Question

A line passes through $(2,0)$. Then which of the following is not the slope of the line, for which it’s intercept between $y = x - 1$ and $y = - x + 1$ subtends a right angle at the origin
(A) $- \dfrac{1}{{\sqrt 3 }}$
(B) $- \sqrt 3$
(C) $\dfrac{1}{{\sqrt 3 }}$
(D) None of these

Answer 1

To solve this question, we first have to form a combined $\left( 1 \right)$ quadratic equation of the pair of straight lines. Then by assuming the slope of the required line as $m$, from the given conditions, we can form the equation of the required line. Then by substituting the required values, we will get a quadratic equation with $m$. Then using the condition that the required line forms a right angle at the origin, with this, we can get the required answer.

Complete step by step answer:
The given equations of lines are,
$y = x - 1$
$\Rightarrow x - y - 1 = 0 - - - \left( 1 \right)$
$y = - x + 1$
$\Rightarrow x + y - 1 = 0 - - - \left( 2 \right)$
Now, the combined equation of the pair of straight lines of equations and $\left( 2 \right)$ is,
$\left( {x - y - 1} \right)\left( {x + y - 1} \right) = 0$
Opening the brackets we get,
$\Rightarrow x\left( {x + y - 1} \right) - y\left( {x + y - 1} \right) - 1\left( {x + y - 1} \right) = 0$
$\Rightarrow {x^2} + xy - x - xy - {y^2} + y - x - y + 1 = 0$
Now, simplifying the equation we get,
$\Rightarrow {x^2} - {y^2} - 2x + 1 = 0 - - - \left( 3 \right)$
Now, it is given that the line passes through $\left( {2,0} \right)$.
Let the slope of the line be $m$.
Therefore, the equation of the line is,
$\left( {y - 0} \right) = m\left( {x - 2} \right)$
[The general form of a line is, $\left( {y - k} \right) = m\left( {x - h} \right)$, where, $h =$x-coordinate of passing point and $k =$y-coordinate of passing point]
$\Rightarrow y = mx - 2m$
$\Rightarrow 1 = \dfrac{{mx - y}}{{2m}}$
Now, substituting this in equation $\left( 3 \right)$, we get,
${x^2} - {y^2} - 2.1.x + {1^2} = 0$
$\Rightarrow {x^2} - {y^2} - 2.\left( {\dfrac{{mx - y}}{{2m}}} \right).x + {\left( {\dfrac{{mx - y}}{{2m}}} \right)^2} = 0$
Now, simplifying the equation, we get,
$\Rightarrow {x^2} - {y^2} - \left( {\dfrac{{m{x^2} - yx}}{m}} \right) + \left( {\dfrac{{{m^2}{x^2} - 2mxy + {y^2}}}{{4{m^2}}}} \right) = 0$
$\Rightarrow 4{m^2}{x^2} - 4{m^2}{y^2} - 4{m^2}{x^2} + 4mxy + {m^2}{x^2} - 2mxy + {y^2} = 0$
$\Rightarrow {m^2}{x^2} + \left( {1 - 4{m^2}} \right){y^2} + 2mxy = 0$
Therefore, the coefficient of ${x^2}$ is ${m^2}$ and of ${y^2}$ is $\left( {1 - 4{m^2}} \right)$.
Now, given, the line subtends the right angle with the intercepts of the two lines.
Therefore, ${m_1}{m_2} = \dfrac{b}{a} = - 1$
[Where, $a =$coefficient of ${y^2}$ and $b =$coefficient of ${x^2}$]
Substituting, $a$ and $b$, we get,
$\dfrac{{{m^2}}}{{1 - 4{m^2}}} = - 1$
$\Rightarrow {m^2} = - 1 + 4{m^2}$
Simplifying, we get,
$\Rightarrow 3{m^2} = 1$
$\Rightarrow {m^2} = \dfrac{1}{3}$
Now, using square root on both sides, we get,
$\Rightarrow m = \pm \sqrt {\dfrac{1}{3}}$
Therefore, the slopes of the line can be, $\sqrt {\dfrac{1}{3}}$ and $- \sqrt {\dfrac{1}{3}}$.
Analyzing the options, we can see that only $- \sqrt 3$ is not a slope of the line, i.e., option (B).

Note:
The equations of a pair of straight lines can be separated by simple middle term factorization in order to get to know the individual pairs of lines. The equation of a pair of straight lines is used to derive the relations of the two lines with any other line that can’t be derived using the individual equations of the lines or would be very complex. The pair of straight lines have their own properties for slope and intercepts, which are different from those of normal straight lines.