Question

find the orthocentre of the triangle whose sides are given by 4x- 7y+10=0,x+y=5 and 7x +4y=15​

Answer 1

$4x - 7y + 10 = 0 \: - - - - (1) \\ x + y - 5 = 0 \: - - - - (2) \\ 7x + 4y - 15 = 0 \: - - - - (3) \\ on \: solving \: eq \: (1) \: and \: (3) \\ 7x + 4( \frac{4x + 10}{7} ) - 15 = 0 \\ \\ \\ \\ \\ \\ \\ 49x + 16 x + 40 - 105 = 0 \\ 65x = 65 = > x = 1 \\ from \: eq \: (1) \\ 4 - 7y = - 10 \\ - 7y = - 14 \\ y = 2 \\ point \: of \: instersection \: is \: (1) \: (2) \\ here \: equation \: (1) \: \: (3) \\ perpendicular \: so \: the \: line \: formed \: right \: angle \: triangle \\ hence \: the \: orthocentre \: of \: right \: angle \: triangle \: is \: the \: point \: of \: which \:90 \: degree \: angle \: formed$