Let M be the foot of perpendicular drawn from point A(- 2,3) to the line 3x-y- 1 = 0 …(i)
Slope of the line 3x-y – 1 = 0 is -3/-1 = 3.
Since AM ⊥ to line (i), slope of AM = -1/3
∴ Equation of AM is y-3 = -1/3 (x +2)
∴ 3(y – 3) = – 1(x + 2)
∴ 3y – 9 = -x – 2
∴ x + 3y – 7 = 0 …………(ii)
The foot of perpendicular i.e., point M, is the point of intersection of equations (i) and (ii).
By (i) x 3 + (ii), we get 10x -10 = 0
∴ x = 1
Substituting x = 1 in (ii), we get 1 + 3y – 7 = 0
∴ 3y = 6
∴ y = 2
∴ The co-ordinates of the foot of the perpendicular Mare (1,2).
