Question

The joint equation of pair of straight lines passing through origin and having slopes $\left( {1 + \sqrt 2 } \right)$ and $\left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)$ is
A. ${x^2} - 2\sqrt 2 xy + {y^2} = 0$
B. ${x^2} - 2\sqrt 2 xy - {y^2} = 0$
C.${x^2} + 2xy - {y^2} = 0$
D. ${x^2} + 2xy + {y^2} = 0$

Answer 1

Hint: The equation of line passing through origin is given by $y = mx$ .
Then, form the equations of the lines with slopes $\left( {1 + \sqrt 2 } \right)$ and $\left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)$ and bring them in the format $y - mx = 0$ .
Now, multiply the L.H.S. and R.H.S. of both equations and form a linear equation.

Complete step-by-step answer:
The equation of a line with slope m is given by $y = mx + c$ .
Here, both the lines pass through origin. So, $c = 0$ .
Thus, equation of line passing through origin is $y = mx$ .
Here, slope of first line is given as $\left( {1 + \sqrt 2 } \right)$ .
So, equation of first line is $y = \left( {1 + \sqrt 2 } \right)x$ , i.e. $y - \left( {1 + \sqrt 2 } \right)x = 0$ . … (1)
Also, slope of the second line is given as $\left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)$ .
So, equation of second line is $y = \left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)x$ , i.e. $y - \left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)x = 0$ . … (2)
Now, we are asked to write the joint equation of both lines.
Thus, the joint equation can be written as $\left\{ {y - \left( {1 + \sqrt 2 } \right)x} \right\}\left\{ {y - \left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)x} \right\} = 0$ .
Now, we have to simplify the above equation
 \[
  \therefore {y^2} - \left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)xy - \left( {1 + \sqrt 2 } \right)xy + \left( {1 + \sqrt 2 } \right)\left( {\dfrac{1}{{1 + \sqrt 2 }}} \right){x^2} = 0 \\
  \therefore {y^2} - \left( {\dfrac{1}{{1 + \sqrt 2 }} + \left( {1 + \sqrt 2 } \right)} \right)xy + {x^2} = 0 \\
  \therefore {y^2} - \left( {\dfrac{{1 + {{\left( {1 + \sqrt 2 } \right)}^2}}}{{1 + \sqrt 2 }}} \right)xy + {x^2} = 0 \\
  \therefore {y^2} - \left( {\dfrac{{1 + \left( {1 + 2 + 2\sqrt 2 } \right)}}{{1 + \sqrt 2 }}} \right)xy + {x^2} = 0 \\
  \therefore {y^2} - \left( {\dfrac{{1 + 3 + 2\sqrt 2 }}{{1 + \sqrt 2 }}} \right)xy + {x^2} = 0 \\
  \therefore {y^2} - \left( {\dfrac{{4 + 2\sqrt 2 }}{{1 + \sqrt 2 }}} \right)xy + {x^2} = 0 \\
  \therefore {y^2} - \left( {\dfrac{{4 + 2\sqrt 2 }}{{1 + \sqrt 2 }}} \right)xy + {x^2} = 0 \\
  \therefore {y^2} - 2\left( {\dfrac{{2 + \sqrt 2 }}{{1 + \sqrt 2 }}} \right)xy + {x^2} = 0 \\
  \therefore {y^2} - 2\left( {\dfrac{{2 + \sqrt 2 }}{{1 + \sqrt 2 }} \times \dfrac{{1 - \sqrt 2 }}{{1 - \sqrt 2 }}} \right)xy + {x^2} = 0 \\
  \therefore {y^2} - 2\left( {\dfrac{{2 - 2\sqrt 2 + \sqrt 2 - 2}}{{1 - 2}}} \right)xy + {x^2} = 0 \\
  \therefore {y^2} - 2\left( {\dfrac{{ - \sqrt 2 }}{{ - 1}}} \right)xy + {x^2} = 0 \\
  \therefore {y^2} - 2\sqrt 2 xy + {x^2} = 0 \\
 \]
Thus, the combined equation is \[{y^2} - 2\sqrt 2 xy + {x^2} = 0\] .
So, option (A) is correct.

Note: Alternate method:
The given lines have slopes $\left( {1 + \sqrt 2 } \right)$ and $\left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)$ .
So, ${m_1} = \left( {1 + \sqrt 2 } \right)$ and ${m_2} = \left( {\dfrac{1}{{1 + \sqrt 2 }}} \right)$ .
Now, ${m_1}{m_2} = \left( {1 + \sqrt 2 } \right)\left( {\dfrac{1}{{1 + \sqrt 2 }}} \right) = 1$ … (1)
Also, \[{m_2} = \dfrac{1}{{1 + \sqrt 2 }} = \dfrac{1}{{1 + \sqrt 2 }} \times \dfrac{{1 - \sqrt 2 }}{{1 - \sqrt 2 }} = \dfrac{{1 - \sqrt 2 }}{{1 - 2}} = \sqrt 2 - 1\]
So, ${m_1} + {m_2} = \sqrt 2 + 1 + \sqrt 2 - 1 = 2\sqrt 2 $
Now, equation of line 1 is $y - {m_1}x = 0$ and equation of line 2 is $y - {m_2}x = 0$ .
So, we can write the equation of the pair of lines as $\left( {y - {m_1}x} \right)\left( {y - {m_2}x} \right) = 0$ .
 $
  \therefore {y^2} - {m_1}xy - {m_2}xy + {m_1}{m_2}{x^2} = 0 \\
  \therefore {y^2} - \left( {{m_1} + {m_2}} \right)xy + {m_1}{m_2}{x^2} = 0 \\
 $
Now, substitute ${m_1} + {m_2} = 2\sqrt 2 $ and ${m_1}{m_2} = 1$ .
 \[\therefore {y^2} - 2\sqrt 2 xy + {x^2} = 0\]
Thus, the combined equation is \[{y^2} - 2\sqrt 2 xy + {x^2} = 0\] .