Question
Two vertices of a triangle are (3,-2) and (-2,3) and its orthocentre is (-6,1). Then its third vertex is
- (1,6)
- (-1,6)
- (1,-6)
- None of these
Hint:
The line passing through the orthocenter and the vertex is always perpendicular to it's opposite side.
The correct answer is: (-1,6)
Given That:
Two vertices of a triangle are (3,-2) and (-2,3) and its orthocenter is (-6,1). Then its third vertex is:
>>> Let the third vertex be (x, y) and I be the Orthocenter
>>> Then Slope of (3,-2) and (-2,3) becomes = 
>>> Also, the slope of the IC is = 
>>> We know the lines AB and IC are perpendicular. Then,
= 1
->> x+6=y-1
x-y+7=0
>>>Slope of AC is = 
>>> Slope of IB = 
>>> We know that the line IB is perpendicular to AC.
= -2
->> y+2 = -2x+6
2x+y-4=0
>>> By solving the above highlighted equations, we get the coordinates of third vertex as (-1,6).
>>> Therefore, the coordinates of the third vertex is (-1,6).
2x+y-4=0 and the x-y+7=0 are the equations that pass through the third vertex.
Related Questions to study
If in triangle 
, circumcenter 
and orthocenter 
then the co-ordinates of mid-point of side opposite to is :
>>> The orthocenter, centroid and circumcenter of any triangle are collinear. And the centroid divides the distance from orthocenter to circumcenter in the ratio 2:1.
>>> Also, the centroid (G) divides the medians (AD) in the ratio 2:1.
>>> ∴D(h, k)=(1,
)
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>>> The orthocenter, centroid and circumcenter of any triangle are collinear. And the centroid divides the distance from orthocenter to circumcenter in the ratio 2:1.
>>> Also, the centroid (G) divides the medians (AD) in the ratio 2:1.
>>> ∴D(h, k)=(1,)
A triangle ABC with vertices A(-1,0), B(-2,3/4)&C(-3,-7/6) has its orthocentre H. Then the orthocentre of triangle BCH will be
A triangle ABC with vertices A(-1,0), B(-2,3/4)&C(-3,-7/6) has its orthocentre H. Then the orthocentre of triangle BCH will be
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>>> acosθ=6 ----(1)
>>> a(sin(30−θ))=4 ----(2)
>>> a = 
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>>> acosθ=6 ----(1)
>>> a(sin(30−θ))=4 ----(2)
>>> a =
If the point 
lies between the region corresponding to the acute angle between the lines x-3y=0 and x-6y=0 then
>>> L11 
L22 <0
>>> (1+cos
)2 -6sin -6sincos -3sin-3sincos+18sin2 < 0
If the point lies between the region corresponding to the acute angle between the lines x-3y=0 and x-6y=0 then
>>> L11 L22 <0
>>> (1+cos)2 -6sin -6sincos -3sin-3sincos+18sin2 < 0
If 
be any point on a line then the range of for which the point ' P ' lies between the parallel lines x+2y=1 and 2x+4y=15 is
((1+
)+2(
) -1).(
) < 0
If be any point on a line then the range of for which the point ' P ' lies between the parallel lines x+2y=1 and 2x+4y=15 is
((1+)+2() -1).() < 0
is any point in the interior of the quadrilateral formed by the pair of lines and the two lines 2x+y-2=0 and 4x+5y=20 then the possible number of positions of the points ' P ' is
is any point in the interior of the quadrilateral formed by the pair of lines and the two lines 2x+y-2=0 and 4x+5y=20 then the possible number of positions of the points ' P ' is
If the point 
,lies in the region corresponding to the acute angle between the lines 2y=x and 4y=x then - .....
u ≡ x - 2y = 0 and v ≡ x - 4y = 0
>>> S(x, y) ≡ x² - 6xy + 8y² = 0
>>> ( a - 2 )( a - 4 ) < 0
If the point ,lies in the region corresponding to the acute angle between the lines 2y=x and 4y=x then - .....
u ≡ x - 2y = 0 and v ≡ x - 4y = 0
>>> S(x, y) ≡ x² - 6xy + 8y² = 0
>>> ( a - 2 )( a - 4 ) < 0
Consider A(0,1) and B(2,0) and P be a point on the line 4x+3y+9=0, co-ordinates of P such 
is maximum is
Hence the point is (
, 
).
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Hence the point is (, ).
Assertion (A): The lines represented by 
and x+ y=2 do not form a triangle
Reason (R): The above three lines concur at (1,1)
Both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
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Reason (R): The above three lines concur at (1,1)
Both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
In Bohr’s hydrogen atom, the electronic transition emitting light of longest wavelength is:
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P1,P2,P3, be the product of perpendiculars from (0,0) to 
respectively then:
P1 = 1;
P2 = 
;
P3 = 
;
>>> Therefore, we can say that P1>P2>P3.
P1,P2,P3, be the product of perpendiculars from (0,0) to respectively then:
P1 = 1;
P2 = ;
P3 = ;
>>> Therefore, we can say that P1>P2>P3.
If θ is angle between pair of lines 
, then 
>>> 
= 2.
>>> tan = 
>>> 
= 10.
If θ is angle between pair of lines , then
>>> = 2.
>>> tan =
>>> = 10.
If the pair of lines 
intersect on the x-axis, then 2fgh=
If the pair of lines intersect on the x-axis, then 2fgh=
If the pair of lines intersect on the x-axis, then ac=
If the pair of lines intersect on the x-axis, then ac=
represents a pair of perpendicular lines then its point of intersection is
