Question

The lines $x+y-1=0,\left( m-1 \right)x+\left( {{m}^{2}}-7 \right)y-5=0$ and $\left( m-2 \right)x+\left( 2m-5 \right)y=0$ are
This question has multiple options
(a) concurrent for three values of m
(b) concurrent for one value of m
(c) concurrent for no value of m
(d) are parallel for $m=3$

Answer 1

In this question, if the equations of the lines are given by {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,$$$${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0,$$$${{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0. Now, from the condition of the lines to be concurrent the determinant of the coefficients of the given lines should be zero given by $\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right|$. Now, on equating the determinant to zero we get a cubic equation which on further simplification gives the possible values of m. Then, for the lines to be parallel the coefficients should be proportional which is given by $\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}$

Complete step by step answer:
Now, given line equations in the question are
x+y-1=0,\left( m-1 \right)x+\left( {{m}^{2}}-7 \right)y-5=0$$$$,\left( m-2 \right)x+\left( 2m-5 \right)y=0
Now, on comparing these lines equations with the general equations {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,$$$${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0,{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0 we get,
${{a}_{1}}=1,{{b}_{1}}=1,{{c}_{1}}=\left( -1 \right),{{a}_{2}}=\left( m-1 \right),{{b}_{2}}=\left( {{m}^{2}}-7 \right),{{c}_{2}}=\left( -5 \right),{{a}_{3}}=\left( m-2 \right),{{b}_{3}}=\left( 2m-5 \right),{{c}_{3}}=0$
As we already know that condition for concurrency for three given lines {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,$$$${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0,{{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0is given by

{{a}_{1}} &{{b}_{1}} &{{c}_{1}} \\\ {{a}_{2}} &{{b}_{2}} &{{c}_{2}} \\\ {{a}_{3}} &{{b}_{3}} &{{c}_{3}} \\\ \end{matrix} \right|=0$$ Now, let us substitute the respective values in the above formula then we get, $$\Rightarrow \left| \begin{matrix} &1 &1 &-1 \\\ &m-1; &{{m}^{2}}-7 &-5 \\\ &m-2; &2m-5 &0 \\\ \end{matrix} \right|=0$$ Now, this can be further written in other form as $$\Rightarrow 1\left( \left( {{m}^{2}}-7\times 0 \right)-\left( \left( 2m-5 \right)\left( -5 \right) \right) \right)-1\left( \left( m-1 \right)\times 0-\left( m-2 \right)\left( -5 \right) \right)-1\left( \left( m-1 \right)\times \left( 2m-5 \right)-\left( \left( {{m}^{2}}-7 \right)\times \left( m-2 \right) \right) \right)=0$$Now, on multiplying each of the respective terms we get, $$\Rightarrow -\left( -10m+25 \right)+\left( -5m+10 \right)-\left( 2{{m}^{2}}-5m-2m+5-\left( {{m}^{3}}-2{{m}^{2}}-7m+14 \right) \right)=0$$ Now, this can be further written in the simplified form as $$\Rightarrow 10m-25-5m+10-\left( -{{m}^{3}}+4{{m}^{2}}-9 \right)=0$$ Now, on further simplification we get, $$\Rightarrow {{m}^{3}}-4{{m}^{2}}+5m+19=0$$ Here, there is no real value of m that satisfies the above equation. Thus, the given lines are concurrent for no value of m. Let us now check the condition for the lines to be parallel As we already know that the condition for the lines to be parallel is given by $$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}$$ Here, we have the values of the coefficients as $${{a}_{1}}=1,{{b}_{1}}=1,{{a}_{2}}=\left( m-1 \right),{{b}_{2}}=\left( {{m}^{2}}-7 \right)$$ Now, on substituting the respective values in the above formula we get, $$\Rightarrow \dfrac{1}{m-1}=\dfrac{1}{{{m}^{2}}-7}$$ Now, on cross multiplication we get, $$\Rightarrow m-1={{m}^{2}}-7$$ Now, on rearranging the terms we get, $$\Rightarrow {{m}^{2}}-m-6=0$$ Now, this can be further written as $$\Rightarrow {{m}^{2}}-3m+2m-6=0$$ Now, on taking the common terms out this can be written as $$\Rightarrow m\left( m-3 \right)+2\left( m-3 \right)=0$$ Now, this can be further written as $$\Rightarrow \left( m+2 \right)\left( m-3 \right)=0$$ Now, on further simplification we get, $$\therefore m=-2,3$$ Thus, for the given lines to be parallel $$m=3$$ **So, the correct answers are “Option C and D”.** **Note:** Instead of considering the determinant to be zero for the lines to be parallel we can also consider the condition that the given 3 lines pass through the same point. So, by simplifying the three equations we get the result but it would be a bit confusing. It is important to note that while finding the determinant and equating it to 0 we should not neglect any of the terms or do incorrect calculation so that it changes the corresponding equation in m and so the result.