The equation of the line parallel to the y-ax | vedclass

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Question

(A) The family of lines passing through the intersection of $ax + by + c = 0$ and $a'x + b'y + c' = 0$ is given by $(ax + by + c) + \lambda(a'x + b'y

Vedclass · Accepted Answer

$x(ab' - a'b) + (cb' - c'b) = 0$

Vedclass · Answer

(A) The family of lines passing through the intersection of $ax + by + c = 0$ and $a'x + b'y + c' = 0$ is given by $(ax + by + c) + \lambda(a'x + b'y + c') = 0$.This can be rewritten as $x(a + \lambda a') + y(b + \lambda b') + (c + \lambda c') = 0$.Since the line is parallel to the $y$-axis,the coefficient of $y$ must be zero.Therefore,$b + \lambda b' = 0$,which gives $\lambda = -\frac{b}{b'}$.Substituting this value of $\lambda$ into the equation:$x(a - \frac{b}{b'}a') + y(b - \frac{b}{b'}b') + (c - \frac{b}{b'}c') = 0$.Multiplying by $b'$:$x(ab' - a'b) + y(bb' - bb') + (cb' - bc') = 0$.$x(ab' - a'b) + (cb' - bc') = 0$.

The equation of the line parallel to the $y$-axis and passing through the point of intersection of the lines $ax + by + c = 0$ and $a'x + b'y + c' = 0$ is:

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