Slope of the line AB: m_1= (2-7)/(1-2)=5m1=2−71−2=5
Slope of CF=perpendicular slope of AB: m_2= -1/5m2=−15
Equation of the line CF is y-5= -1/5(x-3) or 5y-25 =-x+3 or x+5y=28 (1)y−5=−15(x−3)or5y−25=−x+3orx+5y=28(1)
Slope of the line BC: m_3= (5-2)/(3-1)=3/2m3=5−23−1=32
Slope of AE=perpendicular slope of BC: m_4= -1/(3/2)=-2/3m4=−132=−23
Equation of the line AE is y-7= -2/3(x-2) or 3y-21 =-2x+4 or 2x+3y=25 (2)y−7=−23(x−2)or3y−21=−2x+4or2x+3y=25(2) The intersection of CF & AE is the orthocenter of the triangle, which can be obtained by solving equation(1) & (2)
x+5y=28 (1)x+5y=28(1) ; 2x+3y=25 (2)2x+3y=25(2)
2x+10y=56 (1)2x+10y=56(1) obtained by multiplying 2 on both sides
2x+3y=25 (2)2x+3y=25(2) subtracting we get 7y=31:. y=31/7; x=28-5*31/7=41/7:.Orthocenter is at (41/7,31/7)[Ans]
