Question

Find the equation of the line passing through \[\left( {\sqrt 3 , - 1} \right)\] if its perpendicular distance from the origin is \[\sqrt 2 \] .

Answer 1

Hint: In the given question, we are required to find the equation of the line passing through the point \[\left( {\sqrt 3 , - 1} \right)\] given that the perpendicular distance from the origin is \[\sqrt 2 \] . The problem revolves around the basic concepts of coordinate geometry and straight lines. One should be familiar with the different forms of line and there applications.

Complete step-by-step answer:
Let us suppose that the line is given in the slope intercept form as \[y = mx + c\] . So, now we need to find the values of the entities m and c in order to find the whole equation of the straight line.
We have to find the values of m and c according to the conditions given to us in the question itself.
So, we are given that the line passes through \[\left( {\sqrt 3 , - 1} \right)\] . Hence, the point \[\left( {\sqrt 3 , - 1} \right)\] satisfies the equation of the line.
Putting \[\left( {\sqrt 3 , - 1} \right)\] in the equation of the line,
 \[ \Rightarrow - 1 = m\left( {\sqrt 3 } \right) + c\]
Simplifying further, we get,
 \[ \Rightarrow \sqrt 3 m + c = - 1 - - - - - (1)\]
Now, we are also given that the perpendicular distance of the line from the origin is \[\sqrt 2 \] units.
So, Distance of line from origin$ = \left| {\dfrac{{\left( 0 \right) - m\left( 0 \right) - c}}{{\sqrt {{m^2} + 1} }}} \right|$
$ \Rightarrow \left| {\dfrac{c}{{\sqrt {{m^2} + 1} }}} \right|$
So, we have, $\left| {\dfrac{c}{{\sqrt {{m^2} + 1} }}} \right| = \sqrt 2 $
$ \Rightarrow \left| c \right| = \sqrt 2 \left| {\sqrt {{m^2} + 1} } \right|$
Squaring both sides, we get,
$ \Rightarrow {c^2} = 2\left( {{m^2} + 1} \right) - - - - - (2)$
Now, we have questions relating c and m. So, we have to find the values of c and m using these two equations.
Substituting the value of m from equation $(1)$ as \[c = - 1 - \sqrt 3 m\] in equation \[(2)\] , we get,
${\left( { - 1 - \sqrt 3 m} \right)^2} = 2\left( {{m^2} + 1} \right)$
$ \Rightarrow 3{m^2} + 2\sqrt 3 m + 1 = 2{m^2} + 2$
Shifting all the terms from right side of the equation to left side of the equation, we get,
$ \Rightarrow 3{m^2} - 2{m^2} + 2\sqrt 3 m + 1 - 2 = 0$
$ \Rightarrow {m^2} + 2\sqrt 3 m - 1 = 0$
Comparing the equation with standard form of the quadratic equation $a{m^2} + bm + c = 0$ , we get,
$a = 1$,$b = 2\sqrt 3 $ and $c = - 1$.
Now, using the quadratic formula, we get the values of m as $\left( {\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right)$.
Therefore, $m = \left( {\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right)$
Substituting the values of a, b and c. We get,
$m = \left( {\dfrac{{ - 2\sqrt 3 \pm \sqrt {{{\left( {2\sqrt 3 } \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}} \right)$
$ \Rightarrow m = \left( {\dfrac{{ - 2\sqrt 3 \pm \sqrt {12 + 4} }}{2}} \right)$
$ \Rightarrow m = \left( {\dfrac{{ - 2\sqrt 3 \pm \sqrt {16} }}{2}} \right)$
$ \Rightarrow m = \left( {\dfrac{{ - 2\sqrt 3 \pm 4}}{2}} \right)$
$ \Rightarrow m = - \sqrt 3 \pm 2$
So, the possible values of m are: $ - \sqrt 3 + 2$ and $ - \sqrt 3 - 2$ .
Substituting the values of m in equation $\left( 1 \right)$ to find the value of c.
For $m = - \sqrt 3 + 2$, we have,
 \[ \Rightarrow c = - 1 - \sqrt 3 \left( { - \sqrt 3 + 2} \right)\]
 \[ \Rightarrow c = - 1 + 3 - 2\sqrt 3 \]
 \[ \Rightarrow c = 2 - 2\sqrt 3 \]
For $m = - \sqrt 3 - 2$, we have,
 \[ \Rightarrow c = - 1 - \sqrt 3 \left( { - \sqrt 3 - 2} \right)\]
 \[ \Rightarrow c = - 1 + 3 + 2\sqrt 3 \]
 \[ \Rightarrow c = 2 + 2\sqrt 3 \]
So, the equations of the required line are: \[y = \left( { - \sqrt 3 + 2} \right)x + \left( {2 - 2\sqrt 3 } \right)\] and \[y = \left( { - \sqrt 3 - 2} \right)x + \left( {2 + 2\sqrt 3 } \right)\] .
So, the correct answer is “ \[y = \left( { - \sqrt 3 + 2} \right)x + \left( {2 - 2\sqrt 3 } \right)\] and \[y = \left( { - \sqrt 3 - 2} \right)x + \left( {2 + 2\sqrt 3 } \right)\] ”.

Note: In the given problem, we could also have used completing the square method or splitting the middle term method to solve the quadratic equation formed while doing the problem. We can also verify the answer by carrying on the solution in the reverse order and checking if both the equations satisfy the given conditions