The sides AB, BC, CD and DA of a quadrilatera | vedclass

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(C) To find the vertices,we solve the intersection of the given lines:$A = AB \cap DA: x + 2y = 3$ and $5x + y + 12 = 0 \implies A(-3, 3)$$B = AB \cap

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$90$

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(C) To find the vertices,we solve the intersection of the given lines:$A = AB \cap DA: x + 2y = 3$ and $5x + y + 12 = 0 \implies A(-3, 3)$$B = AB \cap BC: x + 2y = 3$ and $x = 1 \implies B(1, 1)$$C = BC \cap CD: x = 1$ and $x - 3y = 4 \implies C(1, -1)$$D = CD \cap DA: x - 3y = 4$ and $5x + y + 12 = 0 \implies D(-2, -2)$Slope of diagonal $AC$ $(m_1)$: $\frac{-1 - 3}{1 - (-3)} = \frac{-4}{4} = -1$Slope of diagonal $BD$ $(m_2)$: $\frac{-2 - 1}{-2 - 1} = \frac{-3}{-3} = 1$Since $m_1 	imes m_2 = (-1) 	imes (1) = -1$,the diagonals are perpendicular.Therefore,the angle between diagonals $AC$ and $BD$ is $90^\circ$.

The sides $AB, BC, CD$ and $DA$ of a quadrilateral are $x + 2y = 3, x = 1, x - 3y = 4$ and $5x + y + 12 = 0$ respectively. The angle between diagonals $AC$ and $BD$ is ......$^o$

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