Show that (2,2), (5,2), (5,5) and (2,5) are t | vedclass

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

Question

(A) Let the vertices be $A(2,2), B(5,2), C(5,5)$ and $D(2,5)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$AB^2 = (5-2)^2 + (2-2

Vedclass · Accepted Answer

Vedclass · Answer

(A) Let the vertices be $A(2,2), B(5,2), C(5,5)$ and $D(2,5)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB^2 = (5-2)^2 + (2-2)^2 = 3^2 + 0^2 = 9 \implies AB = 3$.
$BC^2 = (5-5)^2 + (5-2)^2 = 0^2 + 3^2 = 9 \implies BC = 3$.
$CD^2 = (2-5)^2 + (5-5)^2 = (-3)^2 + 0^2 = 9 \implies CD = 3$.
$DA^2 = (2-2)^2 + (2-5)^2 = 0^2 + (-3)^2 = 9 \implies DA = 3$.
Since $AB = BC = CD = DA = 3$,the quadrilateral is a rhombus.
Now,check the diagonals:
$AC^2 = (5-2)^2 + (5-2)^2 = 3^2 + 3^2 = 9 + 9 = 18 \implies AC = \sqrt{18} = 3\sqrt{2}$.
$BD^2 = (2-5)^2 + (5-2)^2 = (-3)^2 + 3^2 = 9 + 9 = 18 \implies BD = \sqrt{18} = 3\sqrt{2}$.
Since all sides are equal and the diagonals are equal $(AC = BD)$,the quadrilateral $ABCD$ is a square.

Show that $(2,2), (5,2), (5,5)$ and $(2,5)$ are the vertices of a square.

Similar Questions

For given point $A(2, 4)$ and $B(-3, b)$,if $AB = \sqrt{26}$,find the possible values of $b$.

If $A (4, 7)$ and $B (7, 3),$ then $AB = .............$

The point on the $Y-$ axis which is equidistant from $P(3,2)$ and $Q(-1,-5)$ is $\ldots \ldots \ldots \ldots$

The end points of the diameter of a circle are $(2, 4)$ and $(2, 6)$. The radius of the circle is:

If the point $A (2, -4)$ is equidistant from $P (3, 8)$ and $Q (-10, y)$,find the values of $y$. Also,find the distance $PQ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module