The sides $A B, B C, C D$ and $D A$ of a quadrilateral have the equation $x + 2 y = 3, x = 1, x - 3 y = 4, 5 x + y + 12 = 0$ respectively, then the angle between the diagonals $A C$ and $B D$ is:

60^{o}

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45^{o}

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90^{o}

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Solution

The correct option is C $90^{o}$
The given equations of the sides $A B, B C, C D$ and $D A$ of a quadrilateral are, $x + 2 y = 3, x = 1, x - 3 y = 4, 5 x + y + 12 = 0$ respectively.

By solving, we get the co-ordinates of quadrilateral $A B C D$ are,

$A = (- 3, 3), B = (1, 1), C = (1, - 1), D = (- 2, - 2)$

Slope of diagonal $A C = m_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{- 1 - 3}{1 + 3} = \frac{- 4}{4} = - 1$

Slope of diagonal $B D = m_{2} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{- 2 - 1}{- 2 - 1} = \frac{- 3}{- 3} = 1$

$∵ m_{1} \times m_{2} = - 1$

So, the diagonals are perpendicular.

i.e., The angle between the diagonals $A C, B D$ is $90^{\circ}$

Option C is correct.

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Similar questions

Let ABCD be a quadrilateral with diagonals AC and BD. Prove the following statements (Compare these with the previous problem);

(a) AB + BC + CD > AD;

(b) AB + BC + CD + DA > 2AC;

(d) AB + BC + CD + DA > AC + BD.

In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that

(i) AB + BC + CD + DA > 2AC

(ii) AB + BC + CD > DA

(iii) AB + BC + CD + DA > AC + BD.

A B^{2} + C D^{2} = B C^{2} + D A^{2}

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