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Question

(B) The equation of any line passing through the intersection of $x - y - 4 = 0$ and $3x + y - 7 = 0$ is given by $(x - y - 4) + \lambda(3x + y - 7) =

Vedclass · Accepted Answer

$4x + 8y - 1 = 0$

Vedclass · Answer

(B) The equation of any line passing through the intersection of $x - y - 4 = 0$ and $3x + y - 7 = 0$ is given by $(x - y - 4) + \lambda(3x + y - 7) = 0$.This simplifies to $(1 + 3\lambda)x + (\lambda - 1)y - (4 + 7\lambda) = 0$.Since this line is parallel to $x + 2y = 1$,the ratio of the coefficients of $x$ and $y$ must be equal to the ratio of the coefficients in the given line:$\frac{1 + 3\lambda}{\lambda - 1} = \frac{1}{2}$.Cross-multiplying gives $2 + 6\lambda = \lambda - 1$,which implies $5\lambda = -3$,so $\lambda = -\frac{3}{5}$.Substituting $\lambda = -\frac{3}{5}$ into the family equation:$(x - y - 4) - \frac{3}{5}(3x + y - 7) = 0$.Multiplying by $5$: $5x - 5y - 20 - 9x - 3y + 21 = 0$.This results in $-4x - 8y + 1 = 0$,or $4x + 8y - 1 = 0$.

The equation of the line passing through the intersection of the lines $x - y = 4$ and $3x + y = 7$ and parallel to the line $x + 2y = 1$ is:

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