Question

What is the locus of first degree equation in \[x,y,z\]?
A. Straight line
B. Plane
C. Sphere
D. Curved surface

Answer 1

Hint: As in this question, locus point of an equation is given and apart from that, no further information is provided. So, to solve this type of question, we will take options to find out the answer. Here, we will compare the equation which will be formed by every option and compare with the said equation and whichever equation fits the criteria will be the correct option.

Formula Used:
1. Equation of straight line:
$\dfrac{{x - a}}{\alpha } = \dfrac{{y - b}}{\beta } = \dfrac{{z - c}}{\gamma }$
2. Equation of plane:
$\alpha x + \beta b + \gamma y = \delta $
3. Equation of sphere:
${(x - a)^2} + {(y - b)^2} + {(z - c)^2} = {r^2}$

Complete Step-by-step solution
Let us assume the equation as $ax + by + cz = d$
Now, we will compare the assumed equation with each equation formed by the given options.
Checking the equation of option A:
The equation of straight line is
$\dfrac{{x - a}}{\alpha } = \dfrac{{y - b}}{\beta } = \dfrac{{z - c}}{\gamma }$
The above equation does not match with the assumed equation.
So, option A is incorrect.

Checking the condition of option B:
The equation of plane is
$\alpha x + \beta b + \gamma y = \delta $
The above equation matches with the assumed equation.
So, option B is correct.

Checking the condition of option C:
The equation of sphere is
${(x - a)^2} + {(y - b)^2} + {(z - c)^2} = {r^2}$
The above equation does not match with the assumed equation.
So, option C is incorrect.

Checking the condition of option D:
The equations for curved surface can’t be delivered easily as it depends on the surface.
So, option D is also incorrect.
Hence the correct option is B.

Note:A locus is a collection of points that satisfy a particular requirement or circumstance for a shape or figure. For this type of question, it is important to know the equations of different shapes in all forms.