Question ⤵☞Two opposite angular points of a square ABCD are A(-1,

1.	Question ⤵☞Two opposite angular points of a square ABCD are A(-1, 2) and C(3,-2). Find the coordinates of the remaining angular points of the square.__________________________NOTE ☟☞No Spam ☞Full Explanation Solution.
Answer» REQUIRED Answer is in attachment! • Formula USED in solution: $\star\;{\bf {Distance\: Formula}}\implies \sf\sqrt{(X_2-X_1)^2+(Y_2-Y_1)^2}$ $\star\;{\bf {Pythagoras\: theorem}}\sf \implies\:H^2=B^2+P^2$ $\sf Here\begin{cases}\sf H^2\: refers\:to\:square\:of\: hypotenuse.\\\sf P^2 \: refers\:to\: square\:of\: perpendicular.\\\sf B^2\: refers\:to\: square\:of\:base.\end{cases}$ • IDENTITY used: $\star\;{\bf {(A+B)^2=}}\sf A^2+B^2+2AB$ $\star\;{\bf {(A-B)^2=}}\sf A^2+B^2-2AB$

Question ⤵☞Two opposite angular points of a square ABCD are A(-1, 2) and C(3,-2). Find the coordinates of the remaining angular points of the square.__________________________NOTE ☟☞No Spam ☞Full Explanation Solution.

Answer»

REQUIRED Answer is in attachment!

• Formula USED in solution:

$\star\;{\bf {Distance\: Formula}}\implies \sf\sqrt{(X_2-X_1)^2+(Y_2-Y_1)^2}$

$\star\;{\bf {Pythagoras\: theorem}}\sf \implies\:H^2=B^2+P^2$

$\sf Here\begin{cases}\sf H^2\: refers\:to\:square\:of\: hypotenuse.\\\sf P^2 \: refers\:to\: square\:of\: perpendicular.\\\sf B^2\: refers\:to\: square\:of\:base.\end{cases}$