Question

Find the angle between the diagonals of a parallelogram whose vertices taken in order are A(1, 2) ,B(2, 0),C(1,6), D(0, 4).

Answer 1

The angle between both the diagonals is $\theta = tan^-^1\left | \dfrac{1}{-2} \right |$.

Step-by-step explanation:

We can write the slope of diagonal AC.

$m_1 = \dfrac{y_2-y_1}{x_2-x_1}$

$m_1 = \dfrac{6-2}{1-1}$

$m_1 = \infty$

Similarly, We can write the slope of diagonal BD.

$m_2 = \dfrac{y_2-y_1}{x_2-x_1}$

$m_2 = \dfrac{4-0}{0-2}$

$m_2 = -2$

Let the angle between both the diagonals be \theta.

$tan \theta = \left | \dfrac{m_1-m_2}{1+m_1m_2} \right |$

= $lim_m_1\rightarrow\infty\left | \dfrac{m_1-m_2}{1+m_1m_2} \right |$

= $lim_m_1\rightarrow\infty\left | \dfrac{\dfrac{m_2}{m_1}-1}{\dfrac{1}{m_1}+m_2} \right |$

$tan\theta = \left | \dfrac{1}{-2} \right |$

Thus, $\theta = tan^-^1\left | \dfrac{1}{-2} \right |$

This is the required angle between both the diagonals.

Answer 2

Step-by-step explanation:

. Find the equation of the straight line passing through the point 1,1

parallel to line 4x + 4y + 7 = 0.