Question

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 1

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).

Answer 2

$\mathfrak{\blue{\underline{Given :-}}}$

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

$\mathfrak{\red{\underline{To \: Find :-}}}$

• The Coordinates of fourth vertex, C.

$\mathfrak{\green{\underline{Concept :-}}}$

• Diagonals of Square bisect each other perpendicularly.

• Mid point formula : -

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

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

$\mathfrak{\purple{\underline{Solution:-}}}$

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

Point O (m, n) bisect Diagonal BD .

By Mid-point formula

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

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

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

Point O (m, n) bisects Diagonal BC .

Now,

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

Substituting values of (m, n) :

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

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

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

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

$:\implies \sf{x=-6\;,\;\ y=-3}$

Hence,

$\underline{\underline{\pink{Coordinates \:of \:fourth \:vertex \:are \: (-6, -3).}}}$

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