A vertex of a square is (3, 4) and one diagon | vedclass

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(C) Let the given vertex be $A = (3, 4)$.The equation of the first diagonal $d_1$ is $x + 2y = 1$,which has a slope $m_1 = -\frac{1}{2}$.In a square,t

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$2x - y = 2$

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(C) Let the given vertex be $A = (3, 4)$.The equation of the first diagonal $d_1$ is $x + 2y = 1$,which has a slope $m_1 = -\frac{1}{2}$.In a square,the diagonals are perpendicular to each other.Therefore,the slope of the second diagonal $d_2$ is $m_2 = -\frac{1}{m_1} = -\frac{1}{-1/2} = 2$.The second diagonal passes through the vertex $A(3, 4)$.Using the point-slope form,the equation of the second diagonal is $y - 4 = 2(x - 3)$.$y - 4 = 2x - 6$.$2x - y - 2 = 0$,or $2x - y = 2$.

$A$ vertex of a square is $(3, 4)$ and one diagonal lies on the line $x + 2y = 1$. Find the equation of the second diagonal which passes through the given vertex.

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