Q. 304.3( 3 Votes )

In the tria

Answer :


Let AD be the altitude of triangle ABC from vertex A.


Accordingly, AD is perpendicular to BC


Given vertices A (2, 3), B (4, –1) and C (1, 2)


Slope of line BC = m1


m1 = (- 1 - 2)/(4 - 1)


m1 = -1


Let slope of line AD be m2


AD is perpendicular to BC


m1 × m2 = -1


-1 × m2 = -1


m2 = 1


The equation of the line passing through point (2, 3) and having a slope of 1 is


y – 3 = 1 × (x – 2)


y – 3 = x – 2


y – x = 1


Equation of the altitude from vertex A = y – x = 1


Length of AD = Length of the perpendicular from A (2, 3) to BC


Equation of BC is


y + 1 = -1 × (x – 4)


y + 1 = -x + 4


x + y – 3 = 0 …………………(1)


Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by



Comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we obtain


A = 1, B = 1 and C = -3


Length of AD =


The equation and the length of the altitude from vertex A are y – x = 1 and


√2 units respectively.

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