JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

1 Easy

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Analyzing +4 | -1

Vertices of a variable triangle are

(3,4),(5 \cos \theta, 5 \sin \theta)

and

(5 \sin \theta,-5 \cos \theta)

, where

\theta \in R

. Locus of its orthocentre is

View answer and explanation

Correct option: D

x^2+y^2-6 x-8 y-25=0

Option D (

x^2+y^2-6 x-8 y-25=0

) is correct because it represents the locus of the orthocenter derived from the given vertices and their relationships to the circumcenter and centroid.

More Questions from 3D GEOMETRY SYSTEM

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The mid point of the line joining the points

\left( { - 10,\;8} \right)

and

\left( { - 6,\;12} \right)

divides the line joining the points

\left( {4,\; - 2} \right)

and

\left( { - 2,\;4} \right)

in the ratio The equation of the plane perpendicular to the

y z

-plane and passing through the points

(1,-2,4)

and

(3,-4

5)

is The value of

p

for which the lines

2 x+y-3=0

3 x-y-2=0

and

x-p y+5=0

may intersect at a point is If the vertices of a triangle at

O\left( {0,0} \right),A\left( {a,0} \right)

and

B\left( {0,a} \right).

Then, the distance between its circumcentre and orthocentre is The coordinates of the foot of the perpendicular drawn from the point

A(1,0,3)

to the join of the points

B(4,7,1)

and

C(3,5,3)

are The area of the triangle formed by the three lines

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}, \frac{x}{2}=\frac{y}{1}=\frac{z}{3}

and

\frac{x-1}{1}=\frac{2-y}{2}, z=3