Question

(9. If A (0, - 1), B (2, a), C (0, b) and D (- 2. 1) are vertices of a square, find the values ​​of a and 6.​

Answer 1

Answer:

The values of a are 1,-3

The values of b are 3,-1

Step-by-step explanation:

Given:

Vertices of the given square are

A (0, - 1), B (2, a), C (0, b) and D (- 2. 1)

$\text{The distance between two points }(x_1,y_1)\text{ and }(x_2,y_2)\text{ is}$

$\boxed{d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

Then,

$AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$AB=\sqrt{(0-2)^2+(-1-a)^2}$

$AB=\sqrt{4+(1+a)^2}$

$CD=\sqrt{(0+2)^2+(b-1)^2}$

$CD=\sqrt{4+(b-1)^2}$

$AD=\sqrt{(0+2)^2+(-1-1)^2}$

$AD=\sqrt{4+(-2)^2}$

$AD=\sqrt{8}$

since ABCD is a square, we have

AB=BC=CD=AD

$AB=AD$

$\implies\,\sqrt{4+(1+a)^2}=\sqrt{8}$

$\implies\,4+(1+a)^2=8$

$\implies\,(1+a)^2=4$

$\implies\,1+a=\pm2$

$\implies\,a=\pm2-1$

$\implies\,a=1,-3$

$CD=AD$

$\implies\,\sqrt{4+(b-1)^2}=\sqrt{8}$

$\implies\,4+(b-1)^2=8$

$\implies\,(b-1)^2=4$

$\implies\,b-1=\pm2$

$\implies\,b=\pm2+1$

$\implies\,b=3,-1$