Question

Find the orthocenter of the triangle with (1, 2) (2, 6) & (3, -4) as vertices.

Answer 1

Answer:

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Step-by-step explanation:

$to \: find : \\ coordinates \: of \: orthocenter \: of \: △ABC \\ \\ so \: here \: let \\ O \: be \: the \: orthocentre \: of \: △ABC \\ such \: that \\ O = (x,y) \\ \\ so \: considering \\ A = (x1,y1) =( 1,2) \\ \\ B = (x2,y2) = (2,6) \\ \\ so \: thus \: then \\ \\ slope \: of \: AB \: (m1)= \frac{y2 - y1}{x2 - x1} \\ \\ = \frac{6 - 2}{2 - 1} \\ \\ = \frac{4}{1} \\ \\ m1= 4 \\ \\ so \: thus \: then \: as \: here \\ \: \: AB \: ⊥ \: CD \: ie \: ⊥ \: OD \\ \\ slope \: of \: AB \times slope \: of \: CD = - 1 \\ \\ m1 \times m2 = - 1 \\ \\ 4.m2 = - 1 \\ \\ m2 = \frac{ - 1}{4} \\ \\ so \: thus \: here \\ \\ C = (x1,y1) = (3, - 4) \\ \\ we \: know \: that \\ slope - point \: equation \: of \: straight \: line \\ is \: given \: by \\ \\ y - y1 = m(x - x1) \\ \\ y - ( - 4) = \frac{ - 1}{4} (x - 3) \\ \\ 4(y + 4) = - 1(x - 3) \\ \\ 4y + 16 = - x + 3 \\ \\ x + 4y = - 13 \: \: \: \: \: \: \: \: \: (1)$

$similarly \: \: here \\ A = (x1,y1) = (1,2) \\ \\ C= (x2,y2) = (3, - 4) \\ \\ so \: thus \: then \\ \\ slope \: of \: AC = \frac{y2 - y1}{x2 - x1} \\ \\ = \frac{ - 4 - 2}{3 - 1} \\ \\ = \frac{ - 6}{2} \\ \\slope \: of \: AC = - 3 \\ \\ now \: since \: here \\ AC \: ⊥ \: BF \: ie \: \: OF \\ \\ slope \: of \: AC \times slope \: of \: BF = - 1 \\ \\ - 3 \times m = - 1 \\ \\ m = \frac{1}{3} \\ \\ thus \: let \: then \\ \\ B = (x1,y1) = (2,6) \\ \\ we \: have \\ \\ y - y1 = m(x - x1) \\ \\ y - 6 = \frac{1}{3} (x - 2) \\ \\ 3(y - 6) = x - 2 \\ \\ 3y - 18 = x - 2 \\ \\ x - 3y = - 16 \: \: \: \: \: \: \: \: (2)$

$applying \: \: (1) - (2) \\ we \: have \: then \\ \\ 4y - ( - 3y) = - 13 - ( - 16) \\ \\ 4y + 3y = - 13 + 16 \\ \\ 7y = 3 \\ \\ y = \frac{3}{7} \\ \\ substitute \: value \: of \: y \: in \: (1) \\ we \: get \\ \\ x = \frac{ - 103}{17}$

$thus \: then \\ we \: have \\ \\ O = (x,y) = ( \frac{ - 103}{7} \: , \: \frac{3}{7} \: )$