The equation of the straight line perpendicul | vedclass

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(D) First,find the point of intersection of the lines $x + 3y - 1 = 0$ and $x - 2y + 4 = 0$.Subtracting the second equation from the first: $(x + 3y -

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$2x - 3y + 7 = 0$

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(D) First,find the point of intersection of the lines $x + 3y - 1 = 0$ and $x - 2y + 4 = 0$.Subtracting the second equation from the first: $(x + 3y - 1) - (x - 2y + 4) = 0$ $\Rightarrow 5y - 5 = 0$ $\Rightarrow y = 1$.Substituting $y = 1$ into $x + 3y - 1 = 0$,we get $x + 3(1) - 1 = 0 \Rightarrow x = -2$.The point of intersection is $(-2, 1)$.The equation of a line perpendicular to $3x + 2y = 0$ is of the form $2x - 3y + \lambda = 0$.Since the line passes through $(-2, 1)$,we substitute these coordinates into the equation:$2(-2) - 3(1) + \lambda = 0$ $\Rightarrow -4 - 3 + \lambda = 0$ $\Rightarrow \lambda = 7$.Thus,the required equation is $2x - 3y + 7 = 0$.

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

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