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If t 1+ t 2 + t 3 = - t 1 t 2 t 3, then orthocentre of the triangle formed by the points [ at 1 t 2, a ( t 1+ t 2)] , [at 2 t 3, a ( t 2+ t3)] and [at 3 t 1, a ( t 3+ t1)], lies on (a) (a,0) (b) ( -a,0) (c) (0,a) (d) (0, -a) Explain pls.

If t1+ t2 + t3 = - t1 t2 t3, then  orthocentre of the triangle formed by the points [at1 t2, a( t1+ t2)] , [at2 t3, a( t2+ t3)] and  [at3 t1, a( t3+ t1)], lies on

(a) (a,0)  (b) ( -a,0) (c) (0,a) (d) (0, -a)

Explain pls.

Grade:12th Pass

1 Answers

Aman Bansal
592 Points
9 years ago

Dear Aditi,

orthocenter is the intersection of the 3 altitudes of a circle.

for example, the coordinates of the vertices of the triangle are A(a, b), B(c, d) and C(e, f)

first, get the equation of the line passing through C and is perpendicular to AB. (altitude to AB)

then, get the equation of the line passing through A and is perpendicular to BC. (altitude to BC)

then, find the intersection of these lines. (the altitude to AC will also pass through this point since the altitude of a triangle are concurrent to each other - meaning 3 or more lines intersecting at a point...)

the intersection is the coordinates of the orthocenter :D

centroid is the intersection of the medians of a triangle
(median - line connecting a vertex of a triangle and the midpoint of the side opposite to it.)
the coordinates of the midpoint of a triangle with vertices at (a, b), (c, d), (e, f) are ( (a+c+e)/3, (b+d+f)/3 )

circumcenter - the intersection of the 3 perpendicular bisectors of a triangle
(perpendicular bisector - a line perpendicular to a side of a triangle passing through its midpoint)
this point is equidistant to the vertices of the triangle.

btw, there is another "center" thing related to triangles...

incenter - intersection of the angle bisectors of a triangle
(angle bisector - lines bisecting the angles of a triangle)
the incenter is equidistant from the sides of the triangle

Best Of luck

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Aman Bansal

Askiitian Expert

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