​ If is the angle between the diagonals of a parallelogram ABCD whose vertices are A(0, 2), B(2, – 1), C(4, 0) and D(2, 3). Show that tan theta = 2

Question

If \(\theta\) is the angle between the diagonals of a parallelogram ABCD whose vertices are A(0, 2), B(2, – 1), C(4, 0) and D(2, 3). Show that tan \(\theta\) = 2.

Answer 1

The vertices of parallelogram ABCD are 

A(0, 2), B(2, – 1), C(4, 0) and D(2, 3)

AC and BD are diagonals and is the angle between them. 

Slope of AC, m₁ =\(\frac{0-2}{4-0}=-\frac{1}{2}\)

And, slope of BD, m₂ = \(\frac{3-(-1)}{2-2}=\frac{4}{0}\)

= tan 90° 

∴ tan \(\theta\) = \(|\frac{m_2-m_1}{1+m_1m_2}|\)

= \(|\frac{tan90°-tan α}{1+tan90°.tanα}|\)

[let ‘a’ be the inclination of AC, tan a = –\(\frac{1}{2}\) ] 

= |tan (90° − a)| 

= |cot α| 

= |−2| [∴  tan α = −\(\frac{1}{2}\) ] 

⇒ tan \(\theta\) = 2