Question

Coordinates of the vertices of a quadrilateral are $\left(2,-1\right),\left(0,2\right),\left(2,3\right)$ and $\left(4,0\right)$. The angle between its diagonals will be

• A. $90°$
• B. $0°$
• C. ${\tan }^{-1}\left(2\right)$
• D. ${\tan }^{-1}\left(\frac{1}{2}\right)$

Answer 1

The correct option is C

${\tan }^{-1}\left(2\right)$

Explanation for correct option:

Step-1: Finding slope of both the diagonals.

Given, co-ordinate of the vertices of the quadrilateral are $A\left(2,-1\right),B\left(0,2\right),C\left(2,3\right)$ and $D\left(4,0\right)$

We know that $\mathrm{slope}\mathrm{of}\mathrm{line}=\frac{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}$

Slope of the diagonal $AC$ is $m_1=\frac{\left(3-\left(-1\right)\right)}{\left(2-2\right)}$

$\Rightarrow$ $m_1=\frac{4}{0}$

$\Rightarrow$ $m_1=\infty$

It means slope $AC$ is parallel to $Y-axis$

Similarly,

Slope of the diagonal $BD$ is $m_2=\frac{\left(0-2\right)}{\left(4-0\right)}$

$\Rightarrow$ $m_2=\left(-\frac{1}{2}\right)$

Step-2: Finding angle between diagonals.

Since, one diagonal is parallel to $y-axis$

Therefore, angle between diagonal is ${\tan }^{-1}\left(-\frac{1}{m_2}\right)$

$\Rightarrow$ $\theta ={\tan }^{-1}\left(-\frac{1}{\left(-{\displaystyle \frac{1}{2}}\right)}\right)$

$\Rightarrow$ $\theta ={\tan }^{-1}\left(2\right)$

Hence, correct answer is option $\left(C\right)$.