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(C) Step $1$: Find the intersection point of the lines $x - 2y = 1$ and $x + 3y = 2$. Subtracting the first from the second gives $5y = 1$,so $y = \fr

Vedclass · Accepted Answer

$3x + 4y - 5 = 0$

Vedclass · Answer

(C) Step $1$: Find the intersection point of the lines $x - 2y = 1$ and $x + 3y = 2$. Subtracting the first from the second gives $5y = 1$,so $y = \frac{1}{5}$. Substituting $y$ into the first equation,$x - 2(\frac{1}{5}) = 1$,which gives $x = 1 + \frac{2}{5} = \frac{7}{5}$. The intersection point is $(\frac{7}{5}, \frac{1}{5})$.Step $2$: The required line is parallel to $3x + 4y = 0$,so its slope is $m = -\frac{3}{4}$.Step $3$: Using the point-slope form $y - y_1 = m(x - x_1)$,we get $y - \frac{1}{5} = -\frac{3}{4}(x - \frac{7}{5})$.Step $4$: Multiplying by $20$ to clear denominators: $20y - 4 = -15(x - \frac{7}{5}) \Rightarrow 20y - 4 = -15x + 21$.Step $5$: Rearranging the terms gives $15x + 20y = 25$,which simplifies to $3x + 4y = 5$ or $3x + 4y - 5 = 0$.

The equation of the straight line passing through the intersection of the lines $x - 2y = 1$ and $x + 3y = 2$ and parallel to $3x + 4y = 0$ is

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