If AD, BE and CF are the altitudes of a triangle ABC whose vertex A is the point (-4, 5). The coordinates of the points E and F are (4,1) and (-1, -4) respectively, then equation of BC is

Slope of AF=5−(−4)−4+1=9−3=−3∴ Slope of CF=13∴ Equation of CF is x−3y−11=0-----(1) Slope of AE=5−1−4−4=4−8=−12∴ Slope of BE=2∴ Equation of BE is 2x−y−7=0----(2) Solving (i) and (ii), we get point O=(2,−3) Equation of AC is x+2y−6=0----(3) Solving (i) and (iii) we get point C(8,−1) Also slope of AO is −43∴ Slope of BC is 34∴ Equation of BC is 3x−4y−28=0