Question

The equation of a line of the system $2x + y + 4 + \lambda \left( {x - 2y - 3} \right) = 0$ which is at a distance $\sqrt {10}$ units from point A(2,−3) is
A.3x+y+1=0
B.3x-y+1=0
C.y-3x+1=0
D.y+3x-2=0

Answer 1

Hint : In this question, we need to determine the equation of the line such that the line is $\sqrt {10}$ units from point A(2,−3) and belongs to the family of the lines $2x + y + 4 + \lambda \left( {x - 2y - 3} \right) = 0$. For this, we will properties of the equation of the line to determine the point of intersection and then use the formula for the distance between the lines.

Complete step-by-step answer :
The equation for the family of the lines is $2x + y + 4 + \lambda \left( {x - 2y - 3} \right) = 0$ which involves two equations of the lines $2x + y + 4 = 0 - - - - (i)$ and $x - 2y - 3 = 0 - - - - (ii)$ . So, if we solve these two equations, we will get the point of intersection of the lines.
From the equation (i), we get
$\Rightarrow 2x + y + 4 = 0 \\\ y = - 2x - 4 - - - - (iii) \\\$
Substituting the value of ‘y’ in the equation (ii), we get
$\Rightarrow x - 2y - 3 = 0 \\\ x - 2( - 2x - 4) - 3 = 0 \\\ x + 4x + 8 - 3 = 0 \\\ 5x + 5 = 0 \\\ x = - 1 \\\$
Hence, the x-coordinate of the point of intersection of the lines is -1.
Substitute the value of the x-coordinate in the equation (iii), we get
$\Rightarrow y = - 2x - 4 \\\ = - 2( - 1) - 4 \\\ = 2 - 4 \\\ = - 2 \\\$
Hence, the x-coordinate of the point of intersection of the lines is -2.
So, the coordinates of the point of intersection of the lines is (-1,-2).
Now, the general equation of the line passing through the point (-1,-2) is given by
$\Rightarrow y - ( - 2) = m\left( {x - ( - 1)} \right) \\\ \Rightarrow y + 2 = m\left( {x + 1} \right) \\\ \Rightarrow y + 2 - mx - m = 0 \\\ \Rightarrow mx - y + (m - 2) = 0 - - - - (iv) \\\$
The distance of the line $ax + by + c = 0$ from the point $(m,n)$ is given by the formula
$d = \left| {\dfrac{{am + ny + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$
According to the question, the distance of the line $mx - y + (m - 2) = 0$ is $\sqrt {10}$ units from point A(2,−3). So, substituting the values in the formula $d = \left| {\dfrac{{am + ny + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$ .
$\Rightarrow d = \left| {\dfrac{{am + ny + c}}{{\sqrt {{a^2} + {b^2}} }}} \right| \\\ \sqrt {10} = \left| {\dfrac{{(m)(2) + ( - 1)( - 3) + (m - 2)}}{{\sqrt {{m^2} + {{( - 1)}^2}} }}} \right| \\\ \sqrt {10} = \left| {\dfrac{{2m + 3 + (m - 2)}}{{\sqrt {{m^2} + 1} }}} \right| \\\$
Squaring both sides of the equation, we get
$\Rightarrow {\left( {\sqrt {10} } \right)^2} = {\left| {\dfrac{{2m + 3 + (m - 2)}}{{\sqrt {{m^2} + 1} }}} \right|^2} \\\ 10 = \dfrac{{{{\left( {3m + 1} \right)}^2}}}{{{m^2} + 1}} \\\ 10{m^2} + 10 = 9{m^2} + 6m + 1 \\\ {m^2} - 6m + 9 = 0 \\\ {m^2} - 3m - 3m + 9 = 0 \\\ m(m - 3) - 3(m - 3) = 0 \\\ m = 3 \\\$
Hence, the value of ‘m’ is 3.
Substituting the value of m in the equation (iv) to determine the equation of the line.
$\Rightarrow mx - y + (m - 2) = 0 \\\ \Rightarrow 3x - y + (3 - 2) = 0 \\\ \Rightarrow 3x - y + 1 = 0 \\\$
Hence, the equation of the line is $3x - y + 1 = 0$ .

So, the correct answer is “Option B”.

Note : The constant term in the standard equation of the line should always be on the right hand side of the equation and so, its value should be taken carefully with the sign convention.