Let us name the corners of the triangle as #A(4,3), B(7,4) & C(2,8).#
From Geometry , we know that the altitudes of a trangle are concurrent at a point called the Orthocentre of the triangle.
Let pt. #H# be the orthocentre of #DeltaABC,# and, let three altds. be #AD, BE, and CF,# where the pts. #D,E,F# are the feet of these altds. on sides #BC, CA, and, AB,# respectively.
So, to obtain #H#, we should find the eqns. of any two altds. and solve them. We select to find the eqns. of #AD and CF.#
Eqn. of Altd. AD :-
#AD# is perp. to #BC#, & slope of #BC# is #(8-4)/(2-7)=-4/5,# so, slope of #AD# has to be #5/4#, with #A(4,3)# on #AD#.
Hence, eqn. of #AD : y-3=5/4(x-4),# i.e., #y=3+5/4(x-4)..........(1)#
Eqn. of Altd. CF :-
Proceeding as above, we get , eqn. of #CF : y=8-3(x-2)........(2)#
Solving #(1) & (2), 3+5/4(x-4)=8-3(x-2)#
#rArr 12+5x-20=32-12x+24 rArr 17x=64 rArr x=64/17#
BY #(2)#, then, #y=8-3*30/17=46/17.#
Hence, the Ortho centre #H=H(64/17,46/17).#
Hope, you enjoyed this! Enjoy Maths.!
