If the vertices of a quadrilateral are (0, -1 | vedclass

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

Question

(B) Let the vertices be $A(0, -1), B(2, 1), C(0, 3),$ and $D(-2, 1).$Calculate the side lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y

Vedclass · Accepted Answer

Square

Vedclass · Answer

(B) Let the vertices be $A(0, -1), B(2, 1), C(0, 3),$ and $D(-2, 1).$Calculate the side lengths using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$AB = \sqrt{(2-0)^2 + (1-(-1))^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$$BC = \sqrt{(0-2)^2 + (3-1)^2} = \sqrt{4+4} = 2\sqrt{2}$$CD = \sqrt{(-2-0)^2 + (1-3)^2} = \sqrt{4+4} = 2\sqrt{2}$$DA = \sqrt{(0-(-2))^2 + (-1-1)^2} = \sqrt{4+4} = 2\sqrt{2}$Calculate the diagonals:$AC = \sqrt{(0-0)^2 + (3-(-1))^2} = \sqrt{0+16} = 4$$BD = \sqrt{(-2-2)^2 + (1-1)^2} = \sqrt{16+0} = 4$Since all sides are equal $(AB=BC=CD=DA)$ and diagonals are equal $(AC=BD)$,the quadrilateral is a square.

If the vertices of a quadrilateral are $(0, -1), (2, 1), (0, 3),$ and $(-2, 1)$,then it is a

Similar Questions

The points $A(2, 1)$,$B(3, -2)$ and $C(a, b)$ are vertices of the rectangle $ABCD$. If the point $P(3, 4)$ lies on the line $CD$,then $5a + 10b = $

Let the points of intersection of the lines $x-y+1=0$,$x-2y+3=0$,and $2x-5y+11=0$ be the midpoints of the sides of a triangle $ABC$. Then the area of the triangle $ABC$ is .... .

The diagonals of a parallelogram $PQRS$ are along the lines $x + 3y = 4$ and $6x - 2y = 7$. Then $PQRS$ must be a

The perimeter of a triangle is $16 \text{ cm}$, one of the sides is of length $6 \text{ cm}$. If the area of the triangle is $12 \text{ cm}^2$, then the triangle is:

The angular points of a triangle are $A(-1, -7)$, $B(5, 1)$, and $C(1, 4)$. The equation of the bisector of the angle $\angle ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module