Find the height of the triangle ABC, dropped from the vertex A to the side BC, if A (2,1), B (-3,4), C (1,1)

Let AK be the height dropped. From the condition of perpendicularity of vectors x (AK) x (BC) + y (AK) y (BC) = 0 we obtain the equation of the straight line passing through points A and K: 4 (x – 2) + (-3) (y – 1) = 0 or 4x – 3y = 5.

The equation of the straight line passing through points B and C is defined as (x – x (C)) / (x (B) – x (C)) = (y – y (C)) / (y (B) – y ( C)) or 3x + 4y = 7.

The intersection point of the straight lines 4x – 3y = 5 and 3x + 4y = 7 is the point K. The coordinates of the point K (41/25; 13/25).

The quantity AK = (((x (K) – x (A)) ^ 2 + (y (K) – y (A)) ^ 2) ^ 1/2 = 3/5.



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