JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

1 Easy

1 of 1

Analyzing +4 | -1

Three distinct points

A,\;B

and

C

given in the two dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point

\left( { - 1,\;0} \right)

is equal to

\frac{1}{3}

. Then , the circumcentre of the triangle

ABC

is at the point

View answer and explanation

Correct option: A

\left( {\frac{5}{4},\;0} \right)

Option A (

\left( {\frac{{5}}{{4}},\;0} \right)

) is correct because the circumcenter of the triangle lies at the center of the circle defined by the given condition.

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