Statement (A) : The line 2x + y + 6 = 0 is pe | vedclass

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(A) The slope of the line $2x + y + 6 = 0$ is $m_1 = -2$.The slope of the line $x - 2y + 5 = 0$ is $m_2 = \frac{1}{2}$.The product of the slopes is $m

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$A$ and $R$ are both independently true and $R$ is the correct explanation for $A$.

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(A) The slope of the line $2x + y + 6 = 0$ is $m_1 = -2$.The slope of the line $x - 2y + 5 = 0$ is $m_2 = \frac{1}{2}$.The product of the slopes is $m_1 	imes m_2 = (-2) 	imes (\frac{1}{2}) = -1$,which confirms the lines are perpendicular.Checking if the second line $x - 2y + 5 = 0$ passes through $(1, 3)$: Substituting $x=1$ and $y=3$ into the equation gives $1 - 2(3) + 5 = 1 - 6 + 5 = 0$. Since the result is $0$,the point $(1, 3)$ lies on the line.Both statements are true,and the condition of perpendicularity is correctly explained by the product of slopes being $-1$.

Statement $(A)$ : The line $2x + y + 6 = 0$ is perpendicular to the line $x - 2y + 5 = 0$ and the second line passes through $(1, 3)$.
Reason $(R)$ : The product of the slopes of perpendicular lines is $-1$.

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Statement $(A)$ : The line $2x + y + 6 = 0$ is perpendicular to the line $x - 2y + 5 = 0$ and the second line passes through $(1, 3)$.Reason $(R)$ : The product of the slopes of perpendicular lines is $-1$.