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

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Solution

The correct option is C
The four vertices on solving are A(–3, 3), B(1, 1), C(1, –1) and D(–2, –2). $m_{1} = s l o p e o f A C = - 1, m_{2} = s l o p e o f B D = 1; ∴ m_{1} m_{2} = - 1$ .
Hence the angle between diagonals AC and BD is $90^{\circ}$

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Similar questions

A B, B C, C D

D A

x + 2 y = 3, x = 1, x - 3 y = 4, 5 x + y + 12 = 0

A C

B D

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 C ⊥ B D

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

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