Question

Equation of the hour hand at \[4'O\] clock is:
A). \[x-\sqrt{3}y=0\]
B). \[\sqrt{3}x-y=0\]
C). \[x+\sqrt{3}y=0\]
D). \[\sqrt{3}x+y=0\]

Answer 1

Hint: First of all we will find the angle at \[4'O\] clock that is the value of \[\theta \] and through this we can find out the slope. As the line is passing through origin \[(0,0)\] hence we can find out the equation of the line and through that we can check out which option is correct in the given options.

Complete step-by-step solution:
A line is the locus of the two points such that every point of the path joining two points lies on that locus.
A straight line or line is an endless one dimensional figure that has no width and it does not have any curve, it can be horizontal, vertical or slanted.
General Equation of a Straight Line:
The general equation of a straight line is given as \[ax+by+c=0\] where,
\[\text{a,b,c}\] are constants , \[\text{x,y}\] are variables
\[\text{Slope}\] intercept form of an equation of line:
A straight line has \[\text{slope}\] \[\text{(m=tan}\theta \text{)}\], where \[\theta \] is the angle formed by the line with positive \[\text{x-axis}\], and \[\text{y-intercept}\] as \[\text{c}\] which can be written as \[\text{y=mx+c}\].
If the straight line is formed by positive \[\text{x-axis}\]and have the \[\text{Slope}\]\[\text{(m=tan}\theta \text{)}\] and the line is passing through the point \[\text{(}{{\text{x}}_{1}}\text{,}{{\text{y}}_{1}})\] then the equation will be :
\[y-{{y}_{1}}=m(x-{{x}_{1}})\]
Now according to the question:
As we know that the clock can make total angle of \[{{360}^{\circ }}\] in \[12\] hours as shown in fig \[(1)\] hence in one hour it can make angle \[\dfrac{{{360}^{\circ }}}{12}={{30}^{\circ }}\]
               

At \[4'O\] clock, the hour hand makes the angle of \[{{120}^{\circ }}\] in clockwise direction:
With the help of this we can find out the slope:
Slope \[m=\tan \theta \] where \[\theta ={{120}^{\circ }}\]
\[\Rightarrow m=\tan ({{120}^{\circ }})\]
\[\Rightarrow m=\tan ({{90}^{\circ }}+{{30}^{\circ }})\]
As we know that \[\tan ({{90}^{\circ }}+\theta )=-\cot \theta \]
\[\Rightarrow m=-\cot {{30}^{\circ }}\]
\[\Rightarrow m=-\sqrt{3}\]
As the line is passing through the origin \[O(0,0)\]
Hence from the equation of line \[(y-{{y}_{1}})=m(x-{{x}_{1}})\]
Where \[{{x}_{1}}=0\] , \[{{y}_{1}}=0\] and \[m=-\sqrt{3}\]
\[\Rightarrow (y-0)=-\sqrt{3}(x-0)\]
\[\Rightarrow y=-\sqrt{3}x\]
\[\Rightarrow \sqrt{3}x+y=0\]
Hence option \[(4)\] is correct.

Note: We must keep one thing in mind that in the calculation of \[\text{Slope }\]it should be noted that the angle with \[\text{x-axis}\], simply means the positive direction of \[\text{x-axis}\] and if a line is perpendicular to \[\text{x-axis}\], then slope cannot be defined, in this condition \[\theta \] will be \[{{90}^{\circ }}\] that means \[\tan {{90}^{\circ }}=\infty \]