Question

The equation of a straight line which cuts off intercepts on x-axis twice that on y-axis and is at a unit distance from the origin is
A. \[x - 2\sqrt y + \sqrt 5 = 0\]
B. \[x + 2y - \sqrt 5 = 0\]
C. \[x + 2y + \sqrt 5 = 0\]
D. \[x - 2y - \sqrt 5 = 0\]

Answer 1

Hint: We will first assume the equation of a line \[{\rm{y}} = {\rm{mx}} + {\rm{c}}\]. We know that we get the value of x-intercept by substituting \[{\rm{y}} = 0\] in the equation and the value of y-intercept by substituting \[{\rm{x}} = 0\] in the equation of line. Also, we will use the formula to find the distance between origin and a straight line given by,
Let the equation of line be \[{\rm{y}} = {\rm{mx}} + {\rm{c}}\].
\[Dis\tan ce = \dfrac{{\left| c \right|}}{{\sqrt {1 + {m^2}} }}\]

Complete step-by-step answer:
We have been asked to find the equation of a line which cuts off the intercept on x-axis twice that on y-axis and is at a unit distance from the origin.
Let us suppose the equation of a line to be \[{\rm{y}} = {\rm{mx}} + {\rm{c}}\].
We know that we get x-intercept by substituting \[y = 0\] in the equation of line and y-intercept by substituting \[x = 0\].
So, x-intercept is given by,
\[\begin{array}{l}0 = mx + c\\ \Rightarrow - c = mx\\ \Rightarrow \dfrac{{ - c}}{m} = x\end{array}\]
And y-intercept is given by,
\[\begin{array}{l}y = m \times 0 + c\\ \Rightarrow y = c\end{array}\]
According to question,
\[ \Rightarrow \dfrac{{ - c}}{m} = 2c\]
Taking \[\left( {\dfrac{{ - c}}{m}} \right)\] to right side of equation, we get
\[2c + \dfrac{c}{m} = 0\]
Taking 'c' as common, we get
\[\begin{array}{l}c\left( {2 + \dfrac{1}{m}} \right) = 0\\ \Rightarrow c = 0{\rm{ and 2 + }}\dfrac{1}{m} = 0\\ \Rightarrow \dfrac{1}{m} = - 2\\ \Rightarrow {\rm{m = - }}\dfrac{1}{2}\end{array}\]
We know the distance between the origin and a line having equation y=mx+c is given by,
\[Distance = \dfrac{{\left| c \right|}}{{\sqrt {1 + {m^2}} }}\]
According to question,
\[\dfrac{{\left| c \right|}}{{\sqrt {1 + {m^2}} }} = 1\]
When \[{\rm{c}} = 0\], we get
\[\begin{array}{l}\dfrac{0}{{\sqrt {1 + {m^2}} }} = 1\\ \Rightarrow \sqrt {1 + {m^2}} = 0\end{array}\]
On squaring both sides, we get
\[1 + {m^2} = 0\]
\[{m^2} = - 1,\] which is not possible because the value of m is an imaginary number.
So, 'c' cannot be equal to zero.
When, \[m = \dfrac{{ - 1}}{2}\] we get
\[ \Rightarrow \dfrac{{\left| c \right|}}{{\sqrt {1 + {{\left( {\dfrac{{ - 1}}{2}} \right)}^2}} }} = 1\]
\[\begin{array}{l} \Rightarrow \dfrac{{\left| c \right|}}{{\sqrt {1 + \dfrac{1}{4}} }} = 1\\ \Rightarrow \dfrac{{\left| c \right|}}{{\sqrt {\dfrac{{1 + 4}}{4}} }} = 1\\ \Rightarrow \dfrac{{\left| c \right|2}}{{\sqrt 5 }} = 1\\ \Rightarrow \left| c \right| = \dfrac{{\sqrt 5 }}{2}\\ \Rightarrow c = \pm \dfrac{{\sqrt 5 }}{2}\end{array}\]
So, the equation of line are as follows:
Case I:
\[\begin{array}{l}For m = -\dfrac{1}{2}{\rm{ and c = }}\dfrac{{\sqrt 5 }}{2}\\y = {\rm{ - }}\dfrac{1}{2}x + \dfrac{{\sqrt 5 }}{2}\end{array}\]
On taking \[LCM\] of terms, we get
\[\begin{array}{l}{\rm{y = }}\dfrac{{ - x + \sqrt 5 }}{2}\\ \Rightarrow 2y = - x + \sqrt 5 \\ \Rightarrow 2y + x - \sqrt 5 = 0\end{array}\]
Case II:
\[\begin{array}{l}{\rm{When }} m = \dfrac{{ - 1}}{2}{\rm{ and c = }}\dfrac{{ - \sqrt 5 }}{2}\\y = {\rm{ }}\dfrac{{ - 1}}{2}x - \dfrac{{\sqrt 5 }}{2}\end{array}\]
On taking \[LCM\] of the terms, we get

\[\begin{array}{l}{\rm{y = }}\dfrac{{ - x - \sqrt 5 }}{2}\\ \Rightarrow 2y = - x - \sqrt 5 \\ \Rightarrow 2y + x + \sqrt 5 = 0\end{array}\]

Let us draw the figure for the lines as shown below,

Therefore, the correct options are B and C.

Note: The general mistake that we make is using the formula of distance between origin and the line without a modulus function which gives us only one value of 'c' and our answer is incomplete. So, be careful and remember that formula includes the modulus function i.e. \[\dfrac{{\left| c \right|}}{{\sqrt {1 + {m^2}} }}\]
We get \[{\rm{c}} = 0\] during the calculation but for this value of 'c' we get the value of m an imaginary number. So, we could not take this value of 'c'.