The acute angle between the diagonals of a pa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The vertices of the parallelogram are $A(2, -1), B(0, 2), C(2, 3)$,and $D(4, 0)$.The diagonals are $AC$ and $BD$.The slope of diagonal $AC$ $(m_1)

Vedclass · Accepted Answer

$	an ^{-1} 2$

Vedclass · Answer

(C) The vertices of the parallelogram are $A(2, -1), B(0, 2), C(2, 3)$,and $D(4, 0)$.The diagonals are $AC$ and $BD$.The slope of diagonal $AC$ $(m_1)$ is $\frac{3 - (-1)}{2 - 2} = \frac{4}{0}$,which is undefined (vertical line).The slope of diagonal $BD$ $(m_2)$ is $\frac{0 - 2}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}$.Since one diagonal is vertical,the angle $	heta$ between the diagonals is given by $	an 	heta = |\frac{1}{m_2}| = |\frac{1}{-1/2}| = 2$.Therefore,$	heta = 	an^{-1} 2$ or $	heta = \cot^{-1}(\frac{1}{2})$.Checking the options,$	an^{-1} 2$ is equivalent to $\cot^{-1}(\frac{1}{2})$. However,looking at the provided options,the correct value is $	an^{-1} 2$.

The acute angle between the diagonals of a parallelogram whose vertices are $A(2, -1), B(0, 2), C(2, 3)$ and $D(4, 0)$ is

Similar Questions

$A$ line $4x+y=1$ passes through the point $A(2,-7)$ and meets the line $BC$,whose equation is $3x-4y+1=0$,at the point $B$. The equation of the line $AC$ such that $AB=AC$ is

The points $(0, 0)$,$(a, 0)$,and $\left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$ are vertices of:

The area of the triangle formed by the lines $x = 0, y = 0$ and $\frac{x}{a} + \frac{y}{b} = 1$ is

Let $A \equiv (3, 2)$ and $B \equiv (5, 1)$. An equilateral triangle $ABP$ is constructed on the side of $AB$ remote from the origin. The orthocentre of triangle $ABP$ is:

The intersection of three lines $x-y=0$,$x+2y=3$,and $2x+y=6$ forms a:

Vedclass Test Series

Exam Paper Generator

Online Exam Module