The equation of the straight line perpendicul | vedclass

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(A) The given line is $5x - 2y = 7$. The slope of this line is $m_1 = \frac{5}{2}$.Any line perpendicular to this line will have a slope $m_2 = -\frac

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$2x + 5y + 17 = 0$

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(A) The given line is $5x - 2y = 7$. The slope of this line is $m_1 = \frac{5}{2}$.Any line perpendicular to this line will have a slope $m_2 = -\frac{1}{m_1} = -\frac{2}{5}$.Thus,the equation of the required line is of the form $2x + 5y = \lambda$ $(i)$.Now,we find the point of intersection of the lines $2x + 3y = 1$ (ii) and $3x + 4y = 6$ (iii).Multiplying (ii) by $3$ and (iii) by $2$,we get:$6x + 9y = 3$$6x + 8y = 12$Subtracting the two equations,we get $y = -9$.Substituting $y = -9$ into (ii): $2x + 3(-9) = 1$ $\Rightarrow 2x - 27 = 1$ $\Rightarrow 2x = 28$ $\Rightarrow x = 14$.The point of intersection is $(14, -9)$.Since the line $(i)$ passes through $(14, -9)$,we have:$2(14) + 5(-9) = \lambda$$28 - 45 = \lambda$$\lambda = -17$.Substituting $\lambda = -17$ into $(i)$,we get $2x + 5y = -17$,which is $2x + 5y + 17 = 0$.

The equation of the straight line perpendicular to $5x - 2y = 7$ and passing through the point of intersection of the lines $2x + 3y = 1$ and $3x + 4y = 6$ is

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