If u = a_1x + b_1y + c_1 = 0,v = a_2x + b_2y | vedclass

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(A) Given $u = a_1x + b_1y + c_1 = 0$ and $v = a_2x + b_2y + c_2 = 0$.Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \lambda$ (where $\l

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The same straight line $u$

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(A) Given $u = a_1x + b_1y + c_1 = 0$ and $v = a_2x + b_2y + c_2 = 0$.Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \lambda$ (where $\lambda$ is a constant),we have $a_1 = \lambda a_2$,$b_1 = \lambda b_2$,and $c_1 = \lambda c_2$.Substituting these into the equation $u + kv = 0$:$(a_1x + b_1y + c_1) + k(a_2x + b_2y + c_2) = 0$$(\lambda a_2x + \lambda b_2y + \lambda c_2) + k(a_2x + b_2y + c_2) = 0$$(\lambda + k)(a_2x + b_2y + c_2) = 0$Since $a_2x + b_2y + c_2 = 0$ represents the line $v$,the equation $u + kv = 0$ simplifies to a multiple of the original line $v$ (or $u$),which is the same straight line.

If $u = a_1x + b_1y + c_1 = 0$,$v = a_2x + b_2y + c_2 = 0$ and $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,then the curve $u + kv = 0$ is

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