Question

plot A(5,3),B(-2,3)& D(5,-4) are three vertices of a square ABCD plot these points on a graph paper and hence find the coordinates of vertex C. also find the area of square

Answer 1

Need to FinD :-

• The vertex C of the square.
• The area of the square .

$\red{\frak{Given}}\begin{cases} \textsf{ Points given are , A(5,3), B(-2,3) and D(5,-4) .}\\\\\sf ABCD \ is \ a \ square .\end{cases}$

We need to find the third vertex C of the square . Let us firstly take any two vertices of the square .Say (5,3) and (-2,3) . We can find the side length of the square using " Distance Formula" . As ,

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

On substituting the respective values we will get the distance as , 7 units . Therefore the side length of the square is 7 u. So we know that the area of the square is square of side. So that ,

$\sf\longrightarrow Area = side^2 \\\\\\\sf\longrightarrow Area = (7u)^2 \\\\\\\sf\longrightarrow \underline{\underline{\red{Area = 49\ unit^2 }}}$

Now for finding the third coordinate C , we will use Midpoint Formula .Let the coordinate be (x , y ) . As we know that the diagonals of a square bisect each other , therefore ;

$\sf\longrightarrow \bigg( \dfrac{ x + 5}{2},\dfrac{y+3}{2}\bigg) =\bigg( \dfrac{ 5-2}{2},\dfrac{ 3-4}{2}\bigg)\\\\\\\sf\longrightarrow \bigg( \dfrac{ x + 5}{2},\dfrac{y+3}{2}\bigg) = \bigg( \dfrac{ 3}{2},\dfrac{ -1}{2}\bigg)\\\\\\\bf\longrightarrow \bigg( \dfrac{ x + 5}{2},\dfrac{y+3}{2}\bigg) = ( 1.5, -0.5)$

On comparing ,

$\sf\longrightarrow \dfrac{x+5}{2}= 1.5 \\\\\\\sf\longrightarrow x + 5 = 1.5 * 2 \\\\\\\sf\longrightarrow x + 5 = 3 \\\\\\\sf\longrightarrow x = 3 - 5 \\\\\\\sf\longrightarrow \bf{ x = -2}$

Similarly ,

$\sf\longrightarrow \dfrac{y+3}{2}= -0.5 \\\\\\\sf\longrightarrow y + 3 = 2 * -0.5 \\\\\\\sf\longrightarrow y + 3 = -1 \\\\\\\sf\longrightarrow y = -3-1 \\\\\\\sf\longrightarrow \bf{ y = -4}$

Hence ,

$\sf\longrightarrow\underline{\underline{\red{ Coordinate = (-2,-4)}}}$