find orthocentre of triangle whose vertices are (-5,-7),(13,2),(-5,6).
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104
A(-5, -7) B=(13,2) C(-5, 6) Let the Orthocenter H be (x,y)
See that A and C have the same x coordinate. AC is parallel to y axis.
Let us shift the origin O(0,0) to A(-5,-7). Hence new origin is O' = A itself.
We work now in the new coordinate system (X= x+5, Y=y+7).
New coordinates A (0,0), B = (13--5, 2--7) = (18, 9)
C = (-5--5, 6--7) = (0, 13)
AC ⊥ X axis. So BE⊥ AC will be : Y = 9. So ordinate of H = 9.
Slope of AB: (9-0)/(18-0) = 1/2.
=> Slope of CD = -2.
Equation of CD ⊥ AB, passing through C: (Y - 13) / (X - 0) = - 2
Y + 2 X - 13 = 0
Substitute Y = 9, X = 2.
Hnce H(2, 9) in the new coordinate system.
In the original coordinate system : Orthocenter H = (2-5, 9-7)
H = (-3, 2)
You can also use the formula for orthocenter in terms of the coordinates of the vertices.
See that A and C have the same x coordinate. AC is parallel to y axis.
Let us shift the origin O(0,0) to A(-5,-7). Hence new origin is O' = A itself.
We work now in the new coordinate system (X= x+5, Y=y+7).
New coordinates A (0,0), B = (13--5, 2--7) = (18, 9)
C = (-5--5, 6--7) = (0, 13)
AC ⊥ X axis. So BE⊥ AC will be : Y = 9. So ordinate of H = 9.
Slope of AB: (9-0)/(18-0) = 1/2.
=> Slope of CD = -2.
Equation of CD ⊥ AB, passing through C: (Y - 13) / (X - 0) = - 2
Y + 2 X - 13 = 0
Substitute Y = 9, X = 2.
Hnce H(2, 9) in the new coordinate system.
In the original coordinate system : Orthocenter H = (2-5, 9-7)
H = (-3, 2)
You can also use the formula for orthocenter in terms of the coordinates of the vertices.
kvnmurty:
click on red heart thanks above pls
great sir
can you see my answers
when we transform the coordinates by making A as (0,0)., B(x2, y2) and aligning C(x3, 0) along the X-axis... the orthocenter is easily found: x = x2 ... y = x2 (x3 - x2) / y2
hmm now next time i use this concept .
Answered by
84
orthocentre :- it is the point of intersection of altitude of ∆
now,
step 1:- first find equation of two any side .
equation , which passing through (-5 , -7) and (13,2)
(y+5) =(2+7)/(13+5)(x+5)
2y+10 = x + 5
2y -x + 5 = 0 --------------(1)
and
equation which is passing through (13,2) and (-5,6)
(y-6) =(6-2)/(-5-13)(x + 5 )
(y -6) =2/(-9) ( x + 5)
2(x + 5 ) +9(y - 6 ) =0
2x +9y -44 =0 --------------------(2)
step2 :-
find equation of altitude by equation of side .
equation of altitude which is perpendicular to equation (1) and passing (-5 , 6)
e.g ( y - 6) + 2( x + 5) =0
2x +y + 4 =0 --------------(3)
in the same way equation of altitude which is perpendicular in equation (2) and passing through (-5 , -7)
e.g 2(y + 7) =9( x + 5 )
9x -2y + 31=0 ----------(4)
solve equation (3) and (4)
x = -3 and y = 2
so, orthicentre is ( -3 , 2)
now,
step 1:- first find equation of two any side .
equation , which passing through (-5 , -7) and (13,2)
(y+5) =(2+7)/(13+5)(x+5)
2y+10 = x + 5
2y -x + 5 = 0 --------------(1)
and
equation which is passing through (13,2) and (-5,6)
(y-6) =(6-2)/(-5-13)(x + 5 )
(y -6) =2/(-9) ( x + 5)
2(x + 5 ) +9(y - 6 ) =0
2x +9y -44 =0 --------------------(2)
step2 :-
find equation of altitude by equation of side .
equation of altitude which is perpendicular to equation (1) and passing (-5 , 6)
e.g ( y - 6) + 2( x + 5) =0
2x +y + 4 =0 --------------(3)
in the same way equation of altitude which is perpendicular in equation (2) and passing through (-5 , -7)
e.g 2(y + 7) =9( x + 5 )
9x -2y + 31=0 ----------(4)
solve equation (3) and (4)
x = -3 and y = 2
so, orthicentre is ( -3 , 2)
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