If p = a_1 x + b_1 y + k_1 = 0,q = a_2 x + b_ | vedclass

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(C) Given that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{k_1}{k_2} = \lambda$ (say). This implies that $a_1 = \lambda a_2$,$b_1 = \lambda b_2$,and $k

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Same as the straight line $p = 0$

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(C) Given that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{k_1}{k_2} = \lambda$ (say). This implies that $a_1 = \lambda a_2$,$b_1 = \lambda b_2$,and $k_1 = \lambda k_2$. Thus,the equation $p = 0$ can be written as $\lambda a_2 x + \lambda b_2 y + \lambda k_2 = 0$,which simplifies to $\lambda(a_2 x + b_2 y + k_2) = 0$,or $\lambda q = 0$. Since $\lambda 
eq 0$,this means $p = 0$ and $q = 0$ represent the same straight line. The equation of the curve is $p + c q = 0$. Substituting $p = \lambda q$,we get $\lambda q + c q = 0$,which is $(\lambda + c) q = 0$. This represents the same straight line as $q = 0$ (or $p = 0$) for any constant $c$ such that $\lambda + c 
eq 0$.

If $p = a_1 x + b_1 y + k_1 = 0$,$q = a_2 x + b_2 y + k_2 = 0$ and $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{k_1}{k_2}$,then the curve $p + c q = 0$ is

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