A straight line passes through a fixed point \[(h,k)\] . The locus of the foot of perpendicular on it drawn from the origin is
1). \[{{x}^{2}}+{{y}^{2}}-hx-ky=0\]
2). \[{{x}^{2}}+{{y}^{2}}+hx+ky=0\]
3). \[3{{x}^{2}}+3{{y}^{2}}+hx-ky=0\]
4). None of these

Hint: To solve this problem first find the equation of the line that passes through the given point and then find the slope of that line and after applying the perpendicularity condition find the slope of another line and substitute its value and you will get the required answer.

Complete step-by-step solution:
A line can be defined as a one dimensional geometric shape. It is measured in respect to only one dimension (i.e. length). The standard forms of the equation of a line are: Slope-intercept form, Intercept form, Normal form
Point slope form is one of the more commonly used forms of a linear equation, and has the following structure: \[y-{{y}_{1}}=m(x-{{x}_{1}})\] where \[m\] is the slope of the line and \[({{x}_{1}},{{y}_{1}})\] is a point on the line. Point slope form is used when one point of the line and the slope are known.
Now, we will learn about perpendicular lines:
Perpendicular lines are formed when two lines meet each other at the right angle. This property of lines is said to be perpendicularity.
Two lines are perpendicular to each other if and only if the product of the slope of the two lines equals minus of the unity.
Now, according to the given question:
Let, the given point be \[P(h,k)\]
And let the point at the foot of the perpendicular be \[Q(x,y)\]
Now, the equation of the line that passes through the fixed point \[(h,k)\] is given as:
\[(y-{{y}_{1}})=m(x-{{x}_{1}})\]
\[\Rightarrow (y-k)=m(x-h)\] \[..........(1)\]
Here the slope is \[m\]
So, the slope of other line that is perpendicular to \[(1)\] is given as: \[\dfrac{-1}{m}\]
And this line passes through the origin \[(0,0)\]
So, the equation of the parallel perpendicular line can be given as:
 \[(y-{{y}_{1}})=m(x-{{x}_{1}})\]
\[\Rightarrow (y-0)=\dfrac{-1}{m}(x-0)\]
\[\Rightarrow y=\left( \dfrac{-1}{m} \right)x\] \[..........(2)\]
Now, foot of the perpendicular is the intersection of \[(1)\] and \[(2)\]
So, the locus is obtained by eliminating the \[m\]
From \[(2)\] , we get \[m=\dfrac{-x}{y}\]
Now, we will substitute the value of \[m\] in the equation \[(1)\]
\[\Rightarrow y-k=\left( \dfrac{-x}{y} \right)(x-h)\]
\[\Rightarrow y(y-k)=(-x)(x-h)\]
\[\Rightarrow {{y}^{2}}-yk=-{{x}^{2}}+xh\]
\[\Rightarrow {{y}^{2}}-yk+{{x}^{2}}-xh=0\]
\[\Rightarrow {{y}^{2}}+{{x}^{2}}-xh-ky=0\]
Hence the correct option from all the above options is \[1\].

Note: If any two lines are perpendicular to the same line, then they both are parallel to each other and never intersect each other. Some examples of perpendicular lines and surfaces in real life are: The corners of the wall intersect each other at right angles, hands of a clock when it strikes exactly three O’clock, the corners of the desk or doors etc.