Question

If the coordinates of the extermities of diagonal of a square are $(2,-1)$ and $(6,2)$, then the coordinates of extremities of other diagonal are

• A. $(\frac{5}{2},\frac{5}{2})$
• B. $(\frac{11}{2},\frac{3}{2})$
• C. $(\frac{11}{2},\frac{-3}{2})$
• D. $(\frac{5}{2},-\frac{5}{2})$

Answer 1

The correct option is C $(\frac{11}{2},\frac{-3}{2})$

$\begin{array}{c}Coordinationofmid-point0\\ =\left(\frac{6+2}{2},\frac{2-1}{2}\right)\\ =\left(4,\frac{1}{2}\right)\\ AB=BC\\ \Rightarrow {\left(x_1-2\right)}^2+{\left(y_1+1\right)}^2={\left(x_1-6\right)}^2+{\left(y_1-2\right)}^2\\ \Rightarrow 8x_1+6y_1=35\to \left(i\right)\\ AO=BO\\ \Rightarrow {\left(2-4\right)}^2+{\left(-1-\frac{1}{2}\right)}^2={\left(x_1-4\right)}^2+{\left(y_1-\frac{1}{2}\right)}^2\\ \Rightarrow 4+\frac{9}{4}=x_{{}_1}^2+y_1^2-8x_1-y_1+16+\frac{1}{4}\\ \Rightarrow x_1^2+y_1^2-8x_1-y_1=-10\to \left(ii\right)\\ fromequation\left(i\right)\\ put\\ x_1=\frac{35-6y_1}{8}inequation\left(ii\right)\\ \Rightarrow {\left(\frac{35-6y_1}{8}\right)}^2+y_1^2-\left(35-6y_1\right)y_1=-10\\ \Rightarrow 4y_1^2-4y_1-15=0\\ \Rightarrow \left(2y_1+3\right)\left(y_1-5\right)=0\\ \Rightarrow y_1=\frac{-3}{2},5\\ y_1=\frac{-3}{2}\to x_1=\frac{35-6\times -\frac{3}{2}}{8}=\frac{11}{2}\\ y_1=5\to x_1=\frac{35-6\times 5}{8}=\frac{5}{8}\\ Theverticesofothertwoverticesare\\ \left(\frac{11}{2},\frac{-3}{2}\right)and\left(\frac{5}{8},5\right)\end{array}$