Question

If the equations y = mx + c and \[x\cos \alpha +y\sin \alpha =p\] represent the same straight line, then
\[\left( a \right)p=c\sqrt{1+{{m}^{2}}}\]
\[\left( b \right)c=p\sqrt{1+{{m}^{2}}}\]
\[\left( c \right)cp=\sqrt{1+{{m}^{2}}}\]
\[\left( d \right){{p}^{2}}+{{c}^{2}}+{{m}^{2}}=1\]

Answer 1

Hint: First we will assume the equations y = mx + c and \[x\cos \alpha +y\sin \alpha =p\] as equations (i) and (ii) respectively. We will leave equation (i) as it is and simplify equation (ii), so that the coefficient of y becomes 1. Now, in the RHS, compare the coefficients of x and the constant terms from the two equations. Then use the identity \[{{\operatorname{cosec}}^{2}}\alpha -{{\cot }^{2}}\alpha =1\] to get the relation between p, c and m.

Complete step-by-step solution:
We have been given the two equations of a single line.
\[y=mx+c.....\left( i \right)\]
 \[x\cos \alpha +y\sin \alpha =p....\left( ii \right)\]
On observing equation (i), we can get some idea that we have to simplify equation (ii) in such a way that the coefficient of y becomes 1. So, let us simplify the equation (ii).
\[x\cos \alpha +y\sin \alpha =p\]
\[\Rightarrow y\sin \alpha =-x\cos \alpha +p\]
Dividing both the sides with \[\sin \alpha \] we get,
\[\Rightarrow y=\dfrac{-x\cos \alpha +p}{\sin \alpha }\]
\[\Rightarrow y=-x\dfrac{\cos \alpha }{\sin \alpha }+\dfrac{p}{\sin \alpha }\]
Now we know that \[\dfrac{\cos \alpha }{\sin \alpha }=\cot \alpha \] and \[\dfrac{1}{\sin \alpha }=\operatorname{cosec}\alpha ,\] we get,
\[\Rightarrow y=-x\cot \alpha +p\operatorname{cosec}\alpha \]
\[\Rightarrow y=x\left( -\cot \alpha \right)+p\operatorname{cosec}\alpha .....\left( iii \right)\]
Since the coefficients of y in equation (i) and (ii) are the same in the LHS, therefore we can compare the RHS. On comparing the coefficients of x and the constant term in the RHS of the equation (i) and (iii), we get, \[m=-\cot \alpha \] and \[c=p\operatorname{cosec}\alpha .\]
Since \[m=-\cot \alpha \] we get
\[\Rightarrow \cot \alpha =-m.....\left( iv \right)\]
And \[c=p\operatorname{cosec}\alpha \]
\[\Rightarrow \operatorname{cosec}\alpha =\dfrac{c}{p}.....\left( v \right)\]
Now, squaring the equations (iv) and (v), we get,
\[\Rightarrow {{\cot }^{2}}\alpha ={{m}^{2}}\]
\[\Rightarrow {{\operatorname{cosec}}^{2}}\alpha =\dfrac{{{c}^{2}}}{{{p}^{2}}}\]
Here, using the trigonometric identity, \[{{\operatorname{cosec}}^{2}}\alpha -{{\cot }^{2}}\alpha =1\] we get,
\[{{\operatorname{cosec}}^{2}}\alpha -{{\cot }^{2}}\alpha =\dfrac{{{c}^{2}}}{{{p}^{2}}}-{{m}^{2}}\]
\[\Rightarrow 1=\dfrac{{{c}^{2}}}{{{p}^{2}}}-{{m}^{2}}\]
\[\Rightarrow \dfrac{{{c}^{2}}}{{{p}^{2}}}=1+{{m}^{2}}\]
On cross multiplying, we get,
\[\Rightarrow {{c}^{2}}={{p}^{2}}\left( 1+{{m}^{2}} \right)\]
Taking the square root on both the sides, we get,
\[\Rightarrow \sqrt{{{c}^{2}}}=\sqrt{{{p}^{2}}\left( 1+{{m}^{2}} \right)}\]
\[\Rightarrow c=p\sqrt{1+{{m}^{2}}}\]
Hence, option (b) is the right answer.

Note: One may note that we cannot compare the coefficients of x and y and the constant term directly in the two equations. So, first, we simplified the second equation such that the coefficient of y became the same and then we compared. One can also solve this question by assigning some value to ‘m’ and ‘c’ and then by putting the same values in the options to check. But this method will only be applicable when the options are present.