In a triangle whose vertices are A (-2,1), B (5,2), and C (3,7), find the length of the altitude from A to line BC.
Thanks very much for the help!
Vertices: (-2,1), (5,2), (3,7)
sides: 5.385164807, 7.810249676, 7.071067812
angles: 42.064326554°, 76.328692868°, 61.606980579° is an acute scalene triangle
area: 18.5 perimeter: 20.266482295
Altitudes:
The distance from (-2,1) to (4.379310345,3.551724138) is 6.870727513 on 2.551724138x -6.379310345y= -11.482758621 or y =0.4x +1.8
The distance from (5,2) to (1.360655738,5.032786885) is 4.737364558 on 3.032786885x +3.639344262y=22.442622951 or y = -0.833333333x +6.166666667
The distance from (3,7) to (3.74,1.82) is 5.232590181 on x +0.142857143y=4 or y = -7x +28
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Vertices: (-2,1), (5,2), (3,7)
sides: 5.385164807, 7.810249676, 7.071067812
angles: 42.064326554°, 76.328692868°, 61.606980579° is an acute scalene triangle
area: 18.5 perimeter: 20.266482295
Altitudes:
The distance from (-2,1) to (4.379310345,3.551724138) is 6.870727513 on 2.551724138x -6.379310345y= -11.482758621 or y =0.4x +1.8
The distance from (5,2) to (1.360655738,5.032786885) is 4.737364558 on 3.032786885x +3.639344262y=22.442622951 or y = -0.833333333x +6.166666667
The distance from (3,7) to (3.74,1.82) is 5.232590181 on x +0.142857143y=4 or y = -7x +28