A(2, 5), B(2, -3), and D(-6, 5) are three vertices of square ABCD. Wha

1.	A(2, 5), B(2, -3), and D(-6, 5) are three vertices of square ABCD. What are the coordinates of the fourth vertex, C?
Answer» Given : A (2, 5) , B (2, -3) , D (-6, 5) are three vertices of square ABCD. To find : COORDINATES of fourth VERTEX, C(x, y) =? Knowledge required : Diagonals of Square bisect each other perpendicularly. Mid POINT formula $\red{\bigstar}\boxed{\sf{(x\;,\;y)=\left(\dfrac{x_1+x_2}{2}\;,\;\dfrac{y_1+y_2}{2}\right)}}$ [ where (x, y) are the coordinates of mid-point of line segment joining points (x₁, y₁) and (x₂, y₂) ] Solution : Let, diagonals AC and BD of square ABCD bisect each other at a point O with coordinates (m, n) then, since, Point O (m, n) bisect Diagonal BD therefore, Using Mid-point formula $\implies\sf{(m\;,\;n)=\left(\dfrac{(2)+(-6)}{2}\;,\;\dfrac{(-3)+(5)}{2}\right)}$ $\implies\sf{(m\;,\;n)=\left(\dfrac{-4}{2}\;,\;\dfrac{2}{2}\right)}$ $\implies\underline{\underline{\red{\sf{(m\;,\;n)=(-2\;,\;1)}}}}$ ALSO, since, Point O (m, n) also bisect Diagonal BC therefore, Using mid-point formula $\implies\sf{(m\;,n)=\left(\dfrac{(2) + (x) }{2} \;,\;\dfrac{(5)+(y)}{2}\right)}$ putting VALUES of (m, n) $\implies\sf{(-2\;,1)=\left(\dfrac{(2) + (x) }{2} \;,\;\dfrac{(5)+(y)}{2}\right)}$ $\implies\sf{-2=\dfrac{(2) + (x) }{2} \;,\;1=\dfrac{(5)+(y)}{2}}$ $\implies\sf{2+x=-4\;,\;\ 5+y=2}$ $\implies\sf{x=-4-2\;,\;\ y=2-5}$ $\implies\boxed{\underline{\underline{\red{\large{\sf{x=-6\;,\;\ y=-3}}}}}}\;\;\;\red{\bigstar}$ therefore, Coordinates of fourth vertex are (-6, -3).

A(2, 5), B(2, -3), and D(-6, 5) are three vertices of square ABCD. What are the coordinates of the fourth vertex, C?

Answer»

Given :

A (2, 5) , B (2, -3) , D (-6, 5) are three vertices of square ABCD.

To find :

COORDINATES of fourth VERTEX, C(x, y) =?

Knowledge required :

Diagonals of Square bisect each other perpendicularly.

Mid POINT formula

$\red{\bigstar}\boxed{\sf{(x\;,\;y)=\left(\dfrac{x_1+x_2}{2}\;,\;\dfrac{y_1+y_2}{2}\right)}}$

[ where (x, y) are the coordinates of mid-point of line segment joining points (x₁, y₁) and (x₂, y₂) ]

Solution :

Let, diagonals AC and BD of square ABCD bisect each other at a point O with coordinates (m, n)

then,

since, Point O (m, n) bisect Diagonal BD therefore,

Using Mid-point formula

$\implies\sf{(m\;,\;n)=\left(\dfrac{(2)+(-6)}{2}\;,\;\dfrac{(-3)+(5)}{2}\right)}$

$\implies\sf{(m\;,\;n)=\left(\dfrac{-4}{2}\;,\;\dfrac{2}{2}\right)}$

$\implies\underline{\underline{\red{\sf{(m\;,\;n)=(-2\;,\;1)}}}}$

ALSO,

since, Point O (m, n) also bisect Diagonal BC therefore,

Using mid-point formula

$\implies\sf{(m\;,n)=\left(\dfrac{(2) + (x) }{2} \;,\;\dfrac{(5)+(y)}{2}\right)}$

putting VALUES of (m, n)

$\implies\sf{(-2\;,1)=\left(\dfrac{(2) + (x) }{2} \;,\;\dfrac{(5)+(y)}{2}\right)}$

$\implies\sf{-2=\dfrac{(2) + (x) }{2} \;,\;1=\dfrac{(5)+(y)}{2}}$

$\implies\sf{2+x=-4\;,\;\ 5+y=2}$

$\implies\sf{x=-4-2\;,\;\ y=2-5}$

$\implies\boxed{\underline{\underline{\red{\large{\sf{x=-6\;,\;\ y=-3}}}}}}\;\;\;\red{\bigstar}$