 2 to make the coefficients of x cancel out.
- After multiplying, we get: 18x + 81y + 99 = 0 and 18x - 4y - 22 = 0.
- Subtracting these equations eliminates the x term: 85y + 121 = 0.
- Solving for y, we find y = -1.4235.
- Substitute this value of y back into either the second or third equation to find x. Let's use the second equation:
2x + 9*(-1.4235) + 11 = 0
2x - 12.8115 + 11 = 0
2x - 1.8115 = 0
2x = 1.8115
x = 0.9057
- Therefore, the intersection point of lines 2 and 3 is (0.9057, -1.4235).
Now that we have the coordinates of the triangle's three vertices, we can proceed to find the equation of the circle inscribed in this triangle.