The equation of the locus of a point equidistant from the points $\left( {{a_1},\;{b_1}} \right)$ and $\left( {{a_2},\;{b_2}} \right)$ is $\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0$, then the value of $c$ is

$\sqrt {\left( {a_1^2 + b_1^2 + c_1^2} \right)} $

JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

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Application +4 | -1

The equation of the locus of a point equidistant from the points

\left( {{a_1},\;{b_1}} \right)

and

\left( {{a_2},\;{b_2}} \right)

\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0

, then the value of

c

View answer and explanation

Correct option: C

\frac{1}{2}\left( {a_2^2 + b_2^2 - a_1^2 - b_1^2} \right)

Option C (

\frac{1}{2}\left( {a_2^2 + b_2^2 - a_1^2 - b_1^2} \right)

) is correct because it represents the constant term in the equation of the locus of a point equidistant from two given points.

Practice All Chapter MCQs →

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

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

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

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

y z

(1,-2,4)

(3,-4

5)

p

2 x+y-3=0

3 x-y-2=0

x-p y+5=0

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

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

A(1,0,3)

B(4,7,1)

C(3,5,3)

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

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