Altitudes of ABC are: ~9.43, ~5.13, ~8.24 units.
Finding the lengths of the altitudes of triangle ABC
We have the equations for the sides of triangle ABC:
1. AB: 2x-y = 0
2. BC: x+3y = 0
3. CA: x-4y = 14
To find the lengths of the altitudes, we need to first find the coordinates of the vertices A, B, and C. We can do this by solving the system of equations:
1. Multiply equation (3) by 4: 4x-16y = 56
2. Add equation (2) to the newly obtained equation: 5x-13y = 56
3. Solve equation (5) for x: x = 8
4. Substitute x = 8 back into equation (2): 8+3y = 0
5. Solve equation (6) for y: y = -8/3
6. Substitute x = 8 and y = -8/3 back into equation (1): 16-8/3 = 0
7. Solve equation (7) for x: x = 8/3
Therefore, the coordinates of the vertices are:
1. A(8/3, 0)
2. B(8, -8/3)
3. C(0, 24/3)
Now, we can find the lengths of the altitudes by using the distance formula. We'll calculate the distances from each vertex to the opposite side:
1. Altitude from A to BC:
- The equation of line BC is x+3y = 0.
- The slope of line BC is -1/3.
- The y-intercept of line BC is 0.
- The equation of the perpendicular line from A(8/3, 0) is 3x-y-8 = 0.
- The intersection point of these two lines is (6,-6).
- Distance from A to point (6,-6): sqrt((8/3-6)^2 + (0-(-6))^2) = sqrt(25/9 + 36) = sqrt(685/9) ≈ 9.43 units.
2. Altitude from B to CA:
- The equation of line CA is x-4y = 14.
- The slope of line CA is 1/4.
- The y-intercept of line CA is -3.5.
- The equation of the perpendicular line from B(8, -8/3) is 4x-y-2 = 0.
- The intersection point of these two lines is (12, 2).
- Distance from B to point (12, 2): sqrt((8-12)^2 + (-8/3-2)^2) = sqrt(16 + 100/9) = sqrt(204/9) ≈ 5.13 units.
3. Altitude from C to AB:
- The equation of line AB is 2x-y = 0.
- The slope of line AB is 2.
- The y-intercept of line AB is 0.
- The equation of the perpendicular line from C(0, 24/3) is 0.5x-y-8 = 0.
- The intersection point of these two lines is (8, 16).
- Distance from C to point (8, 16): sqrt((0-8)^2 + (24/3-16)^2) = sqrt(64 + 4) = sqrt(68) ≈ 8.24 units.