Of altitudes of triangles

Geometry Level 4

A ( 33 , 26 ) A(33,26) , B ( − 2 , 5 ) B(-2,5) and C ( α , β ) C (\alpha, \beta) are the vertices of a △ A B C \triangle ABC situated in the x y xy plane. If the orthocenter of this triangle is H ( 1 , 2 ) H(1,2) , then find the coordinates of vertex C C .

Submit your answer as the value of α + β \alpha + \beta .


The answer is 1.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

L e t A D b e t h e a l t i t u d e f r o m A , B E f r o m B , a n d C F f r o m C . ∴ Y A D = 26 − 2 33 − 1 ( X − 1 ) + 2 = 3 4 X + 5 4 . ⟹ Y B C = − 4 3 ( X + 2 ) + 5 = − 4 3 X + 7 3 . A l s o Y B E = 2 − 5 1 + 2 ( X − 1 ) + 2 = − 1 ( X − 1 ) + 2 = − X + 3 ⟹ Y A C = 1 ( X − 33 ) + 26 = X − 7. ∴ − 4 3 X C + 7 3 = X C − 7. ∴ α = X C = 4. ⟹ β = Y C = X C − 7 = − 4. S o α + β = 1. Let~AD ~be~the~altitude~from~A,~~BE~~from~B,~~and~CF~from~C.\\ \therefore~Y_{AD}=\dfrac{26-2}{33-1}(X-1)+2=\frac 3 4 X+\frac 5 4.\\ \implies~Y_{BC}=-\frac 4 3 (X+2)+5=-\frac 4 3 X+\frac 7 3.\\ Also~Y_{BE}=\dfrac{2-5}{1+2}(X-1)+2=-1(X-1)+2=-X+3\\ \implies~Y_{AC}=1 (X-33)+26= X - 7.\\ \therefore~-\frac 4 3 X_C+\frac 7 3= X_C - 7 .\\ \therefore~~\alpha=X_C=4.\\ \implies~\beta=Y_C=X_C - 7=-4.\\ So~\alpha+\beta=\huge~~~\color{#D61F06}{1}.

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