Orthocentre of the triangle with vertices (4,1), (7,4), (5,-2) is
Answers
Solution :-
given that, the vertices of the triangle are :-
- A = (4, 1)
- B = (7, 4)
- C = (5, -2)
Let us assume that,
- AD ⟂ BC .
- CF ⟂ AB .
- BE ⟂ AC .
so,
→ Slope of AB = (y2 - y1) / (x2 - x1) = (4 - 1) / (7 - 4) = 3/3 = 1
then,
→ Slope of CF = Perpendicular slope of AB
→ Slope of CF = (-1) / Slope of AB
→ Slope of CF = (-1) / 1
→ Slope of CF = (-1)
now, Equation of CF :-
→ y - y1 = m(x - x1)
→ y + 2 = (-1)(x - 5)
→ y + 2 = 5 - x
→ x + y = 3 ----------------- Eqn.(1)
Similarly,
→ Slope of BC = (y2 - y1) / (x2 - x1) = (-2 - 4) / (5 - 7) = (-6)/(-2) = 3
then,
→ Slope of AD = Perpendicular slope of BC
→ Slope of AD = (-1) / Slope of BC
→ Slope of AD = (-1) / 3
→ Slope of AD = (-1/3) .
now, Equation of AD :-
→ y - y1 = m(x - x1)
→ y - 4 = (-1/3)(x - 1)
→ 3y - 12 = -x + 1
→ x + 3y = 13 ----------------- Eqn.(2)
Subtracting Eqn.(1) from Eqn.(2) we get,
→ (x + 3y) - (x + y) = 13 - 3
→ x - x + 3y - y = 10
→ 2y = 10
→ y = 5
putting value of y in Eqn.(1) ,
→ x + 5 = 3
→ x = 3 - 5
→ x = (-2)
therefore, the orthocentre of the triangle are (-2 , 5) .
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