JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

1 Easy

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Analyzing +4 | -1

Let A be a fixed point (0, 6) and

B

be a moving point (2t, 0). Let

M

be the mid-point of

AB

and the perpendicular bisectors of

AB

meets the

y

-axis at

C

.
The locus of the mid-point

P

MC

View answer and explanation

Correct option: C

2{x^2} + 3y - 9 = 0

Option C (

2{x^2} + 3y - 9 = 0

) is correct because it represents the locus of the midpoint P of MC, derived by finding the coordinates of M and C and then eliminating the parameter t.

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