Question
Vertices of a triangle are (4, 3) ,
Then the locus of the orthocenter is
- a straight line
- a circle
- ellipse
- hyperbola
The correct answer is: a circle
Let
be the vertices and
be the centroid and orthocentre of
If
is the origin 
is the circumcentre of
divides OH in the ratio 1 : 2.
Hence 

is 
locus of
is
which is a circle.
Related Questions to study
A line through A(-5, -4) meets the line x+3y+2=0, 2x+y+4=0 and x-y-5=0 at
and
respectively. If
then the equation of the line is
A line through A(-5, -4) meets the line x+3y+2=0, 2x+y+4=0 and x-y-5=0 at
and
respectively. If
then the equation of the line is
If
as well as
are in G.P with the same common ratio, then the points
.
If
as well as
are in G.P with the same common ratio, then the points
.
The range of
for which the points
and
lie on opposite side of the line 2x+3y=6 is
The range of
for which the points
and
lie on opposite side of the line 2x+3y=6 is
The internal bisectors of the a
, having the sides BC=3, CA=5 and AB =4 meet the sides BC, CA and AB in D, E and F respectively, the area of
is
The internal bisectors of the a
, having the sides BC=3, CA=5 and AB =4 meet the sides BC, CA and AB in D, E and F respectively, the area of
is
A straight line
with negative slope passes through the point (8, 2) and cuts the positive coordinate axes at points P and Q As
varies, the absolute minimum value of OP+OQ= where
is the origin
A straight line
with negative slope passes through the point (8, 2) and cuts the positive coordinate axes at points P and Q As
varies, the absolute minimum value of OP+OQ= where
is the origin
Vertices of a variable triangle are (3, 4)
and
where
Locus of its orthocenter is
Vertices of a variable triangle are (3, 4)
and
where
Locus of its orthocenter is
If (0,0), (a, 2), (2, b) form the vertices of an equilateral triangle, where a and b not lie between 0 and 2, then the value of 4(a
b)‐ab equals
If (0,0), (a, 2), (2, b) form the vertices of an equilateral triangle, where a and b not lie between 0 and 2, then the value of 4(a
b)‐ab equals
The lines
and
are concurrent if
The lines
and
are concurrent if
Equation of the line perpendicular to 4x + 7y + 9 = 0 and such that the triangle formed by it with the coordinates axes forms an area of 3.5 sq. units is
Equation of the line perpendicular to 4x + 7y + 9 = 0 and such that the triangle formed by it with the coordinates axes forms an area of 3.5 sq. units is
(0, 0) is the foot of the perpendicular from (4, 2) on a straight line. The equation of the line is
(0, 0) is the foot of the perpendicular from (4, 2) on a straight line. The equation of the line is
The orthocentre of the triangle having vertices at (2, 3) (2, 5) (4, 3) is
The orthocentre of the triangle having vertices at (2, 3) (2, 5) (4, 3) is
Each side of square is length 5. The centre of square is (3, 7) and one of the diagonals is parallel to y = x. Then the coordinates of its vertices are
Therefore, the coordinates of vertices of square are (1, 5) (1, 9) (5, 9) (5, 5).
Each side of square is length 5. The centre of square is (3, 7) and one of the diagonals is parallel to y = x. Then the coordinates of its vertices are
Therefore, the coordinates of vertices of square are (1, 5) (1, 9) (5, 9) (5, 5).
The sides of a rhombus ABCD are parallel to the lines y = x + 2; y = 7x + 3. If the diagonals of the rhombus intersect at the point (1, 2) and the vertex A is on y – axis then A =
The sides of a rhombus ABCD are parallel to the lines y = x + 2; y = 7x + 3. If the diagonals of the rhombus intersect at the point (1, 2) and the vertex A is on y – axis then A =
The quadrilateral formed by the lines
+ y = 0; x +
y = 0;
x + y = 1; x +
y + 1 = 0 is
The quadrilateral formed by the lines
+ y = 0; x +
y = 0;
x + y = 1; x +
y + 1 = 0 is
If 7x – y + 3 = 0; x + y – 3 = 0 are tow sides of an isosceles triangle and the third side passes through (1, 0) then the equation of the third side is
So here we used the concept of the triangles, and the equations of the lines to solve this problem. We know that the equation of a straight line in slope-intercept form is given as y = mx+c, so we used that here. So the equation of the third side is x + 3y - 1 = 0.
If 7x – y + 3 = 0; x + y – 3 = 0 are tow sides of an isosceles triangle and the third side passes through (1, 0) then the equation of the third side is
So here we used the concept of the triangles, and the equations of the lines to solve this problem. We know that the equation of a straight line in slope-intercept form is given as y = mx+c, so we used that here. So the equation of the third side is x + 3y - 1 = 0.